### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Differential Equations MTH 235

MSU

GPA 3.57

### View Full Document

## 18

## 0

## Popular in Course

## Popular in Mathematics (M)

This 126 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 235 at Michigan State University taught by Gabriel Nagy in Fall. Since its upload, it has received 18 views. For similar materials see /class/207298/mth-235-michigan-state-university in Mathematics (M) at Michigan State University.

## Reviews for Differential Equations

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/19/15

ORDINARY DIFFERENTIAL EQUATIONS GABRIEL NAGY Mathematics Department Michigan State University East Lansing MI 48824 DECEMBER 17 2008 ABSTRACT These are the lecture notes for the courses MTH 235 and MTH 340 Fall Semester 2008 Th notes present a basic introduction to the subject of ordinary erential equations descri ing a collection of methods and techniques used to nd solutions to several types of differential equations These equations include rst order scalar equations second order linear equations and systems of inear equation There is one chapter dedicated to power series methods to nd solutions to variable coef cients second order linear equations There is another chapter dedicated to Laplace transform metho s to nd solutions to cons ant coef cients equations With generalized source func7 tions The last chapter provides a brief introduction to boundary value problems and the Fourier transform We end the chapter showing how to use these ideas to solve an initial boundary value problem for a particular partial differential equation the heat equation Date December 17 2008 pangmathmsu du 1 2 G NAGY 7 ODE DECEMBER 17 2008 CONTENTS Chapter 1 First order equations Linear equations Separable equations Applications Nonlinear equations 15 Exact equations HHHH F905 Chapter 2 Second order linear equations Constant coef cients 22 Variable coef cients 23 Complex roots 24 Repeated roots 25 Undetermined coef cients 26 Variation of parameters Chapter 3 Power series solutions 31 Regular points 32 The Euler equation 33 Regularsingular points Chapter 4 The Laplace Transform De nition and properties 42 The initial value problem 43 Discontinuous sources 44 Generalized sources 45 Convolution solutions Chapter 5 Linear systems 51 Overview of linear algebra Variable coef cients systems Nonrepeated eigenvalues Repeated eigenvalues F90 Chapter 6 Boundary value problems 61 Eigenvalueeigenfunction problems 62 Overview of Fourier series 63 Application The heat equation 117 117 125 133 G NAGY r ODE DECEMBER 17 2008 CHAPTER 1 FIRST ORDER EQUATIONS 1I1I LINEAR EQUATIONS lIlIlI Overview A differential equation is an equation where the unknown is a function and both the function and its derivatives may appear in the equation Differential equations are essential in a physical description of nature Many physical theories are centered in a differential equation Newton7s and Lagrange equations for classical mechanics Maxwellls equations for classical electromagnetism Schrodingerls equation for quantum mechanics Einstein7s equation for the general theory of gravitation Precise examples of differential equations are the following 1 Newton7s second law of motion for a single particle The unknown is a vectorvalued A function x R A R3 where mt represents the position of a particle as function of time t The differential equation is d2 m w m we where m is a positive constant representing the mass of the particle and f R X R3 A R3 represents the force acting on the particle which depends on the time t and the position in space m This is the wellknown mass times acceleration equals force77 law of motion The time decay of a radioactive substance The unknown is a scalar valuedfunction u R A R where ut represents the concentration of the radioactive substance at the time t The differential equation is d 3110 ikut where k is a positive constant concentration the faster it decaysI The wave equation describing waves propagating in a medial An example is sound propagating in the air The unknown is a scalarvalued function of two variables u R X R3 A R where ut 1 represents a perturbation in the air density at the The equation says that the higher the material time t and point x zy2 in space The equation is am 22gt v2 6in m 6514222 63w 22 where v is a positive constant describing the wave speed and we have used the notation E meaning partial derivative The heat conduction equation describing the variation of temperature in a solid material The unknown is a scalarvalued function u R X R3 A R where ut 1 represents the temperature at time t and the point x z yz in the solid The equation is amt 2 Ham 2 651mm 6314222 where k is a positive constant representing thermal properties of the materialI The equations in examples 1 and 2 are called ordinary differential equations ODE since the unknown function depends on a single independent variable t in these examples The equations in examples 3 and 4 are called partial differential equation PDE since the unknown function depends on more than a single independent variable with partial derivatives appearing in the equations 4 G NAGY 7 ODE DECEMBER 17 2008 112i Linear equations We start by introducing a more precise de nition of the type of differential equation we are about to stu yi De nition 11 Given afunction f R2 A R a rst order ODE in the unknown function y R A R is the equation 0 ftyt7 11 where we use the notation y t y The ODE in Eq 11 is called linear i the function with values ft y is linear on its second argument y that is there exist functions ab R A R such that 0 W W W t 9 WW 9 W The sign in front of the function a is just a convention since one can always rede ne the function a and write the linear differential equation as y ay We stick here to the convention y fay 12 since later on we want to write the equation as y ay b and we prefer the plus sign in this last equationi EXAMPLE 111 A rst order linear ODE is given by wt 2W 37 Wm 2y 3 This equation is called of constant coef cients since the functions at 72 and bt 3 do not depend on t On the other hand the following equation is called a rst order linear ODE with variable coef cients 2 2 y tyt4t ftyy4t lt1 lt is well known how to nd solutions of rst order linear ODEi ln the following result we rst consider the constant coef cient casei Theorem 12 Constant coe icients Given constants ab E R with a y 0 the linear di erential equation W 7am b 12 has solutions given by b m 7 13 a where c E R is an arbitrary constant Notice the this Theorem says that Eq 12 has in nitely many solutions one for each value of the constant c which is not determined by the equation Since the differential equation contains one derivative of the unknown function y nding the solution involves the computation of an integral and as a result there is a constant of integration that shows up This constant of integration is the origin of the constant c abovei EXAMPLE 112 Find all solutions to the constant coef cient equation y 2y 3 14 We compute the solution as follows Multiply Eq 14 by a nonvanishing otherwise arbitrary function with values Mt and order the terms in the equation as follows My 7 2W 3 15 The function h is called the integrating factor because it is chosen to be any function satisfying the following property W 7 2w My 1 6 G NAGY r ODE DECEMBER 17 2008 5 That is Mt is a factor that transforms the lefthandside in Eql5 into a total derivative Not any function M satis es this property The Eq 16 is indeed an equation for M as it can be seen from the following calculation Eq 16 is equivalent to My 7 2W My Ly gt 72 My gt 2M MC that is the function y does not enter into Eq 16 The later equation above can be solved for M as follows i 72 42gt lnM 72 Integrating in the equation above we obtain lnM 727 50 42gt Mt 616721 where co is an arbitrary constant and 01 600 Since the function M multiplies the original equation as it is seen in Eql5 we can freely choose co 0 that is 01 1 Then we conclude that the function is the integrating factor Using this integrating factor in Eqs 15 16 we conclude that 3 MW 3M ltgt ef gy 36721 42gt e my c 7 E 672 Therefore we conclude that for any constant c the solution y to Eq 14 is given by 06R 32 60 c lt 0 FIGURE 1 The graph of several solutions to Eq 14 corresponding to different values of the constant c The idea used in this example to nd the solution y can be generalized to any constant coef cient equation like Eq 12 This idea is called the integrating factor method We now follow every step performed in example to above to prove Theorem 12 Proof of Theorem 12 Multiply Eq 12 by a nonvanishing otherwise arbitrary function with values Mt and order the terms in the equation as follows My wybw 17 The function M is called the integrating factor because it is chosen to be any function satisfying the following property W WW My 1 8 6 G NAGY 7 ODE DECEMBER 17 2008 That is M is a factor that transforms the lefthand side in Eql7 into a total derivative Not any function M satis es this property The Eq 18 is indeed an equation for M as it can be seen from the following calculation Eq 18 is equivalent to My twy My ly gt avyv y gt aha7 that is the function y does not enter into Eq 18 The later equation above can be solved for M as fo ows M i a 42gt lnM a M lntegrating in the equation above we obtain lnM at co 42gt Mt clean where co is an arbitrary constant ant c1 e00 Since the function M multiplies the original equation as it is seen in Eql7 we can freely choose co 0 t at is c1 1 en we conclude that the function Using this integrating factor in Eqs 17 18 we conclude that is the integrating factor b WY bu gt e quoty be t gt ea y 5 3 eat where in the last equation we used that a f 0 Therefore we conclude that for any constant c the solution y to Eq 12 is given by This establishes the Theorem D Although the differential equation in 12 has in nitely many solutions the associated initial value problem lVP has a unique solution De nition 13 The initial value problem for a linear ODE is the following Given functions ab R A R and constants tg ya 6 R nd a solution y R A R of the problem 9 W 9 W Wu ya 19 The second equation in 19 is called the initial condition of the problem The main result concerning the IV above is that it has a unique solution We rst present this result in the case of constant coefficient equations Theorem 14 IVP Given constants abtg ya 6 R with a f 0 the initial value problem y 7ay 12 Wu ya has the unique solution given by W ya 7 gte quot EXAMPLE 113 Find the unique solution of the initial value problem y 2y 3 y0 l 110 We know that all solutions of the differential equation have the form yt ce2t 7 32 with c any constant The initial condition in 110 xes the value of the constant c as follows 12 ai 3 5 ly0c7 42gt c G NAGY r ODE DECEMBER 17 2008 7 and then the unique solution to the IV above is 52 3 t7 iii 25 2 Following the idea in this example one can proof Theorem 14 Proof of Theorem 14 The general solution of the differential equation in 19 is given in Eq 13 that is yt ce m a The initial condition determines the value of the constant c as follows b b yo We 66 7 gt c yo 7 ew a a lntroducing this expression for the constant c into the differential equation in 19 we obtain ya yo Egteia0itogt E a a This establishes the Theoremi The results in Theorems 1 2 and 14 can be generalized to variable coef cient differential equations Theorem 15 Variable coef cients Given continuous functions a b R A R and given constants tg ya 6 R the VP y aty W Wu ya 111 has the unique solution 1 z ytiynsbsds 1 12 ltgt ML mom lt gt where the integrating factor function is given by i on 7 wt At 7 asdsi to EXAMPLE 1 14 Find the function y solution of the initial value problem ty 2y 42 yl 2 In order to nd the solution write the equation above in a way it is simple to see the functions a and b in Theorem 15 In this case we obtain y y4t gt at 7 bt 74 113 We now nd an integrating factor as we did in the constant coef cient case That is nd a function n such that 2 2 2 My MyMy gt My MyMy My gt M ng This latter equation can be solved for n as follows Since lni Mn we have that 2 2 2 lth 7 gt mamas 7 7 ds gt 1nit71ni1 7 2lnt 71n1l 1 1 3 Recalling that lnl 0 and 21nt lnt2 we have that M lnWilnt2 gt implant 8 G NAGY 7 ODE DECEMBER 17 2008 Since the function M will multiply the differential equation in Eq 113 we can freely choose M1 1 that is HMIE Once the integrating factor function M is found one knows that for any constant co holds integrating factor 2 t2y Ey t24t gt t2y 2m 43 gt t2y 4 This is the main idea behind the integrating factor method To convert the lefthand side of the differential equation into a total derivative Then we can integrate as follows C t2ytt4co gt yttit2i The initial condition implies that 2y1601 a c01 We follow the main steps performed in this example to proof Theorem 15 Proof Theorem 15 We follow closely the proof of Theorem 12 Multiply the differential equation in Eq 111 by a function M and reorder terms as follows M y aw Mb Find the function M that satis es the equation My wyzyw gt Myzy gt WWIquot As in the proof of Theorem 12 the function y does not enter into the last equation above which is the equation that de nes the function M The solution can be computed as follows M t Mwlt MWHMw ma l l lntegrating in the equation above we obtain the integrating factor function M as follows t 7 z mwma ampAm e MW to where A is a primitive of ai Using this integrating factor M in the differential equation in Eq 111 we obtain WY 57 and integrating in the variable t results in the expression 2 z z MSMSHds 8b8d8 gt tMt yo 5b5d37 to to to where we used the property that Mto 1 Therefore the solution function y is given by 1 z ytiy sbsdsi 0 Jo mmgtml This establishes the Theoremi D G NAGY r ODE DECEMBER 17 2008 9 EXAMPLE 115 Consider the previous example Another way to nd the solution is simply to use Eq 112 First nd the integrating factor function u as follows Atlt ds 2lnt 7 lnl 21nt lnt2 a uteAt gt Then we compute the solution as follows yt tl222s24sds 1 2 1 3 727214sds 2 14 7272t 71 12 SEPARABLE EQUATIONS Nonlinear differential equations are in general more complicated to solve than linear differential equations However if the equation have some particular forms then there exist wellknown techniques to nd solutions One of these cases is when the differential equation is separable De nition 16 Given functions hg R A R a rst order di erential equation on the unknown y R A R is called separable it has the form My Mt 90 Therefore a differential equation y t ftyt is separable iff 7 glttgt y 7 hyv 42gt 100579 My EXAMPLE 121 2 1927 WW new y y2 cos2t 0 gt 7cos2t My The functions 9 and h are not uniquely de ned another choice is W cos2t My q 10 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 122 A differential equation that is not separable is y y costi It is wellknown how to nd solutions to separable equations Theorem 17 Separable equations Given continuous functions gh R A R with h f 0 consider the separable equation hyy yt 114 Let GH R A R be any primitive offunctions g and h respectively that is Gt 9t7 Hu M Then for any constant c the solution function y to Eq 114 satis es the equation Ct c 115 EXAMPLE 123 Find the solution of the IV t2 7 7 yoili m yo L no We use Theorem 17 to obtain the solution y We rst have to nd functions 9 h and then compute their antiderivatives G H respectively is gm e ma us hu 17u2 uigi Then Eqi 1 14 implies that the solution y satis es the equation The constant c is xed with the initial condition in Eq 1 16 3 3 y t 7 t 2 7 3 1 2 17 a 7 t c c 3 3 3 3 So we can rewrite the equation above as yg 3W t3 7 Notice that the solution y is given in an implicit form that is we have obtained an equation that the solution satis es equation that does not contain derivatives of y 0 i EXAMPLE 124 In example 123 above we have used Theorem 1 7 to nd the equation that the solution y satis esi We now show step by step how to obtain Eqi 115 in the case of this example We start with the differential equation in 116 72 7 ltgt 17y2t ytt2 fm oio We now integrate in in t on both sides of the equation above l1 y2tlytdtt2dtc y 1 G NAGY r ODE DECEMBER 17 2008 11 where c is any constant since the integrals are inde nite Introduce the substitution u W du yt div then we obtain u3 t3 17u2dut2dtc ltgt u7 ci Substitute back the original unknown y into the last expression above The rest of the calculation follows the one we did in the example 123 above lt1 We now show the proof of Theorem 17 which generalizes the idea used in the exam ple 124 above Proof of Theorem 17 The equation in 114 can be written as hyty t W v hyty tdt9tdtc7 where c is an arbitrary constant lntroduce on the right hand side of the second equation above the substitution u W du Wt Lit which implies hyt yt dt hudu du gt dt cl Since H and G are the primitives of h and g respectively the second equation above says Ct cl Substitute yt back in the place of u and one obtains that the solution function y satis es the equation Hltylttgtgt Cu a This establishes the Theoremi D EXAMPLE 125 Find the solution of the IV Mt fa cos2t 0 y0 1i 117 The differential equation above is separable with 1 90 COSQW My E7 therefore it can be integrated as follows Wt Wt 7cos2t 42gt dt7 cos2t dtci wt l W gt Again the substitution u yt du yt dt implies that du 1 l E 7cos2tdtc 42gt 7 7Es1n2tci Substitute the unknown function y back in the equation above 1 l 2 7m7 s1n2tc 42gt ytmi 12 G NAGY 7 ODE DECEMBER 17 2008 Which is an explicit expression for the solution function y instead of the implicit expressions we have seen in the previous examples The initial condition implies that 1y0 2 So the solution to the IV is ltgt c1i 0726 EXAMPLE 126 Find all the solutions y of the differential equation 4t 7 t3 t 7i y 4 gm The differential equation above is separable with 90 4i i is My 4 yg therefore it can be integrated as follows 4y3t y t 4t 7 t3 gt 4 y3t y tdt 4t 7 t3dt cl Again the substitution u yt du yt dt implies that 4 Substitute the unknown function y back in the equation above and calling 01 460 we obtain y4t16yt 8t t4 C1 4 4 4u3du4tit3dtcoi 42gt 4uuj2t2iticoi 13 APPLICATIONS We brie y discuss in this section one application of differential equations We present a physical situation and we then provide a mathematical description of that situation involving a differential equation We nally solve the differential equation and analyze the physical predictions given by that solution Consider the situation described in Fig 2 The tank contains a volume Vt of water with a amount of salt dissolved in it Water is pouring into the tank at a constant rate Ti having a constant salt concentration 41 while water is also leaving the tank at a constant rate 7 0 with a salt concentration qoti Finally consider that there is a mixing mechanism in the tank such that the salt that enters into the tank is instantaneously mixed in the tank This mixing mechanism then implies that the salt concentration in the tank is homogeneous at every time that is constant in space at every time with value qt Before stating the problems we are interested in solving we recall the physical units of the main elds involved in the problemi Denoting by the units of the eld Ti then we have Volume Mass Til To 7 my 40 V Volume Massi Volume 7 G NAGY r ODE DECEMBER 17 2008 13 instantaneously mixed FIGURE 2 Description of the water tank problemi Problem 1 Assume that the rates Ti 7 0 are known that the incoming salt concentration qi is known and assume that the initial water volume V0 and the initial amount of salt Q0 in the tanks are known then nd the amount of salt in the tank at every time The main unknown functions in this problem are the total volume of water in the tank Vt and the total salt mass in the tank The time rate of change of the total volume of water in the tank depends on the difference between the incoming and out coming water volume time rates Ti and 7 0 that is d 3V0 7 T 7 Tor The time rate of change of the total amount of salt in the tank Q also depends on the difference between the incoming and out coming salt masses time rates Tiqi and T0406 that is d 3Q W11quot Toqo The hypothesis that the salt is instantaneously mixed implies that m qlttgt therefore the unknown functions Q and V must satisfy the equations d 3V0 Ti 7 7 0 Q t Vt d EQW W11quot To These equations can be expressed in a simpler way recalling that the constants T and 7 0 are data of the problem and so the equation for V can be integrated as follows Wt n e Tot V0 118 14 G NAGY 7 ODE DECEMBER 17 2008 Therefore the equation for Q can be expressed as 7 T0 Q t 7 M 7 0 V0 Qt mg Renaming the functions To t b H7 altgt Ti7T0tV07 0 Ti we conclude that the function Q satisfies the linear differential equation Q O 7W Qt be 119 This equation can be integrated using the integrating factor methodi Since the case Ti 7 0 is studied in the next example we now consider only the case 7 Tor We then obtain 2 7 Ti To t l V0 At as d3 At 7Dln ltgt 0 ltgt ltgt THO V0 then the integrating factor Mt 6A0 has the form W WM lyenerd and then Q Q0 A1Msriqid3 Problem 2 Consider the problem 1 and in addition assume that Ti Tor In this case find the explicit expression of the salt mass function i Since Ti 7 0 then Eqi 1 18 implies that the volume of water in the tank remains constant Vt V0 Therefore Eqi 119 for Q is a linear constant coefficient differential equation given by 7 Qt aoQW boy do 707 be Title 0 The solution is given by QtltQ07igte 0 ioy e lt20Qoeqivo requivoi 120 Let us analyze this solution function Q in Eq 120 The graphs of several solutions for different initial data Q0 are plotted in Fig 3 In the case that the initial amount of salt in the tank has the critical value Q0 quO the solution Q remains constant equal to this critical value quOi In the case that the initial amount of salt in the tank Q0 is bigger than the critical value the salt in the tank decreases exponentially towards the critical value In the case that QO is smaller than the critical value the salt in the tank increases and exponentially approaches the critical value 14 NON LINEAR EQUATIONS In this Section we compare few main properties of solutions to linear and to nonlinear differential equations These properties are obtained from Theorems 1 5 and 18 for linear and for nonlinear equations respectively We summarize these properties in the items a c for the linear case and in items iiii for the nonlinear case We end this Section providing examples of these properties G NAGY r ODE DECEMBER 17 2008 15 Q Q0 Q0 Wu P Qo t FIGURE 3 The graph of the function 4 given in Eq 120 for different values of the initial data 0 Recall that the initial value problem for a linear differential equation is the following Given functions ab R A R and constants toyo E R nd the function y R A R solution of y ia y W No 90 121 The main result about solutions to an initial value problem for a linear differential equation is summarized in Theorem 15 We highlight from this Theorem three main properties of solutions to a linear initial value problem a There is an explicit expression of solutions of the initial value problem in Eq 121 b For every initial condition yo 6 R the corresponding solution yt is wellde ned for all t e R c For every initial condition yo 6 R the corresponding solution is unique None of these properties hold for solutions of nonlinear differential equations Recall that a differential equation yt ft is called nonlinear iff the function with values ft y is not linear in the second argument The following statements are true for solutions to nonlinear differential equations i There is no general explicit expression for the solution yt to a nonlinear differential equation ii Changing the initial data yo in Eq 123 may change the domain on the variable t where the solution yt is de ned iii Nonuniqueness of solution to the IV in Eq 123 may happen at points tu E R2 where Bu t u is not continuous The examples LAD143 below address the statements in i iii respectively EXAMPLE 141 Find the solution y to the equation t2 71 Notice that the nonlinear differential equation above is separable so we can nd solutions following the ideas given in Sect 12 That is y f W a ac a So the solution can be known only in an implicit form lt1 16 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 142 Find the solution y to the initial value problem 0 W07 y0 yo This is a nonlinear separable equation so we can again apply the ideas in Sect 114 MOW C 7 C y2t idt l o a W to a yt 60H Using the initial condition in equation above we obtain 1 9090 0 Co So the solution of the initial value problem above is 1 t 7 y l i 7 t yo This solution diverges at t lyo so the domain of the solution y is not the whole real line R but R 7 yo so it depends on the values of the initial data yo EXAMPLE 143 Find the solution y of the initial value problem Mt 9130 y0 0 122 The equation above is also separable Notice that the function ft u ul3 has deriva tive so the function Buf is not continuous at u 0 The solution to the initial value problem in Eq 12 exists but it is not unique One solution is 91t 0 A second solution can be computed as follows y yt dt dtc0 t c so the solution is g 2 m gem 3 The initial condition above implies 2 s 0yltogt gay 500 so the second solution is 92 75 lt1 The main result concerning solutions of an initial value problem for a nonlinear differ ential equation is summarized in the following statement which is given without a proof Theorem 18 Nonlinear equations Let R t1t2 gtlt ubug C R2 be a nonempty open rectangle on the plane and let f R C R2 A R be a function tu gt gt ftu If f and Buf are continuous functions on R and tmyg E R then there exists a smaller open rectangle R C R with tg ya 6 R such that the VP 6 ft7 W Wu ya 123 G NAGY r ODE DECEMBER 17 2008 17 has a unique solution y on the set R C R2 EXAMPLE 1442 Consider the nonlinear lVP 1 t t l 124 y t71gtltt1gtltylttgt72gtltylttgt3gt y 0 yo In this case the function f is given by l t 125 f t71t1u72u3 therefore f is not de ned on the lines t l t 71 u 2 and u 73 on the plane tu E R2 see Fig 4 In the case that the initial data is to 0 yo 1 then Theorem 18 implies that there exists a unique solution on any region R contained in the rectangle R 711 X 73 2 If the initial data for the IV in Eq 124 is t 0 yo 2 then Theorem 18 cannot be appliedl lt1 11 F71 F1 02 2 01 l 7 l 1 t 1 R u3 FIGURE 4 Regions in the plane Where the function f de ned in Eq 125 is not de ne l Both Theorems 1 5 and 18 state that there exist solutions to linear and nonlinear ODE respectively However the former theorem gives more information about the solutions to a reduced type of equations linear problems While the latter theorem provides less informa tion about solutions to Wide type of equations linear and nonlinear 1l5l EXACT EQUATIONS We introduce a particular type of differential equations called exactl It is wellknown how to nd solutions to exact differential equations and this is the main subject of this section De nition 19 Given continuously di erentiable functions MN R2 A R with values Mtu and Ntu at tu E R2 the di erential equation in the unknown function y R A R given by Ntvyt yt Mtvyt 0 is called exact the following equation holds BENt u BuMt u 18 G NAGY 7 ODE DECEMBER 17 2008 for every point t u E R2 where M and N are de ned EXAMPLE 151 The following differential equation is exact mu Mt 2t y2t 0 Indeed we see that Nt u 2tu BENtu 2u BNt ath Mtu2tu2 a BuMtu2u 1 u EXAMPLE 152 The following equation is also exact sinty thymy t 7 y t 7yt cost 7 2teym lndeed writing the equation as follows sint t2eym 7 l yt cost 2teym 0 we can see that Nt u sint t2eu 7 1 BENt u cost 2teu Mt u ucost 2teu BuMt u cost 2teu Therefore we have that BENt u BuMt u lt1 EXAMPLE 153 The following linear differential equation is not exact yt atyt bt7 CLO 0 Writing the equation in the form y aty 7 bt 0 we conclude that Ntu 1 BENtu 0 Mtu atu 7 bt a BuMt u at P lt1 It is wellknown how to nd solution in implicit form to exact differential equations The main idea to nd such solutions is based in the following result due to Henri Poincare around 1880s This result uncover a very useful property shared by certain functions of two variables Lemma 110 Poincar The continuously di erentiable functions M N R2 7gt R satisfy the equation BENO u BuMt u there exists a twice continuously di erentiable function 11 R2 7gt R such that Zn10711 Nt7u7 BMW Mt7 10 Proof of Lemma 110 lt Since 11 is twice continuously differentiable it holds laud BuatiJ hence BEN 616L111 616111 BuM It is not given D G NAGY r ODE DECEMBER 17 2008 19 EXAMPLE 154 In the case of the differential equation presented above 2mmymufmq the associated functions M and N are given by Nt u Ztu Mt u 2t u2 and they satisfy the hypothesis in Lemma 110 that is BEN 2u BuM Therefore this Theorem implies that there exists a function if such that hiJO u Nt u and 81 tu Mtu One can check that if in our example is Mm t2 W lt1 The main result that provides an implicit expression for the solution of an exact differ ential equation is the following Theorem 111 Exact Given the t39 quot 3 t39 functions MN R2 A R any solution y R A R to the exact di erential equation NO M0 0 MO yt 0 126 must satisfy the equation 1W W Cu 127 where cg is a constant and if R2 A R is a function satisfying WWW NOW WWW MO u 128 Proof of Theorem 111 Since the differential equation in 126 is exact Lemma 110 implies that there exists a function if satisfying Eqs 128 Hence we can do the following calculation 0 Ma at Nu m in 6wtyt Baum gut gum e um where cO is an arbitrary constant This establishes the Theorem D EXAMPLE 1552 Find the solution y to the differential equation mu Mt 2t y2 t 0 The rst step is to verify whether the differential equation is exact We know the answer is yes we did this calculation before nevertheless we repeat it here Nt u 2tu BENtu 2u Mt u 2t u2 BuMtu 2a Since the equation is exact Lemma 110 implies that there exists a function if satisfying the equations BENt u ELMO u MW NOW 521M757 MOW We know proceed to compute the function lntegrate in the variable u the equation Bud N keeping the variable t constant 8u tu 2tu tu tu2 gt 20 G NAGY 7 ODE DECEMBER 17 2008 where g is a constant of integration in the variable u but it can depend in the variable t Now integrate in the variable t the equation 811 M that is 2tu2 MW WW7 uQ 9t7 m 2t glttgt 2 we where co is a constant Then we have found that t u W2 t2 col Therefore Theorem 111 implies that the solution y satis es the implicit equation ty2tt2co 0 i The solution y above can also be written in explicit formi lt1 EXAMPLE 1 56 Find the solution y to the equation sint tQEyO 7 l yt yt cost 2mg i The rst step is to verify whether the differential equation is exact Nt u sint 726 BENO u cost 276 Mt u u cost 2teu ELMO u cost 2teui Therefore the differential equation is exact Then Lemma 1 10 implies that there exists a function 1 satisfying the equations 8u t u Nt u 81 t u Mt We know proceed to compute the function lntegrate in the variable u the equation Bud N keeping the variable t constant 8u tu sint t2 u tu usint t2eu gt where g is a constant of integration in the variable u but it can depend in the variable t Now integrate in the variable t the equation 811 M that is ucost 276 Mt u 81 t u u cost 276 gt 9W 0 9t 507 where co is a constant Then we have found that t u usint t2eu col Therefore Theorem 111 implies that the solution y satis es the implicit equation yt sint tQEyO co 0 i The solution y above cannot be written in explicit formi lt1 We nish this Section with the following observation There exists nonexact differen tial equations that can be converted into an exact differential equation by multiplying the original equation by an appropriate function called integrating factor We summarize this observation in the following result G NAGY 7 ODE DECEMBER 17 2008 21 Theorem 112 Integrating Factor Let MN R2 7 R be continuously di erentiable functions with N f 0 If the di erential equation Nt7yt 0 Mt7yt 0 129 is not exact that is BENtu BuMt u and the function N t u does not depend on the variable u then the equation MtNt7yt 0 WWL 90 0 131 is exact where the function h is the solution of the equation BuMt u 7 amt u 130 7 mama 7 swam EXAMPLE 1 57 Find the solution y to the differential equation 3mm 7 y2t t2 tyt Mi 7 0 132 We first verify whether this equation is exact Ntu t2tu 7 6Ntu 2tu Mt u Stu u2 ELMO u 3t 2u therefore the differential equation is not exact We now verify whether the extra condition in Theorem 1 12 holds that is whether the function in 130 is u independent 1 W lauMt u 7 BLNOZ 7 m St M 7 2t 7 u t u tt u 7 1 7 so this function does not depend on the variable ul Therefore Theorem 1 12 implies that the differential equation in 132 can be transformed into an exact equation multiplying it by the function M solution of the equation 7 BuM76N 7 7 imam 7111a 7 where we have chosen in second equation the integration constant to be zero Then multi plying the original differential equation in 132 by the integrating factor n we obtain 32 yt ty2t t3 t2 yt y t 7 0 133 This latter equation is exact since Mm 7 t3 t2u 7 Mm u 7 3t 7 2m M t u 7 3t2u m2 7 61191 t u 7 32 7 2m that is BEN Bull This establishes that the new equation in 133 is exact Therefore the solution can be found as we did in the previous examples in this Section lt1 22 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 1 58 Linear differential equations that are not exact can always be transformed into linear exact equations This was the main subject in Sect 11 which is now a particular case of Theorem 112 Indeed consider the linear differential equation m 7 W W 7 W alttgt 7 0 This equation is not exact as can be seen from the calculation Nt u 1 BENO u 0 Mt u atu 7 bt ELMO u ati However the function 1 7 8 M t 7 8 N t t Mug u mu z W altgt is independent of the variable u Therefore we can compute the integrating factor M solving the equation Mt f atdt a t t e Mt M Therefore the equation ef Myt atef WW 7 btef W 7 0 134 is exact since Na u ef WI 7 amt u atef Wt AM u atef Wu 7 btef WI 7 BuMt u atef W In this Section we have learned that a solution to 134 can be solved obtaining a function t u solution of Bud N and 811 Mi In Sect 11 we did essentially the same calculation to find solutions of Eq 1 34 lt1 We finish this section with the proof of our last result Proof of Theorem 112 Since the original differential equation in 129 is not exact we multiply this equation by a nonzero function u MtNt7y y MWMOW 0 1 35 Introducing the functions NO u MtNt u and MO u MtMt u we see that this new equation is exact iff BENO u ELMO The calculation BENOt u MtNt u Mt tNt u ELMO u Mt8uMt u says that Eqili35 is exact iff WNW u tWtNOvu 7 tauMtv u which is equivalent to mama 7 m auMlttugt 7 wt um and in turns it is equivalent to Mi 1 MO 7 Nay u BuMt u BENO Since the lefthand side does not depend on the variable u the equation above is solvable iff the righthand side does not depend on this variable ui This observation establishes the Theoremi D G NAGY r ODE DECEMBER 17 2008 23 CHAPTER 2 SECOND ORDER LINEAR EQUATIONS 2i CONSTANT COEFFICIENTS Differential equations may involve not only the unknown function and its rst derivative but also higher derivatives The order of an equation is a name to indicate the maximum number of derivatives of the unknown function that appear in the equation De nition 21 The di erential equation in the unknown y R A R given by y LIN y LIN y W 21 is called a second order linear di erential equation with variable coef cients The Eq is called homogeneous i bt 0 for allt E R The Eq is called of constant coe icients i the coe cients a ag and b are constants EXAMPLE 2ilil A second order7 linear homogeneous7 constant coefficients equation is y 5y 6 0 So it is the following equation 2y 7 3y y 0 lt1 EXAMPLE 212 A second order7 linear inhomogeneous7 variable coefficients equation is y 2ty 7 lnt y e33 lt1 Solutions to homogeneous linear equations have the property that a linear combination of solutions is also a solution Here is a more precise statementi Proposition 22 Superposition property If the functions y and y2 are solutions to the homogeneous linear equation y LIN y WW 9 0 2 then the linear combination c1y1t c2y2t is also a solution for any constants c1 c2 6 R Proof of Proposition 22 Verify that the function y clyl c2y2 satis es Eq 22 for every constants c1 c2 that is 5191 5292 a1 5191 Czyzy aot01y1 6292 Clyi l 529 a1 Clyi l 529 l do 5191 l 5292 51 by a1 a0 7391 l 52 9g a1ty a0 7392 0A B We now describe an idea to nd solutions to second order homogeneous equations with constant coefficients We look for solutions which are proportional to exponential functions7 that is solutions of the form yt equot for an appropriate constant Ti EXAMPLE 213 Find solutions to the equation y By 6y 0 23 We try to nd solutions of the form yt eni The derivatives of this function are y t Tequot and y t TQeT i So such a function is solution of the equation above iff T25T6e 0 gt T25r60i 2T4 24 G NAGY 7 ODE DECEMBER 17 2008 That is r must be a root of the polynomial pr r2 5r 6 This polynomial is called the characteristic polynomial of Eq 23 and the second equation in 24 is called the characteristic equation The solutions of the characteristic equation are l T E 75 i M25 7 24 1 T1 27 Therefore we have found two solutions to the Eq 23 given by wt 6 23 yz 6 33 Proposition 22 implies that we have found in nitely many solutions of the form yt cle m czeist c1c2 E R 2 5 lt1 De nition 23 Given a second order linear homogeneous di erential equation with con stant coe cients y my an 07 26 the Characteristic polynomial and the Characteristic equation associated with Eq 6 are respectively given by pr r2 alr ag pr 0 If r1 r2 are the solutions of the characteristic equation and c1 c2 are real constants then the function W 626 cgerg 27 is called the general solution of the E1 6 The solutions in Eq 27 are called general solution77 because these are all the solutions of 26 Every solution of Eq 26 must be given by a particular choice of the constants c1 and c2 in 27 i his statement is the main result in Theorem 24 be owl Notice that there are two free constants in the expression for the solution found in Eq 25 instead of only one constant as was the case in the rst order equations stud ied in the previous chapter Since Eq 23 is a second order equation two integrations are needed to obtain its solutions hence any solution should contain two arbitrary constants one for each integration performedi Also notice that an initial value problem for a second order equation like Eq 23 above iff there are two initial conditions instead of only one EXAMPLE 214 Find the solution y of the initial value problem a my 6 0 gm 1 Mo 71 We know that a solution of the differential equation above is yt cle m cgeigti We now nd the constants c1 and c2 that satisfy the initial conditions above ly0c1c2 c12 fl yO 72c2 7 3c2 02 1 G NAGY 7 ODE DECEMBER 17 2008 25 Therefore the unique solution to the initial value problem is yt 2e 2 7 e731 i EXAMPLE 215 Find the general solution y of the differential equation 2y 7 3y y 0 We look for solutions of the form yt equot with r solution of the characteristic equation r11 1 2T273T10 7 7 13i98 a 1 T2 2 Therefore the general solution of the equation above is yt cle cgetQ i lt1 The characteristic polynomial pr r2 air a0 is a real coef cients second degree polynomial The general expression for its roots is l ri lt7a1i Va 7 4 The number k a 4 7 a0 can be positive zero or negative which implies that the roots of pr can respectively two different numbers real numbers or complex numbers with one the complexconjugate of the other or only one real number The following result presents all solutions to the differential equation y aly a0 0 for each of these cases Theorem 24 Constant coef cients Given real constants a a0 consider the homo geneous linear di erential equation on the unknown y R 7gt R given by y azy on 0 28 Let r r be the roots of the characteristic polynomial pr r2 aJr ag and let cg c be arbitrary constants Then any solution of Eq 8 belongs to only one of the following two cases a If r r then the general solution of E1 is given by yt cue cler ti b If r r E R then the general solution of Eq is given by W Co 610 Furthermore given real constants tg ya and y there is a unique solution to the initial value problem given by Eq 8 and the initial conditions Wu ya Mtg yz Notice that in the rst case in Theorem 24 the roots r r can be real numbers or complex numbers The rst possibility was the case of the examples in this Section The case that ri a i i with a B E R will be studied later on in this Chapter The general solution in the latter case has the form ya 606mm 51407113 26 G NAGY 7 ODE DECEMBER 17 2008 Proof of Theorem 24 The proof has two main parts First we transform the original equation into an equation simpler to solve for a new unknown second we solve this simpler problemi In order to transform the problem into a simpler one we express the solution y as a product of two functions that is yt utvti Choosing v in an appropriate way the equation for u will be simpler to solve than the equation for y Hence y uv y 11 vu y u v Zuv UHui Therefore Eq 28 implies that uHU 2u v UHu a1uv 110 aouv 0 that is u a121gt ua0uvv a1vu0i 29 1 We now choose the function 1 such that 1 1 a1 2 0 7 77 2 10 a1 v ltgt v 2 We choose a simple solution of this equation given by vt e altg Having this expression for 1 one can compute 1 and v and it is simple to check that a2 v a1v7Zlvi 2 11 Introducing the rst equation in 210 and Eq 211 into Eq 29 and recalling that v is nonzero we obtain the simpli ed equation for the function u given by a u iku0 1617am 21 Eq 21 for u is simpler than the original equation 28 for y since in the former there is no term with the rst derivative of the unknown function In order to solve Eq 21 we repeat the idea followed to obtain this equation that is express function u as a product of two functions and solve a simple problem of one of the functions We rst consider the harder case which is when k f 0 In this case let us express ut e k wti Hence 1 Re t w E t 14 u ke w QREVE w E t wh Therefore Eq 21 for function u implies the following equation for function w 0 u 7 cu e 2Ew w w 2Ew 0 Only derivatives of w appear in the latter equation so denoting zt w t we have to solve a simple equation 1 72EI 1t zoei2 IO 6 RI Integrating we obtain w as follows 10 zoe Q g wt 773672 t col 2 k renaming 01 fro2 we obtain wt cle Q co ut CUEEl CleaEli G NAGY r ODE DECEMBER 17 2008 27 We then obtain the expression for the solution y uv given by ya Cog aTl l 6167a71 Since h a4 7 a0 the numbers 1 ri igi 42gt ri lt7a1iva 74a0gt are the roots of the characteristic polynomial r2a1ra0 0 we can express all solutions of the Eq 28 as follows yt coertt cierquot h f 0 Finally consider the case h 0 Then Eq 21 is simply given by u 0 ut co cit 00016 R Then the solution y to Eq 28 in this case is given by yt co cite a1 2I Since h 0 the characteristic equation r2a1 ra0 0 has only one root r r ialQ so the solution y above can be expressed as yt co cit ertt h 0 The furthermore part follows from the fact that the solutions y found above involve two arbitrary constants which are xed uniquely by the initial condition given in the Theoreml This establishes the Theoreml 22 VARIABLE COEFFICIENTS ln this section we present some properties of solutions to second order linear differential equations with variable coef cientsI We rst present the main result that asserts that there exists solution to this type of equations in the case that the equation coef cients are continuous functions This is the generalization of Theorem 28 to the case of variable coef cientsI We present the statement without proof Theorem 25 Variable coef cients Given an interval I t1t2 C R with t lt t2 and constants ta 6 I ya y C R let the functions a ag and b I C R A R be continuous Then there exists a unique solution y z I C R A R to the initial value problem y azt y and y W Wu ya Mtg yz 213 Notice that in the case of constant coef cient equations Theorem 24 provided an explicit formula for the solution function y while in the case of variable coef cients Theorem 25 provides no such formula The latter theorem applies to a wider class of problems than the former one but less is known about their solutions EXAMPLE 221 Find the longest interval I C R such that the initial value problem ti 1W5 3tv 4y tt17 94 27 Mi 1 has a unique solution We rst write the equation above in the form given in Theorem 25 3t 774i it y tily tilyi 28 G NAGY 7 ODE DECEMBER 17 2008 The intervals where the hypotheses in Theorem 25 are satis ed that is where the equation coefficients are continuous are I1 7001 and I2 lool Since the initial condition belongs to I1 the solution is lt1 We now introduce the notion of linearly dependent and linearly independent functionsl De nition 26 A set of two continuous functions y yg I C R A R is called linearly dependent u on the interval I i there exist constants c1 c2 not both zero such that for all t E I holds 519105 02920 0 The set of functions is called linearly independent on the interval I i it is not ld that is the only constants c and c2 that for all t E I satisfy the equation 01910 02920 0 are the constants c c2 0 Therefore a set of two functions is lldl if the functions are proportional to each other The set is llil if the functions are not proportional to each other Notice that the function zero is proportional to every function so any set containing the zero function is lid EXAMPLE 222 The following functions are lldl y1t sint yg t 2 sint The following functions are llil y1t sint y1t sin2t y1t sint Since 0 c1 sint cgtsint cl y2t tsint c1 Cgt sint t E R c2 since 2y1t 7 M t 0i since they are not proportional to each other lt1 We now return to the study of properties of solutions to differential equations The rest of this section is dedicated to answer the following two questions i Suppose you solve the initial value problem in Eq 213 with particular initial condi tions and you call the solution to be y1tl Suppose you also solve the same equation with a different set of initial conditions and you call the second solution y2tl What are appropriate initial conditions such that yl t and y2 t form a li set of solutions ii Suppose you arrange the initial conditions in the initial value problem in Eq 213 such that the solutions y1t and y2t mentioned above do form a li set at t 0 Does this set of solutions y1 y2 remains llil for all values of t where they are defined or does there exist a value oft such that after that t the solutions y1 y2 become a lid set The answer to these questions involves a new idea something that let one decides whether a set of two functions are dependent or independent without having the explicit expressions of these functions The only information one has about these functions is that they are solutions to initial value problems like the one given in Eq 213 A new idea that works G NAGY 7 ODE DECEMBER 17 2008 29 is to introduce a function that helps one answer these questions This function was rst introduced by a polish scientist Josef Wronski in 1821 while studying a different problem7 and it is now called the Wronskian De nition 27 Given an open interval I C R and continuously di ferentiable functions y1 yg I C R 7gt R the function Wylyg M0920 7 0920 is called the Wronskian of functions y and y2 EXAMPLE 223 We now compute the Wronskian of several set of functions y1t sint 1 d Wyly2 t sint2 cost 7 cost2 sint gt y2t 2sint y1t sint I i Wyly2 t sint sint tcost 7 costtsint yg t tsint i sin2 lt1 We can see that the Wronskian detects whether the two functions for a 1d set or not the Wronskian vanishes identically iff the functions form a 1d set This property is the main result of the following statement Lemma 28 Wronskian and ld1i Given an open interval I C R the nonzero and continuously di ferentiable functions y yg I C R 7gt R form a la set i f Wng t 0 for all t C I Proof of Lemma 28 Since the functions y1 y2 form a linearly dependent set7 then there exists a nonzero constant c such that yl cyg There ore7 Wylyz 692 y 7 Cy 92 0 lt Since the Wronskian of the nonzero functions y1 y2 vanishes7 we obtain that M 92 yl t mt 0W yy7yy 42gt 77 42gt ln ln ylyz 1 2 1 2 91 92 91 to 92 to therefore7 calling c y1tOy2to we obtain that yl t cy2t This establishes the Lemma D EXAMPLE 224 Show whether the following two functions form a 1d or li set yl t cos2t 7 2 cos2 t7 yg t cos2t 2 sin2 In this case it is not clear whether these two functions are proportional to each other or not They could be proportional due to a trigonometric identity Actually this is the case7 as we can see by computing their Wronskian Wyly2 t cos2t 7 2 cos2 72 sin2t 4 sint cost 7 7 2 sin2t 4 sint cost cos2t 2 sin2t but sin2t 2 sint cost7 therefore Wyly2 t 0 lt1 30 G NAGY 7 ODE DECEMBER 17 2008 That the Wronskian is a suitable tool to answer the rst question mentioned earlier in this section follows from the following statement Theorem 29 li Let I C R be an open interval a ag I C R A R be two continuous functions and consider the linear homogeneous equation y a1tyagty 0 214 Let ta 6 I and let y be a solution of Eq 214 with the initial conditions 9160 17 y to 0 Let yg be a solution of Eq 14 with the initial conditions 9260 07 y2to 1 Then the solutions y yg form a li set near t tg Proof of Theorem 29 We simply compute the Wronskian at to that is Wywzaol 91050 92050 yl o 92050 1 Since the Wronskian is a continuous function it will be nonzero in a neighborhood oft 0 Then Lemma 28 establishes the Theorem EXAMPLE 225 Find a li set of solutions to the equation y y 2y0 One way is to nd any two solutions and later on verify that they are li for example we look for solutions of the form yt equot Then we obtain the characteristic equation 2 1 T1 1 T 720 a T7ljl8gt a 2 r2 72 So the functions y1t e y2t e Q form a li set of solutions to the equation above Theorem 29 says that there is another set of li solutions obtained solving the two initial value problems mentioned in that Theorem The respective solutions are 1 2 l 7 y1t geL ge 2 y2t eL 7 e 2 The following result answers the second question asked earlier in this Section Theorem 210 Abel Let I C R be an open interval a ag I C R A R be continuous functions and y yg be twice continuously di erentiable solutions of the equation y a1t yagty 0 215 is a solution of the equation Wy1y2t a1tWyzy2t 0 Therefore for any ta 6 I the Wronskian Wyzyg is given by the expression Then the Wronskian W yzyz Wyzyz t Wy1y2tgef g 115 015 This Theorem answers the second question above because it says that the Wronskian of two solutions of 215 is nonzero at a single point to E I iff it is nonzero on the whole interval I where Wyly2 is de ned That is if the two solutions y1 y2 form a li set at a single point to E I then they form a li set on the whole interval I where they are de ned G NAGY 7 ODE DECEMBER 17 2008 31 EXAMPLE 226 Find the Wronskian of two solutions of the equation to 7tt2y t2y07 tgt0v Notice that we do not known the explicit expression for the solutions Nevertheless Theorem 210 says that we can compute their Wronskiani First we have to rewrite the differential equation in the form given in that Theorem namely 2 2 1 T T 1 lt7 T 0 y lt t y t2 t 9 Then Theorem 210 says that the Wronskian satis es the differential equation 2 Wy1y2t T 1 Wy1y2t 0 This is a rst order linear equation for Wylw so its solution can be computed using the method of integrating factors That is rst compute the integral 2 t 7EOlt1gt d8 721nltggt 7 t7to t2 1nltggt 7 40 Then the integrating factor M is given by t2 m which satis es the condition MOO 1 So the solution Wylyz tWy1y2t0 MlttgtWy1y2lttgtwltt0gtWy1y2ltt0gt0 so the solution is is given by t2 Wy1y2t Wy1y2t0et EO If we call the constant c Wy1y2tOtge 0 then the Wronskian has the simpler form Wy1y2t ct2e lt1 Proof of Theorem 210 Recall that both yl and M are solutions to Eq 215 that is yl a1 yl 0 91 0 y a1 9 0 92 0 Now multiply the rst equation above by 7y2 the second equation by yl and then add the resulting equations together One obtains yl y 7 yi w a1ty1y7 yi yz 7 0 Since Wym y1y27 yi yz 7 W m yl y 7 MM we obtain the equation WlIiyz T aid Wywz 0 This is a rst order linear equation and its solution can be found using the method of integrating factors The result is the expression given in the Theoremi This establishes the Theoremi 32 G NAGY 7 ODE DECEMBER 17 2008 23 COMPLEX ROOTS In Sect 21 we stated and proved Theorem 24 which describes all solutions of the constant coef cient equation y a1y aoy07 alyaOER We saw that all solutions fall into two main classes depending on whether the two roots 7 1 T2 of the characteristic polynomial pT 7 2 a1 T a0 are different or equall In the former case 7 1 T2 the roots can be either realvalued or complexvalued while in the latter case 7 1 T2 they must be realvalued These possibilities are sketched in Fig 5 p pr pr pr FIGURE 5 The graph of the three main cases for the characteristic poly nomial pT that is a polynomial having two complexvalued roots two realvalued roots or a single realvalued rootl Theorem 24 says that in the case T1 T2 all solutions to the differential equation above are given by yt clerlt c2eT2 with Cl 02 constantsl While in Sect 21 we concentrated in the case that the roots are realvalued in this Section we describe the solutions in the case that the roots are complex valued and we leave for the next Section the description of the remaining case where the roots are equal 7 1 T2 The main result in this Section is a Corollary from Theorem 24 Corollary 211 Complex roots Fix the constants a au 6 R with a 7 4ag lt 0 and denote the Toots of the chaTacteiistic polynomial pT 7 2 aJT au as follows 1 Ti a i i with a 7 Exilag 7 a Then any solution to the di eTential equation y a y agy 0 can he expTessed as a lineaT combination of the fundamental solutions y t em cos t y2t em sin tl Theorem 24 said that fundamental solutions to the differential equation y a1 yJrao y 0 are given by Q1 eltai gttv Q2 7 emim while the Corollary above says that another set of fundamental solutions is given by y1t e0 cos t y2 t e0 sin tl EXAMPLE 231 Find the general solution of the equation y 7 2y 6y 0A G NAGY 7 ODE DECEMBER 17 2008 33 We rst nd the roots of the characteristic polynomial l T272T60 7 Ti 2i 4724 7 ri71ii We know from Theorem 24 that the general solution of the differential equation above can be written as follows Mt 5151Ti t 52617i t7 517 52 6 RA The main information given in the Corollary above is that this general solution can also be written the equivalent way yt 01 cos3t 02 sinE t 6 cl 02 E R 216 In order to translate from the former expression to the latter one we only need to remember two main relations First that the following equation holds enii Ezeii r second that adding and subtracting Eulerls formula and its complexconjugate formula ei cosh3t i singt e i cosh3t 7 i singt gives the relations cosh3t ei e i singt ei gt 7 e i Therefore it is simple to show that the following relations hold 6 cosh3t 61Ti t 617i t e singt 61Ti t 7 euilvg li z and so the functions 910 5 COSgt7 92 t 5 Singt7 are a fundamental set of solutions to the differential equation above with the general solution given in 2 16 lt1 Proof of Corollary 211 Since the functions Q1 eltai gttv Q2 elta7i gttv are solutions to the differential equation in the Corollary then so are the functions we 7 amt 7 m m 7 gm 7 M It is simple to check that y1t geamem 671131 yg t ememt 7 671131 em cos t em sin t where we have used the relations emim emeim and the Eulerls formulas amt cos t isin t 67ml cos t 7 isin ti This establishes the Corollaryi D 34 G NAGY 7 ODE DECEMBER 17 2008 24 REPEATED ROOTS The last case in Theorem 24 provided the solutions of the differential equation y a1y aoy0 when the equation coefficients satisfy the relation a 4 that is the characteristic polynomial p39r 7 2 air a0 has the repeated root a1 7 1 7 73 Theorem 24 says that a fundamental solution s set to the differential equation above is 910 erltv 920 term EXAMPLE 241 Find the solution to the initial value problem 5 9y 6y y07 y0 17 y 0 3 The characteristic polynomial is p39r 9T2 67 l with root 1 l n E76 i M36 7 36 a T1 73 Therefore Theorem 24 says that the general solution has the form yt cleft3 cgte ESi Since its derivative is given by c t yt i e ES C2lt17 ail3 then the initial conditions imply that 1 90 C17 5 C 51 1 c2 2 g we 7g m So the solution to the initial value problem above is yt 1 mat3 lt1 We now repeat the last part of the proof of Theorem 24 in order to highlight its main idea It turns out that this idea can be generalized from equations with constant coefficients to equations with variable coef cients and it is called the reduction of order method This generalization occupies the last part of this Section Detailed proof of Theorem 24 case b Since a 4 there is only one root to the characteristic polynomial p39r 7 2 air a0 given by 7 1 eel2i This implies that one solution of the differential equation y a1 y a0 y 0 is given by 910 erlt The main idea in the reduction of order method is to look for a second solution to the differential equation as a product of a function times the known solution that is y2t vter1 i lntroducing this function yg into the differential equation one obtains a differential equation for the new function vi First compute y vT1v er y v 2r1vT 1 anti G NAGY r ODE DECEMBER 17 2008 35 Then the differential equation for 1 is computed as follows 0 v 211v 1 a1v 1110 a0 v 1 211 a1 1 7 air ao1 However the equation for 1 above is simpler than the original equation for M since yl er1t is a solution of the original differential equation which in this case means t at Tfa1T1ao07 2T1a10 So the equation for the function 1 is simple and given by 1 0 The solutions are given in terms of two arbitrary constants c1 and c2 as follows 1t c1 Cgt y2t clef cgtenti Since the constants c1 and c2 are arbitrary choosing c1 0 and c2 1 one obtains the following set of solutions to the original differential equation y1t er y2t terlti This establishes part b in Theorem 24 D This same idea used in the proof above the reduction of order method can be generalized to variable coef cient equationsi Theorem 212 Reduction of order Let I C R be an open interval and let the functions p q z I C R A R be continuous If y is a solution to the di erential equation y W M N y 07 then a second solution of this equation is given by y2t 1t y t where the function 1 is a solution to the equation y1t1 Zyt pty1t1 0 217 The reason for the name reduction of order77 is that the equation for the function 1 involves 1 and 1 but does not involve 1 itself So this equation is a rst order equation for the unknown 11 1 i EXAMPLE 242 Find a second solution yg to the differential equation W 2W 7 2y 07 knowing that y1t t is a solution We look for a solution of the form M t t1ti This implies that yt11 yt121i So the equation for 1 is given by 0 t2t121 2tltt11 7 2t1 t3 1 2t2 2t21 2t7 2t 1 t3 1 472 1 Notice that this last equation is precisely Eq 21 since in our case we have 2 2 y1t pt tv 2tv oi 36 G NAGY 7 ODE DECEMBER 17 2008 The equation for v is a first order equation for w 1 given by 1 7 wt cit 4 016 R Therefore integrating once again we obtain that vc2t 3cg 0203 6 R and recalling that yg tv we then conclude that yg 52772 cgti Choosing 02 l and C3 0 we obtain that a set of lii solutions to the original differential equation is given by 9175 92 t lt1 Proof of Theorem 212 The choice of y2 vy1 implies the equations y v y1vy17 y v y12v ylvylquot This information introduced into the differential equation says that 0 W91 2v yi vyi pv y1vyi ml ylv 2y py1v 914pr qyl The function yl is solution to the differential original differential equation yl ayl qy1 07 then the equation for v is given by M v 291 M1 1 07 which is Eq 21 This establishes the Theoremi D UNDETERMINED COEFFICIENTS The method of undetermined coef cients is a trialanderror procedure to find solutions of inhomogeneous differential equations A differential equation of the form y N y N y W 218 is called inhomogeneous iff the function f is not identically zero The method of unde termined coef cients is then given a particular form of the function f propose a clever guess for the solution y guess that depends on some free parameters or yet undetermined coef cientsi lntroduce this guess y into the differential equation and determine the free parameters Before presenting the method of undetermined coef cients is convenient to do several remarks In this Section we will use the following notation My y W M N y Therefore Eqi 218 can be written as Ly f inhomogeneous equation Ly 0 homogeneous equationi G NAGY r ODE DECEMBER 17 2008 37 It is also convenient to recall that the superposition property proven in Proposition 22 is based in the following property of linear differential equations given two functions y1 y2 and two constants c1 c2 holds LCly1 C292 51Ly1 C2Ly2 219 This property is the core of the superposition property of solutions to the differential equa tion Ly 0 This can be seen studying the proof of Proposition 22 which is the following Given two solutions yl and y2 of the differential equation Ly 0 then any linear combi nation clyl C2y2 is also a solution since LC1y1 0292 51Lyl C2Ly2 0 Finally we state two simple results we need later on Their proof is a straightforward consequence of Eq 219 and they are left as exercise for the reader Proposition 213 Given continuous functions p q let Ly y pt y qty If the functions yh and yp satisfy the equations Lyh 07 Mop f7 then the function y yh yp also satis es the equation My fl If the functions ypl ypn with integer n gt 1 satisfy the equations Lypfi7 i17 771 then the function y yp y satis es the equation My f withffzfn We now summarize the undetermined coef cient method in the following steps 1 Find all solutions to the homogeneous equation Lyh 0 2 If the source function f has the form f f1 fn then look for solutions ypl with i l n to the equations Lypl Once the functions ypl are found then construct yp ypl ypn 3 Given the source functions fi guess the solutions functions ypl following the Table 1 below 4 If any guessed function ypl satis es the homogeneous equation Lypl 0 then change the guess to the function tsypl with s gt 1 suf ciently large such that L 75971 0 5 lmpose the equation Lypl to nd the undetermined constants k given in the table below Then compute yp ypl ypn 6 The solution to the original problem is then given by yt yh t ypt EXAMPLE 251 Find all solutions to the inhomogeneous equation yo 7 3y 7 4y 3621 We rst nd all solutions to the homogeneous equation The roots of the characteristic polynomial are obtained from 7 1 4 T273T740 a 7 7 2 7 71 38 G NAGY 7 ODE DECEMBER 17 2008 1205 9172 t Ke quot ke quot Ktm kmtmkm1tm 1ko Kcosbt or Ksinbt k1 cosbt k2 sinbt Ktme quot e quotkmtm k0 Ke quot cosbt or Ke quot sinbt 6 quot k1 cosbt k2 sinbt Ktm cosbt or Ktm sinbt 16th k0 a1 cosbt a2 sinbt TABLE 1 List of particular solutions yp to the equation Lyp f for different source functions So lii solutions to the homogeneous equation y 7 3y 7 4y 0 are given by yh1t 547 yhz t 67 The second step is trivial in our case since the source function 362 cannot be simpli ed into a sum of simpler functions The third step says that the guessed solution for this case is ft 362 91705 16623 The constant k is the undetermined coefficient we must nd Notice that step four is trivial in our case since 62 is not a solution of the homogeneous equation Which are given by yhl and yhz above So the fth step is to introduce the guessed solution into the differential equation and nd the constant k that is 1 22 7 6 74m 3e2t a 76k 3 a k 75 Notice that the guessed solution must be proportional to the exponential 62 in order to cancel out the exponentials in the equation above Then the particular solution to the inhomogeneous differential equation is l ypt 7E 62 The general solution to the inhomogeneous equation is then given by 4t 7t 1 2t yt 016 C26 7 E e EXAMPLE 252 Find all solutions to the inhomogeneous equation y 7 3y 7 4y 3641 We have the same left hand side as in the example above so the rst step nding the solutions to the homogeneous equation is the same as in the example above yh1t e4t7 yhz t 67 The second step is again trivial since the source function 364g cannot be simpli ed into a sum of simpler functions The third step says that the guessed solution in this case is given y ft 364 91705 1664 G NAGY 7 ODE DECEMBER 17 2008 39 However step four tells us that we have to change our guess since yp 16th is a solution of the homogeneous equation y 7 3y 7 4y 0 Step four says we have to change our guess to ypt steal Step ve is the to introduce this guess into the differential equation y 7 1 7 4t 1664 y 7 8 42 1664 7 8 7 3 42 712 7 4 1664 7 3e therefore we get that 3 3 k g 91705 g 7564 The general solution is then 3 yt 5164 026quot g ten EXAMPLE 253 Find all the solutions to the inhomogeneous equation y 7 3y 7 4y 2 sinti Once again the left hand side of the differential equation above is the same as in the previous examples so the solution to the homogeneous equation is known and a li set is given by yh1 t 547 yhz t 67 The second step is still not needed and the third one says that the guess for the particular solution should be ft 2sint 91705 k1 Sint k2 costi Since our guess is not a solution to the homogeneous equation y 7 3y 7 4y 0 we proceed to step ve and we introduce our guess into the differential equation We rst compute y k1 cost 7 k2 sint y 7161 sint 7 k2 cost and then we obtain 7161 sint 7 k2 cost 7 3k1cost 7 k2 sint 7 4k1sint k2 cost 2sint 75161 3kg sint 73161 7 5kg cost 2sinti The last equation must hold for all t E R In particular it must hold for t 7r2 and for t 0 At these two points we obtain respectively 5 75k13k2 2 7617 73k1 751 70 k 7 3 2 7 So the particular solution to the inhomogeneous equation is given by 1 ypt E 75 sint 3 cost and the general solution is l yt 0164 C2671 E 75 sint 3cost i 40 G NAGY 7 ODE DECEMBER 17 2008 26 VARIATION OF PARAMETERS We describe a method to nd a solution to an inhomogeneous differential equation with variable coef cients of the form y W M W y W in the case that one happens to know a set of two liii solutions to the homogeneous equation y W M W y 0A This method is summarized in Theorem 2 14 belowi One can say that there are two new ideas in the proof of this theorem and while the rst idea is a good one the second idea is much better Let us rst present the main result Theorem 214 Variation of parameters Given an open interval I C R let p q f z I C R A R be continuous functions Assume that there exists functions y yg I C R A R that de ne a linearly independent set of solutions to the homogeneous equation y W M W y 07 and denote by Wyzyg the Wronskian of y and y2 If the functions u and U2 are de ned by ya t f t 92 t f t u t fidt u t idt 2 20 1 Wyz yg t 2 Wyz yz t then the function yp ulyl ugyg is a particular solution to the inhomogeneous equation y Wy qty W 221 The result above is more general than the trialand error method studied earlier since Theorem 2 14 applies to any continuous source function Nevertheless this last result requires to know two solutions to the homogeneous equation in order to solve the inhomo geneous onei EXAMPLE 261 Find the general solution of the inhomogeneous equation y 7 5y 6y 2e Since the construction of yp by Theorem 214 requires to know a li set of solution to the homogeneous problem let us nd these solutions rsti Since the equation has constant coef cients we compute the characteristic equation 2 1 T1 37 r75r60 r 5i25724 2 r2 i So the functions yl and y2 in Theorem 214 are in our case given by 910 53 920 52 The Wronskian of these two functions is given by Wood ewe 7 segue Wow 7e We are now ready to compute the functions ul and ugi Notice that Eq 220 the following differential equations u 7 7 92f u 7 91f 1 i 7 i 7 Wy1y2 WylyZ So the equation for ul is the following u1 7e2 2e 7e 5 u1 2e 2 u e3 2e 7e 5 u 72equot G NAGY r ODE DECEMBER 17 2008 41 where we have chosen the constant of integration to be zero Therefore the particular solution we are looking for is given 9p E hXESE 26quot62 Then the general solution to the inhomogeneous equation above is given by yt 0163 0262 6 0102 E R lt1 Proof of Theorem 214 The problem is to nd a solution of an inhomogeneous equation y N y N y W 222 knowing that there exists a li set of solutions yl and y2 to the homogeneous equation y W M N y 0 One can say that the rst idea to solve this inhomogeneous equation comes from the reduction of order method In that case one nds a second solution M to an homogeneous equation if one already knows another solution yl by proposing yg uyli One then hopes that the equation for u will be simpler than the original equation for yg since yl is solution of the homogeneous equation We now adapt this idea to our situation as follows we propose a form of the solution yp of the form yp u1y1 712927 and we hope that the equation for ul and ug will be simpler than the original equation since yl and y2 are solutions to the homogeneous equation We introduce this expression for yp into the original differential equation We must rst compute the derivatives 91 uiyi uiyl 7292 1129 9 7191 27 in uiyl 7292 2729 l 7 29 After some reordering of terms one obtains that Eq 22 has the form f 7191 1292 7129 100191 7292 u1 91pr qy1 uz y pr qy2 The functions yl and y2 are solutions to the homogeneous equations yl pylqy107 y py qy207 so we obtain that u1 and u2 must be solution of a simpler equation that the one above given y f ul yl 7292 2ulyl u y puly1 7292 223 The second idea is the following Look for functions ui and U2 that satisfy the extra equation l 224 Any solution of this extra equation must satisfy that its derivative is also zero that is 1191 ugyzl 0 and this last equation is explicitly given by ul yl u yz ulyl u y 0 225 Using Eqsi 2 24 and Eq 225 into the original Eqi 223 we obtain that a 2 Therefore instead of solving the original equation in 223 we nd solutions ul and ug for the system given by Eqs 2 24 and 226 This system is simple to solve since it is an 42 G NAGY 7 ODE DECEMBER 17 2008 algebraic system of two equations for the two unknowns 11 and From Eq 2124 we compute u and we introduce it into Eq 2126 as follows 91 9192 u2 7 7 u1 u1y17 7 u 7 1 f gt ul9192 9192 92 92 Since Wyly2 yly 7 yiyg we obtain that lntegrating in the variable t we obtain the expressions in Eq 2120 if we choose the in tegration constants to be zero Then y ulyl ugyg is a solution to the inhomogeneous differential equation in 2121 Notice that the constant of integration can always be cho sen to be zero because of the following argument Denote by ul and ug the functions in Eq 2120 and given any real numbers cl and 02 de ne 91u1517 712 24 52 Then the corresponding solution p is given by 9p 71191 11292 9191 U292 5191 C292 9p 6191 6292 That is the two solutions and yp differ by a solution to the homogeneous differential equation so both are solutions to the inhomogeneous equationi One is then free to chose the constants 01 and 02 in any way We choose them to be zero This establishes the Theoremi D EXAMPLE 21612 Find a particular solution to the differential equation t2 H72y3t271 knowing that the functions yl t2 and y2 lt are solutions to the homogeneous equation t2y 7 2y 01 We rst rewrite the inhomogeneous equation above in the form given in Theorem 2114 that is we must divide the whole equation by t2 We now proceed to compute the Wronskian Wmlttgt7ltt2gtlt17 gtlt2tgtltigt 7 a We now use the equation in 2120 to obtain the functions ul and ug that is 1 1 1 2 1 1 u l7lt3 72gtf3 3tgtlt3 72gtf3 1 7t2 7 G NAGY r ODE DECEMBER 17 2008 43 Therefore a particular solution to the inhomogeneous equation above is p ulyl ugyg that is7 1 72 2 1 3 71 W 7 mm gt t gt7 lt7t tgtltt gt 1 1 1 t21 t 7 7 7 t2 7 n 1 6 3 1 3 1 1 2 772 7tlnt2 3t 1 1 t21nt E 7 g y1ti However7 a simpler expression for a solution of the inhomogeneous equation above is 44 G NAGY 7 ODE DECEMBER 17 2008 CHAPTER 3 POWER SERIES SOLUTIONS Sill REGULAR POINTS In this Section we describe an idea to nd solutions of vaiiable coe cients differential equations of the form 131 y QI y 131 y 07 31 where the functions P Q and R are continuous The main idea is to assume that the solution of the equation above admits a power series representation around a point 10 E R of the form yr Eu 7 10 32 In the case that this power series assumption is true we can obtain the solution by simply introducing the expansion 32 into the differential equation in 31 and then obtaining the coefficients an that solve this equation The power series representation of the solution de ned in this way will be de ned in an interval around 10 where the radius of convergence can be determined after the solution is obtained Notice that the assumption we have done in Eq 32 implies that the solution y is wellde ned at 10 since yzo aoi In the case that the power series assumption in Eq 32 is not true when we solve for the coefficients an we obtain a contradiction There exist equations where this power series assumption for their solutions is in fact not true as it can be seen from the following example The equation I722y 6xi2y6y0 33 has two linearly independent solutions given by 1 1 911 172 921 1723 Both solutions above diverge at 10 2 so if one choose to nd solutions of the equation above around 10 2 using a power series expansion for the solution of the form given in 32 one is going to arrive to a contradiction Equations similar to 33 will be studied in later sections It turns out that the guess of the solution given in 32 is correct when the function P satis es one further condition at the particular point 10 that we choose to perform the power series expansion of the solution We state this condition without a proof and we then study several examples This extra condition is that 1310 0 Points 10 satisfying this latter property deserve a special nameI De nition 31 Given an open interval I C R and continuous functions P Q R z I C R A R a point my 6 R is called a regular point of the equation 131 y y 131 y OI Pzg 0 while the point mg is called a singular point i Pzg 0 EXAMPLE 311 The differential equation given in 33 has a singular point at 10 2 since the function 131 I 7 22 vanishes at that point Any other point is a regular pointI lt1 Therefore the power series method to nd solutions to Eq 31 near a regular point 10 of that equation can be summarized as follows a Verify that the point 10 where the power series expansion of the solution is going to be centered at is a regular point of the Eq 3 G NAGY r ODE DECEMBER 17 2008 45 b Propose a power series representation of the solution centered at 10 given by 00 m 2 an z 7 mo 710 c Introduce Eq 32 into the differential equation in 31 and nd a recurrence relation mong the coef cients an d Solve the recurrence relation in terms of free coefficients e If possible add up the resulting power series for the solution function y We follow these steps in the examples below to nd solutions to several differential equa tions We start with a rst order constant coefficient equation and then we continue with a second order constant coef cient equation The last two examples consider variable coef cient equationsi EXAMPLE 312 Find a power series solution around the point 10 0 of the equation y Cy 0 c e K This is a rst order linear differential equation so using the method of integrating factor we nd that the solution is a0 eiwi We are now interested in obtaining such solution with the power series methodi Although this is not a second order equation the power series method still works in this example Propose a solution of the form 00 00 y Zan In y Znan Emil 710 711 Introduce the expressions above into the differential equation 00 oo 0 ZnanznilcZanz7 711 710 00 oo lan1 In E can In 710 710 infty Z nJr lan1 can z n lan1 can 0 n gt 0 710 The last equation is called a recurrence relation among the coef cients ani The solution of this relation can be found by writing down the rst few cases and then guessing the general expression for the solution that is n0 a17cao a17ca0 62 nl 2a27ca1 a2 ao Cs n2 3a37ca2 a373 a0 C4 n 3 4 icag a4 Iaoi One can check that the coef cient an can be written as Cn an 71 a0 46 G NAGY 7 ODE DECEMBER 17 2008 which implies that the solution of the differential equation is given by 00 c 00l Maomw e unmozwwv e n0 0 n lt1 EXAMPLE 313 Find a power series solution around the point 10 0 of the equation y y 0 The equation in this example can also be solved without the use of power series by nding the roots of the characteristic polynomial 7 2 0 which gives us the solutions do 0081 a1 sinzi We now recover this solution using the power series 00 oo 00 y Zan z y Znan 1n71 y Emmi Dan z 2i 710 711 712 Introduce the expressions above into the differential equation 00 oo 0 Z nn 71anz 2 2 an 1 712 710 00 oo 2n lan2 I Zan 1 710 710 fly in 2n lan2 an z n ln lan2 an 0 n gt 0 710 The solution of this relation can again be found by writing down the rst few cases and we start with even values of n that is l n 07 21a2 a2 any l n 2 43a4 7042 a4 I a0 1 n 4 65a5 ia4 a5 iaaoi One can check that the even coef cients agk can be written as 1 k The coef cients an for the odd values of n can be found in the same way that is l 7117 32a3 70391 as igau l n 37 54a5 7amp3 a5 i an 1 n 5 76a7 7 a7 fiali One can check that the odd coef cients a2k1 can be written as 7 416 a2k1 7 an G NAGY 7 ODE DECEMBER 17 2008 47 Therefore the solution of the differential equation is given by mUk2 m Uk 2 1 91a01 2k zka1 mzk One can check that these are precisely the power series representations of the cosine and sine functions respectively a0 cosz a1 sinz lt1 EXAMPLE 314 Find the first four terms of the power series expansion around the point 10 1 of each fundamental solution to the differential equation M7zd7y0 We then look for solutions of the form 00 oo 00 y Zanz7ln y Znanz7ln71 y Znn7lanz7ln72i 710 711 712 It is convenient to rewrite the functionzy as follows zy Znanzz 71 1 Em z 711z 7 1W1 Znanz 71 Z nanz 71 1i 34 The first sum on the righthand side of Eq 34 can start at n 0 since we are only adding a zero term to the sum that is 00 oo Znanz 71 Znanz 71 711 710 while in the second sum of the same equation we can relabel the sum index as follows 00 Znanz 7 Dn l 1an1z 71 so we obtain the expression zy Znanz 71 1an1z 71 n 0 710 M8 nan n 1an1z 71 i H o n In a similar way relabel the index in the expression for y so we obtain 9 01 2n 1an2gtI 71 710 48 G NAGY 7 ODE DECEMBER 17 2008 Then the differential equation implies that 0 2n lan2z 71 7 Znan n lan1 I 71 7 Z anz 71 710 n 0 710 n 2n lan2 7 n lan1 7 nan 7 an I 7 1 1M8 The recurrence relation for the coef cients an with n gt 0 is then given by 0 n 2n lan2 7 n lan1 7 n Dan n 1 n 2an2 7 an1 1 an which can be rewritten as follows 71 1F 2an2 1 an1 1 an 0 i 315 We can solve this recurrence relation for the rst four coef cients a a n0 2a27a17a00 gt a2 1 o7 a a n1 3a37a27a10 gt a3 1 07 a a 712 4a4ia3 a20 gt a4Zl gel Therefore the rst terms in the power series expression for the solution y of the differential equation are given by y7aoalltz71gtltgtltz71gt2ltgtltz71gt3ltgtltx71gt4 which can be rewritten as 7 1 2 1 3 1 4 yiao1 171 6954 6954 1 2 1 3 1 4 a1z712z71 2954 4I71 So the rst four terms on each fundamental solution are given by 7 7 1 2 1 3 1 4 y1712z7l 6z7l 6z7l w7ltz71gt ltz71gt2 ltz71gt3 ltz71gt4 lt1 EXAMPLE 315 Find the rst three terms of the power series expansion around the point 10 2 of each fundamental solution to the differential equation y 7 z y 0 We then look for solutions of the form y Zanz 7 2 i 710 G NAGY 7 ODE DECEMBER 17 2008 49 It is convenient to rewrite the function my as follows 00 my Zanzz 7 2 710 Zanltz72gt2lltz72gt n H 0 M8 7 am 7 2W1 7 221 7 2 36 n 0 710 We now relabel the rst sum on the righthand side of Eq 36 in the following way Zanz 7 2n1 Zaltn71z 7 2 710 711 We do the same type of relabeling on the expression for y 00 y 7 lanz 7 2 2 712 00 2n 1an2z 7 2 i 710 Then the differential equation above can be written as follows 0 2n 1an2z 7 2 7 2anz 7 2 7 Zaltn71z 7 2 710 710 711 00 0 7 2mm 7 2 7 Z 717 2n lan2 7 2a 7 altn71 z 7 2 711 So the recurrence relation for the coef cients an is given by a2 7 a0 0 n 2n lan2 7 Zan 7 altn71 0 n 91 We can solve this recurrence relation for the rst four coef cients n0 a27a00 a2a07 n1 32a372a17a00 a37 a0 a1 2 4 3 72 7 0 7 7 n a4 a2 a1 a4 6 12 Therefore the rst terms in the power series expression for the solution y of the differential equation are given by 7 7 7 2 E E 7 3 E E 7 4 y7aoa1z 2aoz 2 lt6 3gtz 2 lt6 12gtz 2 which can be rewritten as l l yaolz722gx7236I724 1 3 1 4 a1z72 z72 z72 So the rst three terms on each fundamental solution are given by l y1lz722gz723 50 G NAGY 7 ODE DECEMBER 17 2008 1 3 1 4 yz 2 171EI72 32 THE EULER EQUATION In this Section we study how to nd solutions of certain differential equations with singular points for example an equation like this one Ii229 612y 6y0 We are interested in nding the solution values for for z arbitrary close to the singular point 10 2 Therefore we are not interested in a power series expansion of the solution around a regular point which in the example above is any point different from 10 2 In the previous Section we have stated without proof that the solutions de ned arbitrary close to ID 2 are given by l l I 7 z 7 y1lt 172 92 1723 In this Section we show how to obtain these solutions We start by introducing the type of equations we are interested in De nition 32 Given real constants pg qg and my the di erential equation on the function y given by 1 Io29 o1oyqoy 0 is called an Euler equation with singular point mg The main result concerning solutions of an Euler equation is the following Theorem 33 Euler equation Given real constants pg qg and my consider the Euler equation on the unknown y R 7gt R given 9km2 y puriruy quy 0 37 Let r r be the roots of the Euler characteristic polynomial pr rr 7 l pgr qg and let cg c be arbitrary constants Then any solution of Eq 7 belongs to only one of the following two cases a If r r then the general solution of Eq 7 is given by yt cgz 7 mg c1z 7 zgy i b If r r E R then the general solution of Eq is given by yt cg c ln z 7 Iglgt1 7 zgyti Furthermore given real constants I f 10 ya and y there is a unique solution to the initial value problem given by Eq 7 and the initial conditions yzz ya WEI yz In order to understand the main idea behind the proof of this Theorem it is useful to recall the central idea to nd solutions to constant coef cient equations of the form y aly aoy0 Since the derivatives of the exponential function are proportional to the same exponential that is emy re one can see that introducing this exponential y e into the equation above converts the differential equation for y into an algebraic equation for r as follows r2a1raoe 0 ltgt r2a1rao0i G NAGY 7 ODE DECEMBER 17 2008 51 This is the central idea to nd solutions to a constant coefficient equation that the expo nential function disappears from the equation leaving only an algebraic equation for the coefficient Ti One can later on show that the solutions obtained in this way are in fact all possible solution as it was shown in the proof of Theorem 24 We now follow the same type of idea to find solutions of an Euler equation IiIo29 oI Ioy4090A We now recall that the power function y z 7 10V has the property that y Tz 7 IOV I I 7 zoy Tz 7 10V A similar property holds for the second derivative y TT 7 lz 7 IOT 2 I 7 zo2y TT 7 lz 7 10V Hence introducing this power function into the Euler equation above transform this differ ential equation for y into an algebraic equation for T as follows TT1pOTqOIIOT 0 42gt TT7 lpOTqo 0 The constant T must be a root of the Euler characteristic polynomial pT TT1p07 qoi And we again have different cases depending whether this polynomial has two different roots or only one These cases are described in Theorem 33 Notice that this Theorem asserts more than that The Theorem asserts that these are the only solutions of the Euler equationi EXAMPLE 321 Find the general solution of the Euler equation I2yH4zy2y0i We look for solutions of the form 17 which implies that anyT 7 Hymn 7 TltT 71w therefore introducing this function y into the differential equation we obtain TT14T2IT 0 42gt 7 7 147 20i The solutions are computed in the usual way 1 T1 1 T23T270 7 T 773i 978 7 2 7 2 72 So the general solution of the differential equation above is given by 01171 02 1 2 EXAMPLE 322 Find the general solution of the Euler equation 12y 7 SIyJrlSy 0 We look for solutions of the form 17 which implies that z yz TIT 12 y z TT 7 1 17 therefore introducing this function y into the differential equation we obtain TT13T13IT 0 42gt TT14T130i The solutions are computed in the usual way 7 1 72 3239 1 2 77 7 T 4T13 0 7 T 2 4am 36 7 T igigi 52 G NAGY 7 ODE DECEMBER 17 2008 So the general solution of the differential equation above is given by c1 z 23i c2 172 311 l 38 lt1 In the case that the roots of the Euler characteristic polynomial are complex valued there is a particular linear combination of the solution which results in a realvalued set of fundamental solutions For example the solution of the example above given in Eq 38 can also be expressed as follows yt c3 1 2 cos lnlzl3 c4 1 2 sin lnlzl3 3 9 where the constants c3 c4 are different form the constants c1 c2 present in 38 The steps needed to obtain Eq 39 are summarized in the proof of the Corollary belowl Corollary 34 Complex roots Fix the constants pg qg r0 6 R with 4qg7 170712 gt 0 and denote the roots of the Euler characteristic polynomial pr rr 7 l par 10 as follows 7 l l riai i with 17PHT 4qgpg12 Then any solution to the Euler di erential equation I 7 zg2 y pg 1 7 10 y 10 y 0 can he expressed as a linear combination of the fundamental solutions y1z I 7 zgu cos ln z 7 Igl 7 I 7 zgu sinlnlz 7 Igl l Proof of Corollary 34 For simplicity consider the case 10 0 Since the functions 911 Imam 295 1a7i 7 are solutions to the differential equation in the Corollary then so are the functions y1ltzgt7 wzgt 921 m 7 gm 7 am Properties of the complexvalued power function imply the following equations 1 1 911 ixulrl 17m y21 Exam 7 I m 35 gnaw3 elnlrl 3 ra alum2 7 elnqirza i 1aei lnlxl e7i lnlxl 1aei lnlrl 7 e7i lnlxl i 1 cos I sin lnlzl 1 cos ln zl I sinlnlzl where we have used the relations emim emeim the de nitions 1113 eln fl ei ln flh and the Eulerls formulas em cos t isin t e m cos t 7 isin tl We have then concluded that yl 1 cos ln zl 7 y2z 1 sin ln zl G NAGY 7 ODE DECEMBER 17 2008 53 This establishes the Corollary e now consider the case where the Euler characteristic polynomial has a repeated rooti EXAMPLE 323 Find the general solution of the Euler equation 12y 7 31y4y 0 We look for solutions of the form 17 then the constant r must be solution of the Euler characteristic polynomial 7 7 7l737 40 gt T274T40 a TT 2 Therefore the general solution of the Euler equation in this case is given by WE 6112 0212111010 33 REGULAR SINGULAR POINTS In Sect 32 we have introduced the Euler equation and we have seen that contains one singular point In the same Section we have seen how to nd solutions to this Euler equation that are wellde ned arbitrary close to the singular point As an example recall that the Euler equation 122y 6ri2y 6y0 has its singular point at 10 2 and that a pair of fundamental solutions are given by 1 1 911 my 921 W7 which are wellde ned arbitrary close to ID 2 although not at 10 2 In this Section we introduce a large set of differential equations containing singular points similar to the singular points ofthe Euler equation We will call them regularsingular points where the name tries to suggest that they are singular points which happen to be not so singulari We show how to nd some of the solutions to this broad set of equations that are wellde ned arbitrary close to the regularsingular pointsi De nition 35 Given continuous functions P Q and R assume that the di erential equa tion Pltzgt yquot Qltzgt y Rltzgt y 0 has a singular point my 6 R that is Pzg 0 The singular point mg is called a regular singular point the following limits are nite I 7 my lim 1 7 zg2 131 P 7 17060 P 7 and in addition the functions in the limits above have convergent power series expansions around 10 that is there exists coe cients pn and 47 for n gt 0 such that 1 10QI m n I luVRW m n Wpnz71g Wqnz710i EXAMPLE 331 The singular point of every Euler equation is a regularsingular point In order to see this given real constants pg 40 IO consider the general Euler equation lim x7gtxg 1 10V 9 Polt Io y 409 0 In this case the functions P Q and R introduced in the Def 35 above are given by PW I 0 1001 i 10 RW 40 54 G NAGY 7 ODE DECEMBER 17 2008 Therefore we obtain z 7 IO z 7 IO2 Rz l 1 1520 Pa 100 520 Pz 10 hence both conditions given in the Def 315 are satis ed by the coef cients of every Euler equationi EXAMPLE 31312 Find the regularsingular points of the differential equation 1712y 721yaa1y0 where a is a real constant We rst nd the singular points of this equation which are given by the zeros of the coef cient function that multiplies y that is IO 1 Pz1712 17z1z0 11 711 Let us start with the singular point 10 11 We then have I 7 1 7 z 71721 z 712 Rx 7 I 712aa 1 Pz 7 17z1z Pz 17z1z 21 7171aa139 1 I 7 7 1 z which implies that both functions above have power series expansions around 10 11 Furthermore the following limits are nite namely 7 7 2 lim 1 1mm 1 lim I 1 Rm 01 71 Pz 71 Pz We conclude that 10 1 is a regularsingular point of the differential equation above We now perform the same calculation with the point 11 71 that means I 1 7 z 1721 I 12 Rz z 12aa 1 Pz 7 17z1z PI 17z1z 7 21 7z1aa139 1 i I 7 7 1 7 z which implies that both functions above have power series expansions around 11 711 Furthermore the following limits are nite namely 2 hm ltz1gtQltzgt 1 hm ltz1gt Rltzgt 0 771 Pz 771 Pz Therefore the point 11 71 is also a regularsingular point of the differential equation above lt1 EXAMPLE 31313 Find the regularsingular points of the differential equation 122z71y 3z71y2y 01 The singular point of the differential equation above are given by ID 72 and 11 11 In the rst case we have that lim W 11m W 11m ioo 0H4 Pz 0H4 z 22z 71 0H4 z 2 G NAGY 7 ODE DECEMBER 17 2008 55 so 10 72 is not a regularsingular point In the second case 11 l we obtain as 71gtQltzgt 7 z 71gt2ltz 71gt z 71gt2Rltzgt 7 M 71gt Pltzgt ltz2gtltz 71gt Pltzgt ltz2gt2ltz 71gt 2z 7 l 7 2z 7 l H2 ltz2gt2 which implies that both functions above have power series expansions around 11 1 Furthermore the following limits are nite namely 11QI 7 I12PI 7 1311 Pm 0 71131 Pa 0 Therefore the point 11 71 is also a regularsingular point of the differential equation above lt1 EXAMPLE 334 Find the regularsingular points of the differential equation my 7 z lnlzl y 31 y 0 The singular point is ID 0 Although the following limits are nite 1 ram 7 m ash norm hm WW I z 170 E 1 lim 7 1 li i 170 75m 1 2 2 I R z z 31 hm A hm hm 312 0 70 Pz 170 z 170 the point 10 0 is not a regularsingular point since the function zQP dies not have a power series expansion around zero since IQW 7 PW 7 zlnhzh and the log function does not have a Taylor series at 10 0 lt1 In the last part of this Section we study differential equations haVing regularsingular points We present a method to nd solutions that are wellde ned arbitrary close to a regularsingular point The main idea of the method is the following Near a regular singular point the coefficients of the differential equation are close to the coefficients of a particular Euler equationi It is then possible that the solutions to the original differential equation are close to solutions of this latter Euler equation near that regularsingular pointi Recall that at least one solution to the Euler equation have the form MI I 7 mo 310 where the constant 7 is the root of the Euler characteristic polynomial and IQ is the regular singular point of the original equationi Therefore we look for solutions of the original differential equation haVing the form 00 91I 7 10V 2 an r 7 onl 311 710 It is simple to see that this expression of the solution contains the idea that is close to yez for values of I close to 10 since in this latter case we have t at 91ye1a0a11 1039 anye1 for 1210 where the symbol a 2 b with a b E R means that la 7 bl is close to zero 56 G NAGY 7 ODE DECEMBER 17 2008 It is important to remark that not every solution of a differential equation containing regularsingular points will have the form given in Eq 311 However it can e shown that at least one solution will have this formi A similar situation occurs with the solutions to the Euler equation where there are cases with solutions containing logarithmic terms which means they do not have the form given in Eq 310 See the comment at the end of Example 335 A summary of the method to nd solutions to differential equations having regular singular points is the following 1 Propose that the solution y admits a power series expansion of the form 00 m Zan z 7 more 710 2 Introduce this power series expansion into the differential equation and nd a re currence relation for the coef cients an and the constant T 3 First nd the solutions for the constant T Then introduce this result for T into the recurrence relation for the coef cients a Only then solve this latter recurrence relation for the coef cients ani EXAMPLE 335 Find the solution y near the regularsingular point 10 0 of the differential equation 12y 7zz3yz3y0i We propose a solution of the form 00 2 an zltn i 710 The rst and second derivatives are given by 00 oo yz Tan zltnr71 y z Tn T 71an zltnri2h 710 710 In the case T 0 we had the relation 220 nan zm l 221 nan 1n 1 but in our case T can be nonzero so we cannot change in this way the starting value of the summation index n We now introduce these expression into the differential equation It is convenient to do this step by step We start with the term I 3y which has the form I 3 y I 3 Zan 10 710 00 oo 2 an Iltnr1 Z 3 Iltnrgt n70 710 Z an1 10 E San zltn i 31 711 710 G NAGY 7 ODE DECEMBER 17 2008 57 We continue with the term containing y that is 711 71 31 nTa z 7 lt M lt2 gtZlt gt7 W D 710 7 Tan zltnTD 7 Z 3n Tan 10 710 710 00 oo 7 T 7 lan1 10 7 Z 3n Tan zltnrh 3 13 711 710 Then we compute the term containing y as follows 00 12 y 12 Tn T 71an zltnr72 710 00 ZnTnT7 Dan zltnrh 3 14 710 As one can see from Eqsi3i123il4 the guiding principle to rewrite each term is to have the power function 10 labeled in the same way on every termi For example in Eqsi3i123il4 we do not have a sum involving terms with factors z r 1 or factors z r1i Then the differential equation can be written as follows 0 Zn Tn T 7 lanznr 7 2n T 7 lan1 10 710 7 Z 3n Tan 10 Z an1 10 E San zltnrh 710 711 n In the equation above we need to split the sums containing terms with n gt 0 into the term n 0 and a sum containing the terms with n gt 1 that is 00 0 Tn T 71an 7 n T 7 lan1 7 3n Tan an1 San 10 711 7 7 7 l 7 ST Slaozf and this expression can be rewritten as follows 00 0 ZnTnT 71 7 3nT 3lan 7 nT7171an1x T TT 7 l 7 ST Slaozr and then 0 ZnTnT 71 7 3nT7 1an 7 nT7 2an1 10 711 TT 7 l 7 ST Slaozr hence 0 TT 7 l 7 3T Slaozr ZTl7 71nT 7 3an 7 n T 7 2an1 zltnrh 711 58 G NAGY 7 ODE DECEMBER 17 2008 The recurrence relation is given by the equations TT7l7339r30 315 nr7lnr73an7nr72an1 0 316 The way to solve the recurrence relation given in Eqs 315 316 is the following First solve Eq 315 for the exponent 7 which in this case has two solutions Ti second introduce the rst solution T into the recurrence relation in Eq 316 and solve for the coef cients an the result is a solution y of the original differential equation then introduce the second solution 7 into Eq 316 and solve again for the coef cients an the new result is a second solution y Let us follow this procedure in the case of the equations above 1 T 3 T274T30 a Ti 4jl6712 a 71 Introducing the value T 3 into Eq 316 we obtain n 2nan 7 n lan1 0 One can check that the solution y obtained form this recurrence relation is given by 2 l 1 yr a0131 1112E13 Introducing the value T 1 into Eq 316 we obtain nn 7 2an 7 n 7 lan1 0 One can also check that the solution y obtained form this recurrence relation is given by 2 l l 97a21I2 13114EI5 397 3 4 15 One last comment We have found that the solution y is not linearly independent of the solution y This Example shows an important property of the method described in this Section to nd solutions of equations near a regularsingular point a property that we already mentioned earlier in this Section The method does not provide all solutions of a differential equation near a regularsingular point it only provides at least one solution near a regularsingular point It can be shown the following result If the roots of the Euler characteristic polynomial T T differ by an integer then the second solution y the solution corresponding to the smaller root has a more complicated expression than the one given in Eq 311 involving logarithmic terms We do not study this type of solutions in these notes lt1 a2zglgzl12izg yy 1 G NAGY r ODE DECEMBER 17 2008 59 CHAPTERAT THE LAPLACE TRANSFORM 41 DEFINITION AND PROPERTIES We now introduce a new idea to nd solutions to differential equations that is radically different from all the methods we have seen so far The idea is centered around the Laplace transform of a function We rst introduce the de nition of the Laplace transform and then present several examples In the following Sections we explain how to use such transform to nd solutions to differential equations The Laplace transform method is useful to solve inhomogeneous differential equations but only with constant coef cients De nition 41 The Laplace transform of a function f R A R is another function f R A R given by 00 4e4V dt an H for those values ofs 6 Eur where the integral converges We will often use an alternative notation for f the Laplace transform of f namely lfl f This notation emphasizes that the Laplace transform de nes a map from the set of piecewise continuous functions into the same set Recalling that a scalarvalued function of a single variable is a map denoted by t gt gt ft the same notation can be used with the Laplace transform that is f gt gt In this Section we are interested to study the properties of this map L In particular we will show in Theorem 43 that this map is linear and in Theorem 45 that this map is onetoone and so invertible on the appropriate sets It can be seen from this de nition that the Laplace transform of a function f is an improper integralI As we review in the examples below in general the improper integral in q 41 does not converge for every 8 6 Row The interval I where the Laplace transform of a function f is wellde ned for s E I depends on the particular function We see below that e 1 with a gt 0 is wellde ned for s gt a while L sinat is wellde ned for s gt 0 We now compute the Laplace transforms of simple functionsi EXAMPLE 4Ilil Compute 1I The function l is the simplest function we can use to nd its Laplace transformI Following the de nition we obtain MUm amp O lim eindt mac 7 1 in etude lo lim 7 e m 71gt naoo l 7 for sgt0I s The second line in the equation above is obtained by the de nition of an improper integral while the last line comes from the taking the limit it A 00 and using the fact that lim e75 0 for s gt 0 naoo 60 G NAGY 7 ODE DECEMBER 17 2008 Therefore we conclude that EXAMPLE 4 12 Compute e 1 where a E R Again following the de nition we obtain ea Eistea dt O lim 7lt57ugttdt 71700 O l 7 7s7atn 71320 Hide lo lim 1 6757 1 71gt n7oltgt7 s 7 a 1 for sgtai 57a We have repeated all the steps we have done on the previous example that is the second line in the equation above is obtained by the de nition of an improper integral while the last line come from the taking the limit n 7gt 00 and using the fact that lim eiltsiagtn 0 for s gt a n7gtoo Therefore we conclude that Sgt0n EXAMPLE 4 13 Compute te 1 where a E R In this case the calculation is more complicated than above since we need to integrate by parts tea eistte quot dt 0 lim teicia dt 71700 O te5 gt l 1 675 dt s 7 a O l l 7 757at n 7 7 757at n 37ate l0 87a2 lol Notice that the rst term on the righthand side of the last line above vanishes since l7 s 1 a 1 7 57a n 1 7 57a nh7gtnolois7ane 07 37a16 tl 001 Regarding the other term 1 7 57a n 1 Sill nl o me 07 W Hefm G NAGY r ODE DECEMBER 17 2008 61 Therefore we conclude that EXAMPLE 4 14 Compute sinat where a E R In this case we need to compute sinat lim 6 sinat dti ngtoo O The de nite integral in the last line above can be computed as follows Integrating by parts twice it is not dif cult to check that the following formula holds n l 6 sinat dt if 751 sinat O s which implies that a n is 7 1 is 1 82 0 e s1nat dt 7 is e s1nat Hence it is not dif cult to see that 2 2 00 lt8 Ta gt eistsinat dt 2 8 gt 0 O s 32 n 7 i2 751 cosat o s n O n 3320 eistsinat dt n 7 i2 751 cosat o s n O which is equivalent to sinat s gt 0 a s2a2 lt1 We here present a short list of Laplace transforms They can be computed in the same way of the examples above 1 1 fltsgt s gt 0 W fltsgt s gt mama f0 t fltsgt s gt 0 flttgt sinltatgt fltsgt 54a 3 gt 0 N cosltatgt fltsgt Su s gt 0 W sinhltatgt fltsgt s gt 0 N coshltatgt fltsgt 3 gt 0 N We 1 s gt maxa0 f0 sinazt fltsgt s gt mama f0 cosltbtgt M S a s gt mama 37a2b2 62 G NAGY 7 ODE DECEMBER 17 2008 We have seen how to compute the Laplace transform of several functions We now answer the following question How big is the set of functions that have a wellde ned Laplace transform It turns out that this is a big set of functions7 containing all functions that can be bounded by an exponential This is the main concept is the following result Theorem 42 Suf cient conditions If the function f R A R is piecewise continuous and there exists positive constants k and a such that lle lt he then the Laplace transform of f exists for all s gt a Proof of Theorem 42 We show that f is wellde ned7 that is nite7 by bounding its absolute value by a nite number One can check that the following inequalities hold7 MM 0 O lt name 0 lt e sthea dt h e quot O Therefore7 we can bound by the function Ma gkqaqsk sgta Since is bounded by a nite number for s gt a7 then is wellde ned This bound establishes the Theorem D We conclude this Section presenting three important properties of the Laplace transform7 which are summarized in the following three Theorems The rst one says that the Laplace transform of a linear combination of functions is the linear combination of their Laplace transforms Theorem 43 Linear combination If the Laplace transforms f and g of the functions f and g are wellde ned and a b are constants then the following equation holds hfh a b M Proof of Theorem 43 Since integration is a linear operation7 so is the Laplace transform7 as this calculation shows7 uwm wmowo O a e ft dt 12 e gt dt 0 O au wuu This establishes the Theorem D EXAMPLE 415 Compute 3t2 5 cos4t From the Theorem above we know that 3t2 5 cos4t 3 t2 5 cos4t gtogt m 33 8242 G NAGY r ODE DECEMBER 17 2008 63 Therefore 20s3 632 96 L 32 5 4t l T COSlt s3 32 16 lt1 The second Theorem relates f with L transform to solve differential equations Theorem 44 Derivatives If the f and f are wellde ned then the following relation holds This result is crucial to use the Laplace Llfl SLlfl f0 Furthermore f is wellde ned then it also holds lf l S2 lfl Sf0 fOl Proof of Theorem 44 The rst formula in the Theorem is proved by an integration by parts namely fl lim n eTSEft dt 332 New Mammal lim lawful 7 160 3 e ft dt naoo O f0 s lfl Since is wellde ned we have used the equation lim e m n 0 we then conclude that lfl s lfl f0A This proves the rst part of the Theorem The furthermore part is obtained using the relation we have just proved namely llyl S lfl fO ss m 7 1007 1 82 lfl 8f0 f 0A This establishes the Theorem D The third and last Theorem of this Section is also crucial to use this transform to solve differential equations It says that the map L acting on the set of piecewise continuous functions is onetoone This property is essential to de ne the inverse of the Laplace transform L l e present this result without the proof Theorem 45 Onetoone Given any piecewise continuous functions f and y it holds Llfl ngl fy As we mentioned before this last result allows to de ne the inverse Laplace transform Given a continuous function with values fs we de ne L 1 as follows ll s f6 gt ftl s 64 G NAGY 7 ODE DECEMBER 17 2008 42 THE INITIAL VALUE PROBLEM We now use the Laplace transform to nd solutions to differential equations with constant coefficients The main idea of the method is the following 1 Compute the Laplace transform of the whole differential equation 2 Use Theorems 43 and 44 to transform the original differential equation for the unknown function y into an algebraic equation for the unknown function y 3 Solve the algebraic equation for y 4 Transform back to obtain the solution function y EXAMPLE 421 Use the Laplace transform to nd the solution yt to the initial value problem y 2y 0 90 37 We have seen in the previous Chapters that the solution to this problem is yt 36 23 We now recover this result using Laplace transforms First compute the Laplace transform of the differential equation y 2y 0 0 By Theorem 43 we know that the lefthand side in the equation above can be expressed as follows ly l 2 lyl 07 Then Theorem 44 implies that s m7yltogt 2 y0 7 82 yy0 In the last equation we have been able to transform the original differential equation for y into an algebraic equation for We can solve for the unknown y as follows 90 3 7 7 9 32 9 82 where in the last step we introduced the initial condition y0 3 From the list of Laplace transforms given in Sect 41 we know that l 3 Lat 73L 72L 7 372z eisia7sH2 e17 M So we arrive at the equation y 3 2 I EXAMPLE 422 Use the Laplace transform to nd the solution yt to the initial value problem y 7 y 7 2y 0 ND 17 y 0 07 First compute the Laplace transform of the differential equation myquot 7 y 7 2y 7 0 7 0 By Theorem 43 we know that the lefthand side in the equation above can be expressed as follows um ly l 2 lyl 07 Then Theorem 44 implies that my 7 sylt0gt 7 m 7 my 7 gm 7 my 7 o G NAGY 7 ODE DECEMBER 17 2008 65 which is equivalent to the equation 82 i 8 i 2 lyl 8 1y0 y 0 1n the last equation we have been able to transform the original differential equation for y into an algebraic equation for Introduce the initial condition into the last equation above that is R77mqme7w We can solve for the unknown y as follows 8 i 1 32 7 s 7 2 i We now use partial fraction methods to nd a function whose Laplace transform is the right hand side of the equation above First nd the roots of the polynomial in the denominator lyl 3273720 7 3i1im 7 that is the polynomial has two real rootsi In this case we obtain 7 S i 1 s 7 2 s 1 i The partial fraction decomposition of the righthand side in the equation above is the fol lowing Find constants a and I such that s 7 1 a b s72s1 lyl A simple calculation shows amp L L s72s17s72 31 as 1 123 7 2 s72s1 saba72b 1 2 W r a b Hence 1 1 2 1 i 7 7 71 3 s 7 2 3 s 1 From the list of Laplace transforms given in Sect 411 we know that 1 1 1 2t 8 7 a 8 7 2 l6 l7 s 1 91 lea may So we arrive at the equation 7 l 2 a 7t m73 kl3 kl L 62 2equot yt 66 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 4213 Use the Laplace transform to nd the solution yt to the initial value problem y 74y 4y07 ND 17 y 0 71 First compute the Laplace transform of the differential equation Lb 7 4y 4y 0 01 By Theorem 43 we know that the lefthand side in the equation above can be expressed as follows 41M 7 4 4M 4 lyl 7 0 Then Theorem 44 implies that 82 41y 7 890 7 40 7 418 1yl7 y0 Mlyl 7 0 which is equivalent to the equation 52 7 4s 4 191 S 7 4 90 yOl This is the algebraic equation for Introduce the initial conditions y0 1 and y 0 1 into the last equation above that 1s 327434 y 8731 We can solve for the unknown y as follows 8 7 3 32 7 4s 4 i We now want to nd a function y whose Laplace transform is the righthand side in the equation above In order to see if partial fraction methods will be needed we now nd the roots of the polynomial in the denominator lyl 7 1 327437470 7 3i 4ix16716 7 3s 21 that is the polynomial has a single real rootsi In this case we obtain 7 s 7 3 7 8 7 2 This expression is already in the partial fraction decompositioni We now rewrite the right hand side of the equation above so that it is simple to see what function is its inverse Laplace transform lyl s 7 2 2 7 3 L y s 7 2s 1 7 s 7 2 1 7 8 7 2 8 7 2 7 1 1 7 s 7 2 s 7 2 From the list of Laplace transforms given in Sect 411 we know that 1 1 at 7 7 2 16173 7 82 we 1 tea 7 71 71 te2tli s7 a s 72gt G NAGY 7 ODE DECEMBER 17 2008 67 So we arrive at the equation lyl 62 l lteml L 62 7 tam yt 62 7 tam lt1 EXAMPLE 424 Use the Laplace transform to nd the solution yt to the initial value problem y 7 4y 4y 38in2t7 ND 17 40 1 First compute the Laplace transform of the differential equation yH 7 4y 4y 3 sin2ti The righthand side above can be expressed as follows i 7 L 32 22 7 32 4 By Theorem 43 we know that the lefthand side in the equation above can be expressed as follows 3 sin2t 3 sin2t 3 4W i 4 4M 4 lyl Then Theorem 44 implies that 6 324 6 2 7 7 7 7 7 S lyl 890 9 0 4 ls lyl y0 Mlyl 82 7 4 which is equivalent to the equation 6 2 5 i 43 4 Llyl 5 7 4 90 l yO This is the algebraic equation for Introduce the initial conditions y0 1 and y 0 1 into the last equation above that is 327434 y 373 6 32 4 We can solve for the unknown y as follows 3 7 3 6 8274s4 32744s24 From the Example above we know that s2 7 4s 4 s 7 22 so we obtain lyl l l 6 fly 372 3722 3722s24 4 2 From the previous example we know that l 1 2t 7 2t 7 e te S 7 2 S 7 2 43 We know use partial fractions to simplify the third term on the righthand side of Eq 42 The appropriate partial fraction decomposition for this term is the following Find constants a b c d such that 6 as b c d W 27 t 7 t 72 372 3 4 8 4 372 872 68 G NAGY 7 ODE DECEMBER 17 2008 Take common denominator on the righthand side above and one obtains the system a c 0 74ab720d0 4a 7 4b 7 4c 0 4d 4b 0 The solution for this linear system of equations is the following 3 77 b 77 7 d 3 a 4 2 C 4 Therefore 3 3 m A 3722324 4 324 4 372 3722 We can rewrite this expression above in terms of the Laplace transforms given on the list in Sect 41 as follows 7 3 s 3 2 3 1 3 1 3722324 1 s222 1 s222 1372 3722 cos2t 7 sin2t ELEM 3 t62 3 2 2 7 L ltcos2t 7 s1n2t e 4te 44 Finally introducing Eqs 43 and 44 into Eq 42 we obtain that the solution y has the form 72E Mt 15 2t62 cos2t sin2t 43 DISCONTINUOUS SOURCES We consider differential equations with piecewise continuous source functions and we use the Laplace transform to obtain their solutions The discontinuous sources are expressed in terms of appropriate combinations of step functionsi Theorem 47 is the main tool to obtain the inverse Laplace transform of these discontinuous functionsi Let us start by recalling that a function u is called a step function at t 0 iff 0 tlt0 ut 1 t 0 45 um mi c m c FIGURE 6 The graph of the step function given in Eq 45 a translation of this step function to the right and to the left respectively by a constant c gt 0 In Fig 6 we plot the step function u together with the functions with values ut 7 c and ut c with c gt 0 where the latter are the shift of the function u to the right and to the G NAGY 7 ODE DECEMBER 17 2008 69 left7 respectively Another useful function needed later on is a unit box in the interval a b wit gt a7 given 0 tlta btut7a7ut7b 7 W 1 alttltb 4 6 0 tgtb In Fig 7 we show a box function constructed as a combination of step functions m 7a m 7b m FIGURE 7 A box function I constructed with translated step functions We are interested to nd the Laplace transform of the step functions Lemma 46 Given any real number 0 the following equation holds 670 ut 7 0 8 Proof of Lemma 46 ut 7 0 Am e 5 ut 7 0 dt 00 e stdt c 7 39 1 7n 05 7 312 6 7 e gt 6705 T This establishes the Lemma D EXAMPLE 431 The Lemma 46 implies the following equations 36725 7 4314 7 2 8 r71 6353 ut 3 We now introduce two important properties of the Laplace transform Theorem 47 Translations Assume that the function f was a wellde ned Laplace transform for s gt a gt 0 and let 5 be a positive constant Then the following equations hold Llu Cft Cl 45133 5 gt a 4 7 e0 ft fs 7 c s gt a c 48 Notice that the inverse form of Eqs 4748 is given the the following equations7 respectively7 lle 138 7 W 7 cft 7 c 49 lms 7 6 7 e ft 410 70 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 432 We have shown in Sect 41 that 3 t 7 cosa 82 a2 Therefore Theorem 47 implies the following equations 725 5 ut72 cosat72 e W s 7 3 7 e cosat 7 8 7 32 a2 8 gt 3 lt1 Proof of Theorem 47 The proof is again based in a change of the integration variable We start with Eq 47 as follows ut 7 cft 7 5 7 Am e s ua 7 cft 7 6 dt 7Cme5 ft7cdt 739 t7c d7 dt O 574776 d7 7 a 0 6477 47 67 g sgtai The proof of Eq 48 is even simpler ectft 675160116 dt 0 7 eltHgt ft dt 0 Rs 7 c s gt or This establishes the Theoremi D EXAMPLE 433 Find the Laplace transform of the function flt t27gt2 4 11 Since the function f vanishes for all t lt l we can express it as follows ft ut71t2 7 2t 2 Now notice that completing the square we obtain t272t2t272t1712 t7121i That is the polynomial is a parabola t2 translated to the right by l and up by one This is a discontinuous functions as it can be seen in Fig 8 So the function f can be written as follows ft ut71t712 ut71i G NAGY 7 ODE DECEMBER 17 2008 71 FIGURE 8 The graph of the function f given in Eq 411 Since we know that t2 2337 then Eq 47 implies lftl Llut1t12l Llu 1l 5 2 751 e Ee ft lt1 EXAMPLE 434 Find the function f such that 6745 ft Notice that l 7 745 me7e 5 i v3 52 W Recall that sinat as2 a27 then Eqi 4 77 or its inverse form Eq 49 imply ft 7 ut 7 4sin3t 7 4 7 ft 7 ut 7 4sin3t 7 4 i lt1 EXAMPLE 435 Find the function such that 7 s 7 l u m7gj qg We rst rewrite the righthand side above as follows7 7 s 7 1 7 1 1 me7773F77 872 1 s7223s7223 ampLL 3732 J23 xP s722 527 We now recall Eq 48 or its inverse form Eqi 4 107 which imply mm 7 62 cost 7 kmsinbgt i 72 G NAGY 7 ODE DECEMBER 17 2008 So7 we conclude that 62 ft cost sin t 1 EXAMPLE 41316 Find a function f such that um17g3 3 We need to rewrite the denominator of the rational function 132 s 7 2 in order to use the partial fraction method to simplify this rational function We first find out whether this denominator has real or complex roots 1 5 1 8i7 71i 18 8772 We are in the case of real roots7 so we can rewrite s2s72 s71s21 The partial fraction decomposition in this case is given by 4 L L s71s27s71 82 7abs2a7b ab0 37132 2aib1i The solution is a 13 and b 7137 so we arrive to the expression u m7 a 1 1 371 36 32 Recalling that WT 1 Sid and Eq 47 we obtain the equation 1 7 1 7 7 g ut 7 2 e 2 7 g ut 7 2 e 2 2 which leads to the conclusion 1 7 7 P2 7 40 ft 3w 2e e 1 i lt1 The last two examples in this Section show how to use the methods presented above to solve differential equations with discontinuous source functions EXAMPLE 4317 Use the Laplace transform to find the solution of the initial value problem y 2y ut 7 47 y0 31 We compute the Laplace transform of the whole equation7 e745 ly l 2 lyl 1110 4 S G NAGY 7 ODE DECEMBER 17 2008 73 From the previous Section we know that 4 74s s y7y02 y1 7 e 7 swam 7ylt0gtes We introduce the initial condition y0 3 into equation above 7 3 745 1 7 72 745 1 y17732e 786 y173 e 16 islts2y We need to invert the Laplace transform on the last term on the righthand side in equation above We use the partial fraction decomposition on the rational function above as follows 1 a b 7 i 7 88 2 s s 2 7abs2a ab07 ss2 2a1i We conclude that a 12 and b 712 so 7 1 L ss272 8 82 We then obtain w 5 w 7 3 e72t rm 7 4 7 rpm 4 67min Hence we conclude that 7 72t l 7 7 72t74 yt 736 2ut 41 e i lt1 EXAMPLE 4318 Use the Laplace transform to nd the solution to the initial value problem 5 y y 1y 9t7 ND 07 WW 0 where the function g has the form sin t 0 lt t lt 7r gt 0 7 m 4112 As it can be seen in Fig 9 the source function y can be written as the following product W 7 W 7 W 7 gt1 sine since ut 7 ut 7 7r is a square function taking value one in the interval 0 7r and zero on the complement Finally notice that the equation sint 7 sint 7 7r implies that the function y can be expressed as follows gt ut sint 7 ut 7 7r sint gt ut sint ut 7 7r sint 7 W The last expression for g is particularly useful to nd its Laplace transform 1 7 1 gt WE 74 G NAGY 7 ODE DECEMBER 17 2008 mamp1 m FIGURE 91 The graph of the sine function a square function ut 7 ut7 7r and the source function 9 given in Eq 4112 Having this last transform is then not dif cult to solve the differential equation As usual Laplace transform the whole equation 5 lm ly l 1 lyl 1917 Since the initial condition are y0 0 and yO 0 we obtain 5 1 lt82s1gt y1 1 lt82s 3 821 We now need to use the partial fraction method to simplify the rational faction on the last equation above We rst nd out whether the denominator has real or complex roots si771iw75 so the roots are complex numbers In this case we rewrite the denominator completing the square 5217er w lyl1 45gt 5 1 1 1 5 5 1 2 2 7 2 2 7 7 77 7 2 7 7 11 31814 131218141 44 S1814 3121 In this case we have the decomposition 1 asb 58bd s21321821521 7 asb321csbdsgt21 ltsgt21lts21gt This equation implies the that a b c and d must be solution of the system ac0 bcd0 5 alcd0 5 b 7d11 4 Here is the solution to this system 12 16 4 a b 07 d i Hence we have found that 4 4s 3 743 1 my 06 17 W G NAGY 7 ODE DECEMBER 17 2008 75 Rewriting the polynomial 1 4s34lts 7 2 we then obtain the expression 1 4 5 a 1 s 1 7 7 7 7 7 4 7 7 lsgt21llt821gt 17 l 1lts gt2 11 1lts 5 11 lt821gt lt821gtl 4 L e E2 cost L equot2 sint 7 4Lcost L smog Therefore introducing the function 7 ht 4eit2 cost e E2 sint 7 4cost sint we have proven that 1 12 2 s 5 1 3 1 We can nally obtain the solution y to the differential equation since y satis es the equation Llyl 1 6 7 LWW Lht e Lhti yt ht ut 7 7rht 7 7r 1 Law We then conclude that 4141 GENERALIZED SOURCES This Section provides an elementary presentation of a particular generalized function the Dirac delta generalized function We introduce Diracls delta as an appropriate limit of a sequence of functions We also introduce the de nition of the integral and the Laplace transform of this generalized function as appropriate limits of the respective integral and Laplace transform of this sequence of functions We then use the Laplace transform meth ods to nd solutions of inhomogeneous differential equations with generalized functions as sources We call this presentation elementary since we do not mention the theory of distributions which should be the most appropriate way to present a complete theory of generalized functionsi Since we restrict ourselves to the Dirac delta generalized function only we have chosen to present it as a limit of a sequence of functions De nition 48 Consider the sequence of functions for n gt 1 0 tlt0 Mt n 0lttlt 0 tgt1 Fix a value t E R and then for every t introduce the limits lim 6 t 6ti The collection of all those limits denoted by 6 is called Dirac s delta generalized function 76 G NAGY 7 ODE DECEMBER 17 2008 We see that the Dirac delta generalized function is the zero function on R 7 0 it is not de ned at t 0 and has the extra property summarized in Theorem 49 Essentially a generalized function is a function for every value of its argument t E R with the possible exception of a nite number of those t where the function can be in nite in a very particular way In Fig 10 can be seen the graph of few members of the sequence above and it is not difficult to see that the elements of the sequence can be expressed as follows 1 w 7 n ut ut 71 413 We remark that there exists different sequences that provide the same limit 6 1 FIGURE 10 The graph of the function 6 given in Eq 413 for n l 2 and 4 Using this limit procedure one can generalize several properties from functions to gen eralized functions For example linear 39 39 and 39 quot 39 of a function by a generalized function as follows 6t 7 c 1301067 7 0 am b6t n13 a6nt b6nt ft6t lim ft6nt n7gtoltgt Il The integral and the Laplace transform of the Dirac delta generalized function can also be de ned a a limiting process name y b b sum lim 6ntdt rm lim many The following result summarizes one of the most important properties of the Dirac delta generalized functioni Theorem 49 Dirac s delta Let c 6 ab C R and f R 7gt R be a continuous function Then the following equations hold b6t7cdt 1 rpm 1 414 b ft so 7 0 dt 165 rm 7 5 emf 415 Proof of Theorem 49 We prove both equations in 415 and then the equations in 414 are just the particular cases f 1 an 5 0 respectively We start with the rst equation G NAGY 7 ODE DECEMBER 17 2008 77 in 415 b b so 7 c ft dt 7 lim w 7 c ft dt a ab 1 nlgroioa nut 7 c 7 ut 7 c7 ftdt c lim dt where in the last line we used that c E abl If we denote by F any primitive of f that is F then we can write I so 7 c ft dt 7 Jigqu 7 1 The second equation in 415 can be proven as follows mm 7 a 7 7113 mm 7 a 7 ut 7c7 7c 7ltcigts a i 67gt lim it 71700 3 l 7 e e hm n7gtoltgt i n The limit on the last line above is a singular limit of the form 9 so we can use the l Hopital 0 7 rule to compute it that is l 7 e 5 lim lim n7gtoltgt n7gtoltgt We therefore conclude that 6t 7 c e i This establishes the Theoreml D The solution of a differential equation with a Dirac s delta generalized function as a source and zero initial conditions will play an important role later on so we give that function a particular namel De nition 410 Given continuous functions p q and a constant c the function yg solution to the initial value problem yf ptyfsqtya 76t67 mo 07 430 0 is called the fundamental solution EXAMPLE 441 Find the function yg the fundamental solution of the initial value problem a my 2y 7 6a yltogt 7 0 Mo 7 0 78 G NAGY 7 ODE DECEMBER 17 2008 We compute the Laplace transform on both sides of the differential equation above Llynl 2 lyl 2 Llyl LWUl 1 v 52 25 2 lyl 17 where we have introduced the initial conditions on the last equation above So we obtain 1 Llyl 32 2s2 The denomination in the equation above has complex valued roots since 1 5i 72i478 therefore we complete squares s2 23 2 s D2 l and then we have to solve the equation 1 My 870271 equotsint 7 Mt equotsint EXAMPLE 442 Find the solution y to the initial value problem y7y7206t73 y0 1 yO 0 The Laplace transform of the differential equation is given by y 7 y 720 6t 7 3 32 7 l y 7 s 720 6 35 where in the second equation we have already introduced the initial conditions We arrive to the equation y 732871 7 20 6 35 732171 cosht 7 20 ut 7 3 sinht 7 3 which leads to the solution yt cosht 7 20 ut 7 3 sinht 7 3 EXAMPLE 443 Find the solution to the initial value problem y4y6t77r76t727r y0 0 yO 0 We again Laplace transform both sides of the differential equation W17 my 7 mm 7 n 7 mm 7 2m 7 lt52 4 my 7 7 where in the second equation above we have introduced the initial conditions We then obtain e77rs e72 m 7 W ut 7 7r sin2t 7 7 ut 7 27r sin2t 7 270 lyl The last equation can be rewritten as follows yt ut 7 7r sin2t 7 70 7 ut 7 27r sin 2t 7 2 G NAGY 7 ODE DECEMBER 17 2008 79 which leads to the conclusion that 2t ut 7 ut 7 2 sin2t 45 CONVOLUTION SOLUTIONS This Section describes a way to express a solution to an inhomogeneous linear differen tial equation in terms of a convolution between the fundamental solution and the source function of the differential equation This particular form to express the solution is useful in mathematical physics to study different properties of solution It is also a widely used in physics and engineering The plan for this Section is to rst present the de nition of a convolution of two functions and then discuss the main properties of this operation with Theorems 413 and 414 containing perhaps the most important results of this Section The latter result uses convolutions to write solutions to inhomogeneous differential equations De nition 411 Assume that f and g are piecewise continuous functions or one of them is a Dirac s delta generalized function The convolution of f and g is a function denoted by f 96 g and given by the following expression 2 f 7 gm 7 f7gt 7 7 d7 0 We now show an example of the convolution of two functions and later on in this section we will explicitly compute the integral de ning the convolution in this example EXAMPLE 451 The convolution of the functions e t and gt sint is given by L f gt e77 sint 7 7 d7 0 In this case it is not dif cult to compute the integral above just by integrating by parts twice which leads us to the formula Ot e77 sint 7 7 d7 e77 cost 7 7 hence we obtain 2 7 e77 sint 7 7 t z 7 e77 sint 7 7 d7 0 o z z 2 e77 sint 7 7 d7 e77 cost 7 7 7 e77 sint 7 7 O o o equot 7 cost 7 0 sint We then conclude that z e77 sint 7 7 d7 equot sint 7 cost 416 O The main properties of the convolution are summarized in the following result Lemma 412 Properties Given piecewise continuous functions f g and h the following equations hold f 96 g g 96 f commutative f 96 g 96 h 96 g 96 h associative 80 G NAGY 7 ODE DECEMBER 17 2008 f g h f g f 96 h distributive iv f 96 0 0 neutral element 1 f 96 6 f identity element Proof of Lemma 412 We only prove properties and i the rest are simple to obtain from the de nition of convolution The rst property can be obtained by a change of the integration variable as follows fgtOtfrgtirdr 7277 472747 fty1df Eewaeoo O a ftA Finally property v is straightforward form the de nition of the delta generalized function na 2 f6t fT6tTdT O ftA This establishes the Lernrnal D We now show the relation between the Laplace transform and the convolution operations The following result says that the Laplace transform of a convolution is the usual product of the individual Laplace transforms Theorem 413 Laplace transform Assume that the functions f and 9 have well de ned Laplace transforms and well de ned convolution including the case that one of them is a Dirac s delta generalized function Then the following equation holds lf gl Llfl lyl EXAMPLE 4 52 Compute the Laplace transform of the convolution z ft e77 sint 7 739 dr 0 The function f above is the convolution of the functions 9t 6quot Mt 51110 that is f g 96 hi Therefore Theorem 41 says that Llfl lyl lhl Since lo we then conclude that r o 0 Then is straightforward to check that may WW we M dr 0 0 lt9 fT dt 0 lw fl w lfl lgl lfwl This establishes the Theoremi D EXAMPLE 453 Use Laplace transforms to compute the convolution z ft 6 7 sint 7 739 dr 0 We have seen in the example above that 1 W m 82 G NAGY 7 ODE DECEMBER 17 2008 A partial fraction decomposition of the righthand side above implies that 7 1 1 17 3 fl l31 321l 1 m 1 m m 7 2 31 32H 32 again 4mm 7 moose This says that ft equot sint 7 costi We then conclude that 0 e77 sint 7 739 d7 equot sint 7 cost 7 Which agrees With 416 in the first example above lt1 We close this Section With an important application of Theorem 41 to differential equa tions Theorem 414 Convolution solutions Given constants a ag and a piecewise con tinuous function g the solution y to the inhomogeneous initial value problem y a y any 90 y0 07 MO 07 418 can be expressed in the convolution form W ya 9W where y is the fundamental solution associated with the equation above that is y5 is the solution of yf a1y auy s 6t7 mo 07 MW 0 Proof of Theorem414 Compute the Laplace transform of Eq 418 that is 82 a1 8 ac lyl lyly Where we have already introduced the initial conditions of the problemi We then have the equation 1 y 82alsao Where the last equation is obtained from formula 1 L i val 82 a18 a0 a lyal lyl Therefore Theorem 41 implies that y 96 9 This establishes the Theoremi D EXAMPLE 454 Use convolutions to express the solution of the inhomogeneous initial value problem 9 2y 2y Sinat7 90 07 yO 0 We compute the Laplace transform of the Whole equation 32 23 2 L y sinat y 32 82 sinati G NAGY r ODE DECEMBER 17 2008 83 We have seen in Example 441 that 1 1 it y6 m W e s1nt Therefore7 um e sinlttgt usinwn and we then conclude that sin7 sinat 7 7 d7 84 G NAGY 7 ODE DECEMBER 17 2008 CHAPTER 5 LINEAR SYSTEMS 5ili OVERVIEW OF LINEAR ALGEBRA We now review the main ideas from linear algebra we need to study systems of linear differential equations We focus only on matrix algebra linearly dependent and independent vector sets determinants and the 39 of 39 and 39 o a square matrix Sill Systems of algebraic equations We start the review recalling the de nition of an algebraic system of linear equations De nition 51 An n X n algebraic system of linear equation is the following Given constants aij and 12 where indices 2 l n gt 1 nd the constants zj solutions of the system 1111 ainzn 7 b1 51 5 2 anizz annzn 7 bn 53 The system is called homogeneous the sources vanish that is b bn 0 EXAMPLE Silil A 2 X 2 linear system on the unknowns 11 and 12 is the following 211 7 12 0 711 212 3 A 3 X 3 linear system on the unknowns 11 I2 and 13 is the following 11 21213 1 7311 12 313 24 12 7 413 71 lt1 One way to nd a solution to a linear system is by substitution Compute 11 from the rst equation and introduce it into all the other equationsi Then compute 12 from this new second equation and introduce it into all the remaining equationsi Repeat this procedure till the last equation where one nally obtains zni Then substitute back and nd all the ziforiln7i A computational more efficient way to nd the solution is to perform Gauss elimination operations on the augmented matrix of the systemi In order to introduce the latter we rst introduce the coefficient matrix the source vector and the unknown vectors of the n X n system given in Eqs 5 35il which are respectively given by an 39 39 39 ain 1 1 I1 A 7 z z 7 b 7 z 7 m 7 anl am bn In Using this matrix notation the linear system given in Eqs 5 35il can be written in a compact way as follows Am b 54 G NAGY r ODE DECEMBER 17 2008 85 where we have introduced the matrixvector product an quot39 ain I1 a1111 quot 39 ainIn A9 anl quot39 am In an111quot39 ainIn The augmented matrix of the system above is an n X n 1 matrix A111 which contains the coef cients and the source coef cients Therefore the augmented matrix contains the complete information of the linear system EXAMPLE 512 The coef cient matrix source vector and augmented matrix of the rst system given in the Example 511 is the following 211721 1712 MPH 721 1 The same objects for the second system in that example are given by 1 2 1 1 1 2 1 1 1 A 731 3 b 24 A112 731 3 1 24 0 1 74 71 0 74 1 71 The matrixvector product Am for the 2 X 2 case is the following 7 2 71 117 211712 171 21 1121 fwd The matrixvector product Am for the 3 X 3 case is the following 1 2 1 11 11 212 13 Am 73 1 3 12 7311z2313 0 1 74 13 12 7 4 lt1 The matrixvector product Am has two important interpretations we discuss the rst one here while the second interpretation is given in the next Subsection 512 The matrixvector product can be understood as a linear combination of the column vectors of the matrix A that is expressing A 01 an then Am mm anzn EXAMPLE 513 The matrixvector product Am for the 2 X 2 case can be written as follows 7 2 71 11 7 2 71 A 171 21 1121 1711 l 21 The matrixvector product Am for the 3 X 3 case can be written as follows 1 2 1 11 1 2 Am 73 1 3 12 73 11 1 12 3 13 0 1 74 13 0 1 74 86 G NAGY 7 ODE DECEMBER 17 2008 511121 A matrix is a function The matrixvector product leads to the interpretation that an n X n matrix A is a function If we denote by R the space of all n vectors we see that the matrixvector product associates to the n vector m the unique n vector y Ami Therefore the matrix A determines a function A R A Rnl EXAMPLE 51114 Describe the action on R2 of the function given by the 2 X 2 matrix 0 1 A 1 0 55 The action of this matrix on an arbitrary element x E R2 is given below 7 0 1 11 7 12 A 11 ol 1le 1111 A Therefore this matrix interchanges the components 11 and 12 of the vector m It can be seen in Fig 515 that this action can be interpreted as a re ection on the plane along the line 11 121 FIGURE 12 Geometrical meaning of the function determined by the matrix in Ed 515 and the matrix in Ed 516 respectively EXAMPLE 51115 Describe the action on R2 of the function given by the 2 X 2 matrix 0 71 A 1 0 56 The action of this matrix on an arbitrary element x E R2 is given below 7 0 71 11 7 712 A 11 ol 1le i l Ill In order to understand the action of this matrix we give the following particular cases 071170 0711771 0717170 1 0 0 7 1 7 1 0 1 7 1 7 1 0 0 7 71 These cases are plotted in the second gure on Fig 12 and the vectors are called 1 y and 2 respectively We therefore conclude that this matrix produces a ninety degree counter clockwise rotation of the plane lt1 G NAGY r ODE DECEMBER 17 2008 87 5 13 Gauss elimination operations We now review three operations that can be per formed on an augmented matrix of a linear systemi There operations were introduced by Gauss and they have the property that they change the augmented matrix without chang ing the solutions of the linear system associated with that augmented matrix The Gauss elimination operations refers to the following three operations performed on the augmented matrix i Adding to one row a multiple of the another ii lnterchanging two rows iii Multiplying a row by a nonzero numberi These operations are respectively represented by the symbols given in Fig 13 l3 l FIGURE 13 A representation of the Gauss elimination operationsi Cl As we said above7 the Gauss elimination operations change the coef cients of the aug mented matrix of a system but do not change its solution Two systems of linear equations having the same solutions are called equivalent It can be shown that there is an algorithm using these operations such that given any n X n linear system there exists an equivalent system whose augmented matrix is simple in the sense that the solution can be found by inspection EXAMPLE 516 Find the solution to the 2 X 2 linear system given in Example Silil using the Gauss elimination operations Consider the augmented matrix of the 2 X 2 linear system in Example 5 117 and perform the following Gauss elimination operations7 2710H2710H2710H 7123 44 603 6 In the last augmented matrix the solution is easy to read A precise way to de ne the notion of easy to read is captured in the notion is in echelon formi An m X n matrix is in echelon form iff every element below the diagonal vanishesi Matrices with this property are also called upper triangular A matrix is in reduced echelon form iff it is in echelon form and the rst nonzero element in every row satis es both that it is equal to l and it is the only nonzero element in that column EXAMPLE 517 The following matrices are in echelon form7 1 3 2 3 2 g 1 i 0 l 0 4 72 0 0 0 And the following matrices are not only in echelon form but also in reduced echelon form7 10 104 01015 000 88 G NAGY 7 ODE DECEMBER 17 2008 lt1 Summarizing the Gauss elimination operations can transform any matrix into a reduce echelon formi Once the augmented matrix of a linear system is written in reduced echelon form it is not difficult to decide whether the system has solutions or not EXAMPLE 51118 Find the solution to the 3 X 3 linear system given in Example 511 using the Gauss elimination operations Consider the augmented matrix of the 3 X 3 linear system in Example 511 and perform the following Gauss elimination operations 121 1 121 1 121 1 7313124H076l27a0174171 01741710174171076127 10 9 3 10 9 3 100 76 1147 0174171H0174171H01013 123 0034134 00111 00111 131 5141 Linearly dependent and independent set of vectors De nition 52 A set of vectors 121 vk with h gt 1 is called linearly dependent Ha i there exists constants c ck with at least one of them nonzero such that vcquot39vkck 0 57 The set of vectors is called linearly independent l i it is not linearly dependent that is the only constants c ck that satisfy Eq 57 are given by c ck 0 In other words a set of vectors is linearly dependent iff one of the vectors is a linear combination of the other vectorsi When this is not possible the set is called linearly inde pendent EXAMPLE 51119 Show that the following set of vectors is linearly dependent 13 fl 3737 and express one of the vectors as a linear combination of the other two We need to nd constants c1 c2 and c3 solutions of the equation 1 3 71 0 1 3 71 c1 0 2 c1 2 52 2 c3 0 gt 2 2 2 c2 0 3 1 5 0 3 1 5 c3 0 The solution to this linear system can be obtained with Gauss elimination operations as follows 1 3 71 1 3 71 1 0 2 Cl 2037 2 2 2 A 0 74 4 A 0 l 71 A 0 l 71 gt C2 3 1 8 0 l 71 0 0 0 c3free1 n w G NAGY r ODE DECEMBER 17 2008 89 Since there are nonzero constants cl C2 C3 solutions to the linear system above the vectors are linearly dependenti Choosing C3 71 we obtain the third vector as a linear combination of the other two vectors lt1 5151 Determinants We now review the de nition and properties of the determinant of 2 X 2 and 3 X 3 matrices De nition 53 The determinant of a 2 X 2 matrix a a A 11 12 a21 a22 is given by an a12 detA anagg 7 algagli a21 a 2 The determinant of a 3 X 3 matrix an an dis 0431 0432 0433 is given by an a1 dis detA agl agg agg all an a32 ass 0421 0422 0431 0432 0422 0423 0432 0433 a12 0413 0421 0423 0431 0433 EXAMPLE 511110 The following three examples show that the determinant can be a negative zero or positive number 1 2 1 2 2 1 3 4 47672 3 4 8735 2 4 47401 The following is an example shows how to compute the determinant of a 3 X 3 matrix 1 3 71 1 1 2 1 2 1 21 1 131 w 1 3 2 1 2 1 3 1 3 2 172732737473 71371 1 lt1 The absolute value of the determinant of a 2 X 2 matrix A 01 02 has a geometrical meaning It is the area of the parallelogram whose sides are given by 01 and biag that is by the columns of the matrix A see Fig 141 Analogously the absolute value of the determinant of a 3 X 3 matrix A 01 mg as also has a geometrical meaning It is the volume of the parallelepiped whose sides are given by Hi 02 and bias that is by the columns of the matrix A see Fig 141 90 G NAGY 7 ODE DECEMBER 17 2008 FIGURE 14 Geometrical meaning of the determinant The determinant of an n X n matrix A can be de ned generalizing the properties that areas of parallelogram have in two dimensions and volumes of parallelepipeds have in three dimensions One of these properties is the following if one of the column vectors of the matrix A is a linear combination of the others then the gure determined by these column vectors is not n dimensional but n71dimensional so its volume must vanish We highlight this property of the determinant of n X n matrices in the following result Proposition 54 The set of nvectors 1 on with n gt 1 is linearly dependent detv1 on 0 EXAMPLE 5111 Show whether the set of vectors below linearly independent 1 3 73 2 2 2 3 1 7 The determinant of the matrix whose column vectors are the vectors above is given by 1 3 73 2 2 2 1147273147673276 3 1 7 12 7 24 12 0 Therefore the set of vectors above is linearly dependent lt1 516 Eigenvalues and eigenvectors The concepts of linear algebra that we have re viewed so far are needed to understand the following subject which is going to be the central concept in solving systems of differential equations The eigenvalues and eigenvec tors of an n X n matrix De nition 55 A number A and a nonzero nvector v are respectively called an eigenvalue and eigenvector of an n X n matrix A the following equation holds Av A1 The eigenvectors determine particular directions in the space R that remain invariant under the action of the function given by the matrix A G NAGY r ODE DECEMBER 17 2008 91 EXAMPLE 511112 Verify that the pair A1 v1 and the pair A2 v2 are eigenvalue and eigen vector pairs of matrix A given below 1 1 4 7 1 1 7 1 3 A 7 l3 ll 7 A 2 1 2 7 i 3902 7 1 A We just must verify the de nition of eigenvalue and eigenvector given above We start with the rst pair 1 3 1 4 1 MLJHMLFNW A similar calculation for the second pair implies 1 3 71 2 71 AElldld ldhw EXAMPLE 5113 Find the eigenvalues and eigenvectors of the matrix 0 1 ALJA Ga This is the matrix given in Example 514 The action of this matrix on the plane is a re ection along the line 11 12 as it was shown in Fig 121 Therefore this line 11 12 is left invariant under the action of this matrix This property suggest that one eigenvector is any vector on that line for example an A 1 am A So we have found one eigenvalue eigenvector pair g Analogously it is simple to nd a second eigenvector as follows m A l511311311Hgtl31 A So we have found one eigenvalue eigenvector pair These two eigenvectors are displayed on Fig 151 lt1 There exist matrices that do not have eigenvalues and eigenvectors as it is show in the example belowi EXAMPLE 5114 Fix any number 9 6 0270 and then de ne the matrix 7 cost9 7 81116 7 sint9 cost9 It can be veri ed that the action of this matrix on the plane is a rotation counterclockwise by and angle 9 as shown in Fig 519 A particular case of this matrix was shown in Ex ample 515 where 9 7r21 Since eigenvectors of a matrix determine directions which are left invariant by the action of the matrix and a rotation does not have such directions we 59 92 G NAGY 7 ODE DECEMBER 17 2008 FIGURE 15 The rst picture shows the eigenvalues and eigenvectors of the matrix given in 58 The second picture shows that the matrix in Eq 59 makes a counterclockwise rotation by an angle 9 which proves that this matrix does not have eigenvalues or eigenvectors conclude that the matrix given in Eq 59 does not have eigenvectors and so it does not have eigenvalues The last part of this Subsection is dedicated to solve the eigenvalueeigenvector problem Given an n X n matrix A nd if possible all its eigenvalues and eigenvectors that is all pairs A and v f 0 solutions of the equation Av Av This problem is more complicated than nding the solution x to a linear system Am b where A and b are known In the eigenvalueeigenvector problem above neither A nor 1 are known We solve the eigenvalueeigenvector problem for a matrix A as follows a First nd the eigenvalues A b Second nd for each eigenvalue A the corresponding eigenvectors v The following result summarizes a way to solve each step mentioned above Theorem 56 Eigenvalueseigenvectors a The number A is an eigenvalue of an n X n matrix A the following equation holds detA7AI 0 510 7 Given an eigenvalue A of an n X n matrix A the corresponding eigenvectors v are the nonzero solutions to the homogeneous linear system A7AIv0 511 Proof of Theorem 56 The number A and the nonzero vector v are solutions to the equation AvAv ltgt A7AIv0 Since 1 f 0 the last equation above says that the columns of the matrix A 7 A1 are linearly dependent This last property is equivalent by Proposition 54 to the equation detA 7 AI 0 which is the equation that determines the eigenvalues A Once this equation is solved substitute each solution A back into the original eigenvalueeigenvector equation A 7 A01 0 G NAGY r ODE DECEMBER 17 2008 93 Since A is known this is a linear homogeneous system for the eigenvector components It always has solutions since A is such that det AiAI 01 This establishes the Theoremi D EXAMPLE 51115 Find the eigenvalues A and eigenvectors v of the 2 X 2 matrix 1 3 A 3 ll 1 We rst nd the eigenvalues as the solutions of the Eqi 5110 So we start computing 713 10713 A0717A 3 A Alilg 11410 11713 1Ho 1H 3 11 and then we compute its determinant 7 7 717A 3 7 7 27 4 OidetA AI7 3 17A 7A 1 9 A772 We have obtained two eigenvalues so now we introduce A1 4 into 5111 We start 174 3 73 3 A 4Il3 174ll3 73l then we solve for DJ the equation computing 73 3 71 0 7 7 1 7 A 41 7 0 ltgt 3 73 v 7 0 The solution can be found using Gauss elimination operations as follows 73 3H171H171 7117 3 73 3 73 0 0 1 free Al solutions to the equation above are then given by v 1 1 1 1 7 where we have chosen vr 11 A similar calculation provides the eigenvector v associated with the eigenvalue A 72 that is rst compute the matrix v7 1 v4r v2 A 21 then we solve for v the equation 1 W 0 a g 1 13 131 1 The solution can be found using Gauss elimination operations as follows 3 3 1 1 1 1 17 1 7 13 31 A 13 31 A lo 01 free Al solutions to the equation above are then given by 113131121 3 W111 94 G NAGY 7 ODE DECEMBER 17 2008 where we have chosen v 1 We therefore conclude that the eigenvalues and eigenvectors of the matrix A above are given by lt1 It is useful to introduce the following notation Given an n X n matrix A we call the function p detA 7 Al the Characteristic polynomial of the matrix A Therefore the eigenvalues of a matrix A are the roots of its characteristic polynomial ln Example 5i1i15 above the characteristic polynomial is given by pQAilV79 pQA272A78 Since the matrix A in this example is 2 X 2 its characteristic polynomial is of degree two One can show that the characteristic polynomial of an n X n matrix has degree n De nition 57 Since the characteristic polynomial p of an n X n matrix A having eigenvalues M with i l h lt n can always be expressed as follows MMMAN MMW7 we call the number ri the algebraic multiplicity of the eigenvalue Ai EXAMPLE 5116 Find the algebraic multiplicities of the eigenvalues of the matrix Hi i l In order to nd the algebraic multiplicity of the eigenvalues we need rst to nd the eigenvalues We now that the characteristic polynomial of this matrix is given by ma a lfy wiutg The roots of this polynomial are A1 in the following way 4 and A2 72 so we know that p can be rewritten MMM O We conclude that the algebraic multiplicity of the eigenvalues are both one that is A14 r1l lM7z 1 lt1 An n X n matrix A has every algebraic multiplicity coef cients equal to one iff A has n different eigenvaluesi The following results asserts that the corresponding eigenvectors form a linearly independent set Theorem 58 Different eigenvalues If the n X n matrix A has n di erent eigenvalues then A has a set ofn linearly independent eigenvectors If A has a repeated eigenvalue then the number m of linearly independent eigenvectors satis es the inequality m lt n G NAGY r ODE DECEMBER 17 2008 95 Proof of Theorem 58 Let A1 An be the eigenvalues of matrix A all different from each other Let v0 v the corresponding eigenvectors that is Av i Aivi with i l n We have to show that the set v1 v is linearly independent We assume that the opposite is true and we obtain a contradiction Let us assume that the set above is linearly dependent that is there exists constants cl on not all zero such that 511 571100 0 512 Let us name the eigenvalues and eigenvectors such that 01 f 0 Now multiply the equation above by the matrix A the result is 51A1v1 annv 0 Multiply Eq 512 by the eigenvalue An the result is clAnv1 annv 0 Subtract the second from the rst of the equations above then the last term on the right hand sides cancels out and we obtain c117 Anv1 CHQH 7 mam 7 0 513 Repeat the whole procedure starting with Eq 513 that is multiply this later equation by matrix A and also by AnL1 then subtract the second from the rst the result is cm 7 mm 7 A1vlt1gt CHMH 7 wow 7 MAM 7 0 Repeat the whole procedure a total of n 7 1 times in the last step we obtain the equation 510 7 nl 1 n7139 quotA1 A2U1 0 Since all the eigenvalues are different we conclude that 01 0 however this contradicts our assumption that 01 0 Therefore the set of n eigenvectors must be linearly independent The second part of the Theorem is straightforward since there are repeated eigenvalues then the proof above does not hold so the set of n eigenvectors could be linearly dependent This means that the set of linearly independent eigenvectors contains less than or equal to n eigenvectors This establishes the Theorem In the case that the eigenvalues are repeated the Theorem above can be made more precise For this purpose we introduce the following concept De nition 59 The geometric multiplicity of an eigenvalue Ai denoted by 51 is the maximum number of linearly independent eigenvectors corresponding to that eigenvalue Ai EXAMPLE 5117 Find the geometric multiplicities of the eigenvalues of the matrix Adi i l In order to nd the geometric multiplicity of the eigenvalues of the matrix above we need rst to nd both the eigenvalues and the eigenvectors of this matrix This part of the work was already done in the Example 5115 above and the result is 96 G NAGY 7 ODE DECEMBER 17 2008 The eigenvalues of the matrix presented in the previous examples have equal algebraic and geometric multiplicities Hence they have the maximum possible number of linearly independent eigenvectors which in the case of 2 X 2 matrices is two We have seen in Theorem 58 that the maximum number m of linearly independent eigenvectors of an n X n matrix satis es the inequality 0 lt m lt n One can prove an even more precise statementi One can prove that a similar inequality holds for the maximum number of eigenvectors corresponding to each eigenvalue Here is the precise statement which we present without proof Theorem 510 Eigenvalue multiplicities If an eigenvalue A of an n X n matrix A has algebraic multiplicity r and geometric multiplicity s then the following inequality holds 8 lt r We nish this Subsection with an example showing that two matrices can have the same eigenvalues and so the same algebraic multiplicities but different eigenvectors with different geometric multiplicities EXAMPLE 51118 Find the eigenvalues and eigenvectors of the matrix 3 0 l A 0 3 2 0 0 1 We start nding the eigenvalues which are the roots of the characteristic polynomial 37A 0 l A117 17 p 0 37A 2 771732 a A 73 72 0 0 1 7 A 2 7 T2 We now compute the eigenvector associated with the eigenvalue A1 l which is the solution of the linear system 2 0 1 v 0 A71v10 gt 0 2 2 pg 0 0 0 0 v9 0 1 1 v3 2 0 1 1 0 5 v1 7 022H011 171 0 0 0 0 0 0 U2 111 free Therefore choosing 111 2 we obtain that fl 1 72 A11 n1 811 2 In a similar way we now compute the eigenvectors for the eigenvalue A2 3 which are all solutions of the linear system 2 0 0 1 v1 0 A73Iv20 gt 0 0 2 pg 0 0 0 72 Ug 0 G NAGY r ODE DECEMBER 17 2008 After the few Gauss elimination operation we obtain the following 2 0 0 1 0 0 1 1 free 002H000 v 2gtfree 0 0 72 0 0 0 U9 3 1 Therefore we obtain two linearly independent solutions the rst one M with the choice v1 1 v2 0 and the second one 102 with the choice v1 0 v2 1 that is 1 0 W 0 1112 1 A23 2 3221 0 0 Summarizing the matrix in this example has three linearly independent eigenvectorsi lt1 EXAMPLE 51119 Find the eigenvalues and eigenvectors of the matrix 3 1 1 A 0 3 2 0 0 1 Notice that this matrix has only the coefficient an different from the previous example Again we start nding the eigenvalues which are the roots of the characteristic polynomial 1 A 7 1 7 1 7 7 1 71 2 771732 17 A23 7 22 So this matrix has the same eigenvalues and algebraic multiplicities as the matrix in the previous example We now compute the eigenvector associated with the eigenvalue A1 1 which is the solution of the linear system 37 1 pm o 34gt 0 0 2 1 1 0 A71v10 gt 0 2 2 Uglgt 0 0 0 0 0 After the few Gauss elimination operation we obtain the following 1 2 1 1 1 1 1 1 0 0 1 07 022H011H011 v217v 1 0 0 0 0 0 0 0 0 0 111 freer Therefore choosing 111 1 we obtain that 0 v0 71 A11 n1 811 1 In a similar way we now compute the eigenvectors for the eigenvalue A2 31 However in this case we obtain only one solution as this calculation shows 0 1 1 v1 0 21730112 gt 0 0 2 Ug2gt 0 0 0 72 Ug 0 98 G NAGY 7 ODE DECEMBER 17 2008 After the few Gauss elimination operation we obtain the following 2 011 010 free 002H001 v 2gt 0072 000 2 1131 Therefore we obtain only one linearly independent solution which corresponds to the choice 112 1 that is 1 7 7 1 v20 A23 2 3211 0 Summarizing the matrix in this example has only two linearly independent eigenvectors and in the case of the eigenvalue A2 3 we have the strict inequality 1 52 VARIABLE COEFFICIENTS SYSTEMS 52L Systems of linear differential equations We begin the study of rst order sys tems of linear differential equations on several unknown functions The equation is called a system since there is more than one single equation and linear if all the differential equations are linear in the unknown functionsi De nition 511 A system of n linear rst order differential equations on n un knowns with n gt 1 called ngtlt n system is the following Given the functions aij bi de ned on an interval I C R where ij l n nd the functions zj solutions of the equations 1 a11tzla1ntznb1t 5 14 1an1t zla7mtznbnti 515 The system is called homogeneous the source functions satisfy that b bn 0 EXAMPLE 521 The case n 1 corresponds to a single equation in the unknown 11t that is 11 a11t 11 121 The case n 2 is called a 2 X 2 system and is the following Find the unknown functions 11 and 12 solutions of the system 1 1 a11t 11 and b1t7 12 a21t 11 a22t 12 b2t1 lt1 EXAMPLE 522 Find all solutions 11t and 12t of the 2 X 2 constant coefficients homo geneous system 11 11 i 127 12 711 12 G NAGY r ODE DECEMBER 17 2008 99 The solutions solution of this system can be found as follows First add the two equations together second take the second equation from the rst one one then obtain the following two equations respectively 11120 I17I2 211 712 Introduce the unknowns vzl12 wI1I27 then equations above have the form 1 0 1 Cl 11 2w 105262 1 From the de nitions of v and 11 we also have the inverse equations 1 1 11 vw 12 v7w1 So we conclude that the general solution is given in terms of two free constants 01 and 02 as follows 12t l 01 7 5262 i 2 1 11t E 01 5262 7 lt1 5122 Matrix notation It is not dif cult to see that matrix notation is very helpful to express systems of linear equations Consider the n X n linear system given in Eqs 5114 5115 and introduce the matrix of coef cients the source vector and the unknown vector as follows an t 39 quot 1716 121t 11t Am 2 z 7 W 2 7 ma 2 am amt Mt 1 Then the n X n system of linear differential equations given in Eqs 5145115 can be expressed as follows 1 At1 bti EXAMPLE 5213 Use matrix notation to write down the 2 X 2 system given Example 512121 In this case the matrix of coef cients and the unknown vector have the orm 7 1 1 7 110 A l71 ll 1 km 7 so the differential equation can be written as follows 1 A1 ltgt 2 lt1 EXAMPLE 51214 Find the explicit expression for the linear system 1quot A1 b in the case that 7 1 3 7 e 7 11 A713 11 wily 100 G NAGY 7 ODE DECEMBER 17 2008 The 2 X 2 linear system is given by z 1 3 z e 11I13I2etv 1 1 3 7 42gt 12 3 1 12 2e 1 2 311x22e3 i lt1 523i Reduction to rst order A second order linear differential equation for a single unknown function can be transformed into a 2 X 2 linear system for the unknown and its rst derivative More precisely we have the following statement Proposition 512 Reduction to rst order Every solution y to the second order linear equation y pty qty 9t7 5 16 de nes a solution I y and 12 y of the 2 X 2 rst order linear di erential system 1 12 5 17 I 4t11Pt129t 5 18 Conversely every solution I 12 of the 2 X 2 rst order linear system in Eqs 517518 de nes a solution y z of the second order di erential equation in 516 Proof of Proposition 512 Assume that y is solution of Eq 5 16 and introduce the functions 11 y and 12 yi Therefore Eq 51 holds due to the relation 11 y I27 Also Eqi 5 18 holds because of the equation 1 2 y 7W 9 7 PO y 90 740117 pt12 90 The converse statement is proven in a similar way This establishes the Propositioni D EXAMPLE 525 Find the rst order reduction to the differential equation y 2y By sin5t and express it in matrix formi We then introduce the functions 11 y and 12 yi Therefore 11 and 12 satisfy the 2 X 2 differential linear system 11 127 I 7311 7 212 sin5ti Introduce the matrix source vector and unknown vector 7 0 1 7 1 7 11t A 73 72 7 W lsin5tl 7 3 12t 7 then the system above can be written as follows 1 Amb gt L03 712 T lsing5tl G NAGY r ODE DECEMBER 17 2008 101 524 Existence and uniqueness of solutions The main result in this Subsection is Theorem 516 which says that every n X n linear rst order system of equations with continuous coefficients has a set of n linearly independent solutions In order to present this result in a more precise way we need to introduce few concepts De nition 513 Given an n X n matrixvalued function A consider the n X n homogeneous system of linear di erential equations 1 At 92 519 The n solutions 11 t of t of this Eq 519 are called fundamental solutions at t E R i f they form a linearly independent set Given two arbitrary constants c and c2 the vectorvalued function mt clm1t 594 is called the general solution of Eq 519 i the functions 11 and 12 are fundamental solutions of the same equation The n X n matrixvalued function X MD 39 quot 7 is called a fundamental matrix of the Eq 519 att E R i the column vectors in X are fundamental solutions of Eq 519 Finally the function wt detXt is called the Wronskian of the fundamental solutions of the E11 519 The Wronskian function de ned above reduces to the function de ned in Sect 22 in the case where the linear system in Eq 519 is a 2 X 2 system coming from a rst order reduction of a second order linear homogeneous equation like the ones we studied in Sect 22 Proposition 514 The matrix Xt 711 t of is a fundamental matrix of the E11 519 at t E R it holds that wt 0 Proof of Proposition 514 This result follows from Proposition 54 D EXAMPLE 526 Show that the vectorvalued function below are linearly independent for all t E R 3 it 1 e 2 e m heal 1 ball We need to compute the determinant of the matrix 63 67 X 263 7267 7 that is wt 72e2 7 2e2 es 67 2e 72equot which is nonzero for all t E R lt1 The following result establishes that the Wronskian of solution to an homogeneous linear system vanishes at a single point iff it vanishes for all t E R In order to state this result we need to introduce the trace of an n X n matrix A as follows trA ELI aii that is the trace of a square matrix is the sum of the diagonal elements The proof of the following requires concepts from algebra and calculus beyond the scope of these notes so it is not reproduced here 102 G NAGY 7 ODE DECEMBER 17 2008 Proposition 515 Wronskian equation IfA is an n X n continuous matrix and a with i l n are arbitrary solutions of the di erential equation 1quot Atm then the Wronskian WW dethOUv X0 1 gtlttgt 44 is given by the equation t wt we 040 at Am a to Notice that in the case of a constant matrix A the equation above for the Wronskian reduces to the following one wt 100 60 We repeat that this result says that the Wronskian of n solutions of a linear homogeneous differential equation is nonzero at a single point then it is nonzero for all t E R This result says that n solutions that are linearly independent at the initial time to remain linearly independent for every time t E R Theorem 516 Existence and uniqueness If the n X n matrixvalued function A and the nvector b are continuous on an open interval I C R then the linear system 1t At 1 bt 5320 always has a set of n and no more than n linearly independent solutions 931t7791 tl Furthermore the initial value problem given by Eq 520 together with the initial condition 10 my has a unique solution EXAMPLE 527 Show that the vectorvalued functions 94 62 94 at are solutions to the 2 X 2 linear system 3 72 1Am A2 72 We start computing the righthand side in the differential equation above for the function 1 m that is 173 72 2 274 2 2 272 2H m Am 7 iZJLeiZe 721e 71ltegt7m so we conclude that 11 Am Notice that in the third equality in the calculation above we used that the vector v1 is an eigenvector of the matrix A with eigenvalue A1 2 and we also used in the fourth equality that this eigenvalue is the coefficient that appears in the exponential e2 So we can express the solution 1390 as follows 11 t v1 ekl where Av1 Alvl The calculation above is then reproduced as follows A10 Avl e Alvl ekl v1e 1 11 G NAGY r ODE DECEMBER 17 2008 103 Therefore we have also proven that the vector 12 is a solution to the differential equation above since v2 6 2 with 1 3902 2 7 2 being eigenvector and eigenvalue respectively of matrix A as the following calculation shows 3 72 1 1 A l2 721 121 J 121 EXAMPLE 51218 Find the solution to the initial value problem 1 Am 10 A i 717 From the example above we know that th general solution of the differential equation above is 1t 51 62 02 6 Before imposing the initial condition on this general solution is it convenient to write this general solution in terms of the fundamental matrix of the equation above that is 262 equot c mmg2 le e wX a where we introduced 262 equot 01 XO 62 267 0 CJ This notation simpli es the calculation of the constants 01 and C2 which are xed by the initial condition as the following calculation shows m0X0c 0 X07lm0i Since Xo e wm r 13 31 El 1131 We conclude that the solution to the initial value problem above is given by mt 7 621 at i so we obtain that 104 G NAGY 7 ODE DECEMBER 17 2008 5 3 NON REPEATED E1 GENVALUE S In this and the following Section we study homogeneous linear systems with constant coef cientsi That is given an n X n constant matrix A nd all solutions mt to the equation 1 Am 521 We have mentioned in Theorem 516 without a proof that an n X n system of linear differential equations with variable coef cients and source functions always has a set of n linearly independent solutionsi However we have not said how to obtain these solutions This is the subject of this and the following Section in the particular case of Eq 521 We divide the study in two main cases depending on whether the coef cient matrix A has a set of n linearly independent eigenvectors of or not The former case is studied in this Section and the latter case is studied in the next Section We recall that an example of a 3 X 3 matrix having three linearly independent eigenvectors was shown in Example 5118 while an example of a 3 X 3 matrix having only two linearly independent eigenvectors was shown in Example 5119 If the n X n coef cient matrix A has a set of n linearly independent eigenvectors then the solutions to Eq 521 have a simple form as it is shown in the result belowi Theorem 517 Constant coef cients If the n X n constant matrix A has n linearly independent eigenvectors v1 v with corresponding eigenvalues A1 An then the general solution to the E11 521 is given by 10 v1ezt DOOEMK 522 where c cn are arbitrary constants Furthermore introducing the fundamental matrix Xt and the constant vector 0 X 01 5M7 woman 0 523 then the general solution to the E11 521 is given by mt X t of Proof of Theorem 517 We rst verify that every x is solution to the differential Eqi 521 following a simple calculation we already introduced in Example 527 For any i compute Ama Ava em Awakm Du emy may Therefore every vectorvalued function 1 satis es the equation may A1 We must only show that these solutions are linearly independent for all t E R In other words we must show that the matrix XO 1 1t7 44 has nonzero Wronskian wt detXt for all t E R Since at t 0 the matrix M vlt1gt 71W has by assumption linearly independent column vectors Proposition 54 implies that detX0gt 100 y 0 The Proposition 515 then says that wt 100 emA 0 for all t E R The furthermore part of the Theorem is straightforward This establishes the Theoremi D G NAGY r ODE DECEMBER 17 2008 105 EXAMPLE 51311 Find the general solution to the 2 X 2 differential system 1 3 1 7 Am A 7 3 1 1 Following Theorem 51 in order to nd the solution to the equation above we must start nding the eigenvalues and eigenvectors of the coef cient matrix Al This part of the work was already done in Example 5111121 We have found that A has two linearly independent eigenvectors more precisely 1 1 1 1 4 A1 4 v 1 1 1 e 1 71 A2 727 v2 1 g 932 1 e 23 Therefore the general solution of the differential equation is w 1 4 71 2 R icl 1 e c2 1 e c1c26 1 Introduce the fundamental matrix Xt and the constant vector 0 as in Eq 5123 that is 64 7672 Cl X i 64 672 7 C 7 7 then the general solution above can be expressed as follows I t e4t ie Q c m Xlttgtc e l i lt1 The following result provides a simple way to nd the solution of an initial value problem in the case that the coef cient matrix has n linearly independent eigenvectorsi Theorem 518 IVP Assume that the n X n constant matrix A has n linearly indepen dent eigenvectors 01 v with corresponding eigenvalues A An and introduce the fundamental matrix Xt as in Eq Then the unique solution to the initial value problem given by Eq 521 and initial condition 10 mg is given by 71 mt Stmg St X t X 0 1 EXAMPLE 51312 Find the solution to the initial value problem 2 1 3 1Am m04 A3 1 We known from Example 531 that the general solution to the differential equation above is given by mt c1 e4t c2 711 e 2 c1c2 E R1 Introduce the fundamental matrix X and the vector 0 as in Eq 5123 then we obtain 64 7672 C m Xlttgtc Xlttgt c 1 Therefore the initial condition is given by the linear system X0c930 gt All 106 G NAGY 7 ODE DECEMBER 17 2008 The solution of the linear system above can be expressed in terms of the inverse matrix Xltogt 1 31 as follows a BBB n m a my So the solution to the initial value problem is given by mt 3 641 11 e 2 lt1 Proof of Theorem 518 We know that the general solution of Eq 521 has the form mt 01 um em an M EM gt mt Xtc where we have introduced the fundamental matrix Xt and the vector 0 as in Eq 523 The later notation is particularly useful to find solutions to initial value problem since it is simple to express the constant vector 0 in terms of the initial condition vector 10 mg as the following calculation shows m0m0X0c cX071o The inverse matrix above exists since the eigenvectors of matrix A form a linearly indepen dent set Therefore substituting c in the general solution given above we obtain we XlttgtXlt0gt 1mo This establishes the Theorem D We now present a useful particular case of Theorem 517 which is the case where the coefficient matrix A in Eq 521 has n different eigenvalues Then A has n linearly indepen dent eigenvectors Notice that the converse is not true as we have seen in Example 5118 where we had a matrix with three linearly independent eigenvectors but only two different eigenvalues Theorem 519 Different eigenvalues Assume that an n X n matrix A has n dif ferent eigenvalues A1 An Then the matrix A has n linearly independent eigenvectors v0 DUI and the general solution to E11 521 is given by Eg Proof of Theorem 519 If the matrix A has n different eigenvalues then Theorem 58 implies that there exists a set of n linearly independent eigenvectors and so Theorem 517 says that the general solution to Eq 521 is given by Eq 522 When an n X n matrix has n different eigenvalues we call the eigenvalues nonrepeated Nonrepeated eigenvalues of a 2 X 2 matrix with real coefficients are either both real or a complex conjugate pair The following two Subsections are dedicated to explain the phase diagrams corresponding to each of these two cases G NAGY r ODE DECEMBER 17 2008 107 531 Real nonrepeated eigenvalues We present a qualitative description of the solu tions to Eq 521 in the particular case of 2 X 2 systems with matrix A having two real and different eigenvalues The main tool will be the sketch of phase diagrams also called phase portraits The solution at a particular value it is given by a 2vector I105 mt 7 126 so it can be represented by a point on a plane while the solution function for all t E R corresponds to a curve on that plane In the case that the solution vector mt is interpreted as the position function of a particle moving in a plane at the time t the curve given in the phase diagram is the trajectory of the particle Arrows are added to this trajectory to indicate the motion of the particle as time increases Since the eigenvalues are different Theorem 519 says that there always exist two lin early independent eigenvectors v0 and M associated with the eigenvalues A1 and M respectively The general solution to the qu 521 is then given by mt 51 v0 e 52 02 eAZ l A phase diagram contains several curves associated with several solutions that correspond to different values of the free constants 01 and 02 In the case that the eigenvalues are non zero the phase diagrams can be classi ed into three main classes according to the relative signs of the eigenvalues A1 A2 of the coef cient matrix A as fol ows i 0 lt A2 lt A1 that is both eigenvalues positive ii A2 lt 0 lt A1 that is one eigenvalue negative and the other positive iii A2 lt A1 lt 0 that is both eigenvalues negative The study of the cases where one of the eigenvalues vanishes is simpler and is left as an exercise We now nd the phase diagrams for three examples one for each of the classes presented above These examples summarize the behavior of the solutions to 2 X 2 linear systems with coefficient matrix having two real different and nonzero eigenvalues A2 lt All The phase diagrams can be sketched following these simple steps First plot the eigenvectors 02 and 01 corresponding to the eigenvalues A2 and All Second draw the whole lines parallel to these vectors and passing through the origin These straight lines correspond to solutions with one of the coef cients 01 or 02 vanishing Arrows on these lines indicate how the solution changes as the variable it grows If t is interpreted as time the arrows indicate how the solution changes into the future The arrows point towards the origin if the corresponding eigenvalue A is negative and they point away form the origin if the eigenvalue is positive Finally nd the nonstraight curves correspond to solutions with both coefficient 01 and 02 nonzerol Again arrows on these curves indicate the how the solution moves into the future EXAMPLE 533 Case 0 lt A2 lt A1 Sketch the phase diagram of the solutions to the differential equation 1 ll 3 17141 Ailh 9 The characteristic equation for this matrix A is given by 524 A13 A22l One can show that the corresponding eigenvectors are given by 173 27 2 Hv M detA7AI27560 a 108 G NAGY 7 ODE DECEMBER 17 2008 So the general solution to the differential equation above is given by mt 51 v1e 1 c2 v2e 2 ltgt mt 51 eg c2 e23 In Fig 16 we have sketched four curves each representing a solution mt corresponding to a particular choice of the constants 01 and 02 These curves actually represent eight different solutions for eight different choices of the constants 01 and C2 as is described belowi The arrows on these curves represent the change in the solution as the variable t growsi Since both eigenvalues are positive the length of the solution vector always increases as t growsi The straight lines correspond to the following four solutions 01 l 02 0 Line on the rst quadrant starting at the origin parallel to 01 01 0 02 1 Line on the second quadrant starting at the origin parallel to M2 01 71 02 0 Line on the third quadrant starting at the origin parallel to iva 01 0 02 71 Line on the fourth quadrant starting at the origin parallel to ivah FIGURE 16 The graph of several solutions to Eq 524 corresponding to the case 0 lt A2 lt A1 for different values of the constants 01 and 02 The trivial solution x 0 is called an unstable pointi Finally the curved lines on each quadrant start at the origin and they correspond to the following choices of the constants 01 gt 0 02 gt 0 Line starting on the second to the rst quadrant 01 lt 0 c2 gt 0 Line starting on the second to the third quadrant 01 lt 0 02 lt 0 Line starting on the fourth to the third quadrant 01 gt 0 02 lt 0 Line starting on the fourth to the rst quadranti lt1 EXAMPLE 534 Case A2 lt 0 lt A1 Sketch the phase diagram of the solutions to the differential equation 1 3 1 Am A 3 1 i 525 G NAGY r ODE DECEMBER 17 2008 109 We known from Example 531 that the general solution to the differential equation above is given by mt clv1e 1 c2 v2e 2 ltgt mt 51 e c2 711 e Q where we have introduced the eigenvalues and eigenvectors 72 02 711 i In Fig 17 we have sketched four curves each representing a solution mt corresponding to a particular choice of the constants 01 an 02 hese curves actually represent eight different solutions for eight different choices of the constants 01 and C2 as is described belowi The arrows on these curves represent the change in the solution as the variable t growsi The part of the solution with positive eigenvalue increases exponentially when t grows while the part of the solution with negative eigenvalue decreases exponentially when t growsi The straight lines correspond to the following four solutions A14 WEE and A2 51 l 02 0 Line on the rst quadrant starting at the origin parallel to 01 51 0 02 1 Line on the second quadrant ending at the origin parallel to M2 51 71 02 0 Line on the third quadrant starting at the origin parallel to iva 51 0 02 71 Line on the fourth quadrant ending at the origin parallel to ivah FIGURE 17 The graph of several solutions to Eq 525 corresponding to the case A2 lt 0 lt A1 for different values of the constants 01 and 02 The trivial solution x 0 is called a saddle pointi Finally the curved lines on each quadrant correspond to the following choices of the constants 01 gt 0 02 gt 0 Line from the second to the rst quadrant 01 lt 0 02 gt 0 Line from the second to the third quadrant 01 lt 0 02 lt 0 Line from the fourth to the third quadrant 01 gt 0 02 lt 0 Line from the fourth to the rst quadranti 110 G NAGY 7 ODE DECEMBER 17 2008 EXAMPLE 535 Case A2 lt A1 lt 0 Sketch the phase diagram of the solutions to the differential equation 1 79 3 1 Am A 1 1 in 526 The characteristic equation for this matrix A is given by A1 72 detA7I A2560 A2 73 One can show that the corresponding eigenvectors are given by 3 72 1 2 PM MM So the general solution to the differential equation above is given by mt 51 v1e 1 c2 v2e 2 ltgt mt 51 62 c2 e s i In Fig 18 we have sketched four curves each representing a solution mt corresponding to a particular choice of the constants 01 and 02 These curves actually represent eight different solutions for eight different choices of the constants 01 and C2 as is described belowi The arrows on these curves represent the change in the solution as the variable t growsi Since both eigenvalues are negative the length of the solution vector always decreases as t grows and the solution vector always approaches zero The straight lines correspond to the following four solutions 51 l 02 0 Line on the rst quadrant ending at the origin parallel to 01 51 0 02 1 Line on the second quadrant ending at the origin parallel to M2 51 71 02 0 Line on the third quadrant ending at the origin parallel to iva 51 0 02 71 Line on the fourth quadrant ending at the origin parallel to ivah FIGURE 18 The graph of several solutions to Eq 526 corresponding to the case A2 lt A1 lt 0 for different values of the constants 01 and 02 The trivial solution x 0 is called a stable pointi G NAGY r ODE DECEMBER 17 2008 111 Finally the curved lines on each quadrant start at the origin and they correspond to the following choices of the constants c1 gt 0 c2 gt 0 Line entering the rst from the second quadrant c1 lt 0 c2 gt 0 Line entering the third from the second quadrant c1 lt 0 c2 lt 0 Line entering the third from the fourth quadrant c1 gt 0 c2 lt 0 Line entering the rst from the fourth quadrant lt1 532 Complex nonrepeated eigenvalues The complex eigenvalues of an n X n matrix A with real coef cients are always complex conjugate pairs as it is shown below Lemma 520 Conjugate pairs If an n X n matrix A with real coe cients has a complex eigenvalue A with eigenvector 1 then X and T are also an eigenvalue and eigenvector of matrix Proof of Lemma 520 Complex conjugate the eigenvalue eigenvector equation for A and v that is Avv ltgt AvXv since A A D Since the complex eigenvalues of a matrix with real coef cients are always complex con jugate pairs there are an even number of complex eigenvalues and it holds that A A with a similar equation for their respective eigenvectors that is 39U E Hence an eigenvalue and eigenvector pair has the form Ai a i i Mi ai ib 527 where a B E R and a b E R It is simple to obtain two linearly independent solutions to Eq 521 in the case that matrix A has a complex conjugate pair of eigenvalues and eigenvectors These solutions can be expressed both as complexvalued or as realvalued functions Theorem 521 Complex pairs Let Ai a i i be eigenvalues of an n X n matrix A with respective eigenvectors Mi aiib where a B E R and a b E R and n gt 2 Then linearly independent complex valued solutions to E11 521 are given by w vltgt ekt m v eAquot 528 while linearly independent real valued solutions to E11 521 are given by 11 a cos t 7 b sin t e0 12 a sin t b cos t ea 529 Proof of Theorem 521 We know from Theorem 517 that two linearly independent solutions to Eq 521 are given by Eq 528 The new information in Theorem 521 above is the realvalued solutions in Eq 529 They can be obtained from Eq 528 as follows t a i ib emim ema i ib eim ema i ib cos t i isin t emla cos t 7 b sin t i iem a sin t b cos t 112 G NAGY 7 ODE DECEMBER 17 2008 Since the differential equation in 521 is linear the functions below are also solutions 11 m 17 em a cos t 7 b sin t 1 12 f 1 7 m ema sin t b cos ti z This establishes the Theoremi D EXAMPLE 536 Find a realvalued set of fundamental solutions to the differential equation 2 3 1 Am A 73 2 5 30 and then sketch a phase diagram for the solutions of this equation Fist nd the eigenvalues of matrix A above 27m 3 7 2 7 3 Mww 7 t Then nd the respective eigenvectorsi The one corresponding to A is the solution of the homogeneous linear system With coefficients given by 27 3i 2423 3il A A 711 7iil A 1 Therefore the eigenvector 39U is given by 2 v17iv2 a v21 7114 a vltgt 23ii The second eigenvectors is just the complex conjugate of the eigenvector found above that is 0 Notice that m i m Therefore the real and imaginary parts of the eigenvalues and of the eigenvectors are given by 0 71 H7 0 So a realvalued expression for a fundamental set of solutions is given by 931 0033t 7 01 sin3tgt 62 v 12 53 a 0 71 12 s1n3t 0 cos3tgt 62 The phase diagram of these two fundamental solutions is given in Fig 19 belowi There is also a circle given in that diagram corresponding to the trajectory of the vectors i0 sin3t 2 7 cos3t cos3t s1n3t The trajectory of these vectors is a circle since their length is constant equal to one G NAGY r ODE DECEMBER 17 2008 113 FIGURE 19 The graph of the fundamental solutions 11 and 12 of the Eqi 530 In the particular case that the matrix A in 521 is 2 X 2 then there are no other solutions linearly independent to the solutions given in Eq 529 That is the general solution of given by mt 51 11 t 52 12 t where 11 a cos t 7 b sin t em 12 a sin t b cos t emi We now do a qualitative study of the phase diagrams of the solutions for this case We rst x the vectors a and b and the plot phase diagrams for solutions having a gt 0 a 0 and a lt 0 These diagrams are given in Fig 20 One can see that for a gt 0 the solutions spiral outward as t increases and for a lt 0 the solutions spiral inwards to the origin as t increases FIGURE 20 The graph of the fundamental solutions 11 and 12 dashed line of the Eqi 529 in the case of a gt 0 a 0 and a lt 0 respectively Finally let us study the following cases We x a gt 0 and the vector u and we plot the phase diagrams for solutions with two choices of the vector 11 as shown in Fig 21 It can then bee seen that the relative directions of the vectors a and 11 determines the rotation direction of the solutions as t increases 5 4 REPEATED E1 GENVALUE S Consider an n X n matrix A having an eigenvalue A with algebraic multiplicity 7 2 and geometric multiplicity s 1 Since the algebraic multiplicity is bigger than one the 114 G NAGY 7 ODE DECEMBER 17 2008 FIGURE 21 The graph of the fundamental solutions 11 and 12 dashed line of the Eqi 529 in the case of a gt 0 for a given vector u and for two choices of the vector b The relative positions of the vectors a and b determines the rotation directioni eigenvalue A is repeated and since the geometric multiplicity is one then there is only one linearly independent eigenvector 1 associated with this eigenvalue In this case one solution to Eq 521 is given by 11 t DEAR Since there is no second eigenvector associated with A and linearly independent to 1 then Theorem 51 does not hold The following result stated without a proof provides two linearly independent solutions to 5 21 associated with the eigenvalue Al Theorem 522 Repeated eigenvalue Assume that the n X n matrix A has an eigen value A with algebraic multiplicity r 2 and geometric multiplicity s l and let 1 be an eigenvector associated with A Then the algebraic linear system A7AIw v 531 always has in nitely many solutions 11 Furthermore two linearly independent solutions to the di erential equation 139 Am are given by 11 t De 12 t vt w eMi Notice that the matrix A 7 AI is not invertible since A is an eigenvalue of A that is detA 7 AI 0 Nevertheless the Theorem above says that Eq 531 has solutions and property that v is an eigenvector of A is crucial in the proof which is not reproduced here EXAMPLE 541 Find the fundamental solutions of the differential equation 1 76 4 1Am A171 72 As usual we start nding the eigenvalues and eigenvectors of matrix Al The former are the solutions of the characteristic equation 07727Al 1 7A 3 A1 17 A 7A 2 7 1 ltlAgt lt llt l1 2171l 4 2 Therefore there solution is G NAGY r ODE DECEMBER 17 2008 115 There is only one linearly independent eigenvector associated with this eigenvalue and it is computed as the solution of the linear system A 1 0 that is 3 1 lt1 11 j i Ag Hg 5 a 4 2 4 2 Choosing v2 1 then v1 2 and we obtain Following Theorem 52 we now must solve for the vector 11 the linear system AIw vi The augmented matrix for this system is given by 1 g1l2H172l74H172li 724 771 1172li4 00 0 wl wr 42 We have obtained in nitely many solutions given by m m As one could have imagined given any solution 11 the CD w is also a solution for any 5 E R We choose the simplest solution given by Therefore a set of fundamental solutions to the differential equation above is given by 11 e4 12t t 73 equot i 5 32 lt1 The phase diagram of the fundamental solutions in Example 541 above are given in Fig 22 This diagram can be obtained as follows First draw the vectors 1 and w The solution 11 t is easy to draw since it is represented by a straight line in the rst quadrant parallel to v ending in the origin The solution approaches the origin as the variable t grows since the eigenvalue A 71 is negative The solution 711 is the the straight line in the third quadrant ending in the origin and parallel to ill The solution 12 t is more dif cult to drawi One way is to rst draw the trajectory of the timedependent vector 2 7 2 74 m 7 1 t 0 i This is a straight line parallel to v passing through 11 and is represented by a dashed line in Fig 22 As t grows the function e moves from the third quadrant to the second and then to the rst The solution 12 differs from 2 by the multiplicative factor 6quot so for t lt 0 we have that 12 t has bigger length than e t while the opposite happens for t gt 0 In the limit t A 00 the solution 12 t approaches the origin since the exponential factor 6quot decreases faster than the linear factor t growsi The result is then the upper half of the sshaped trajectory in Fig 22 The lower half corresponds to em 2h 116 G NAGY 7 ODE DECEMBER 17 2008 FIGURE 22 The graph of the fundamental solutions 11 12 computed in Example 541 and given in Eq 5327 together With their negativesi Given any vectors 1 and w7 and any constant A7 the phase diagrams of functions W t we 94 vt w e 533 can be obtained in a similar way as the one in Fig 22 The results are displayed in Fig 23 for the case A lt 0 and A gt 07 respectively FIGURE 23 The trajectories of the functions 11 12 given in 533 for the cases A lt 0 and A gt 07 respectively G NAGY r ODE DECEMBER 17 2008 117 CHAPTER 6 BOUNDARY VALUE PROBLEMS 61 EIGENVALUE EI GENFUNCTI ON PROBLEMS In this Section we consider second order linear ordinary differential equations In the rst half of the Section we study boundary value problems for these equations and in the second half we focus on a particular type of boundary value problems called the eigenvalue eigenfunction problem for these equations 611i Comparison IVP and BVP Given real constants a1 and a0 consider the second order linear homogeneous constant coef cients ordinary differential equation y a1y aoy0 61 We now review the initial boundary value problem for the equation above which was dis cussed in Sect 21 where we showed in Theorem 24 that this initial value problem always has a unique solution De nition 61 IVP Given the constants tg ya and y nd a solution y of E1 61 satisfying the initial conditions 9050 9m ytu 91 62 There are other problems associated to the differential equation above The following one is called a boundary value problemi De nition 62 BVP Given the constants tg t1 ya and y nd a solution y of E1 61 satisfying the boundary conditions 9050 ya 96 91 63 One could say that the origins of the names initial value problem77 and boundary value problem77 originates in p ysicsi Newton7s second law of motion for a point particle was the differential equation to solve in an initial value problem the unknown function y was interpreted as the position of the point particle the independent variable t was interpreted as time and the additional conditions in Eq 62 were interpreted as specifying the position and velocity of the particle at an initial time In a boundary value problem the differential equation was any equation describing a physical property of the system under study for example the temperature of a solid bar the unknown function y represented any physical property of the system for example the t A t the 39 A variable t position in space and it is usually denoted by z and the additional conditions given in Eq 63 represent conditions on the physical quantity y at two different positions in space given by to and t1 which are usually the boundaries of the system under study for example the temperature at the boundaries of the bar This originates the name boundary value problemlh We mentioned above that the initial value problem for Eq 61 always has a unique solution for every constants yo and y1 result presented in Theorem 24 The case of the boundary value problem for Eq 61 is more complicated A boundary value problem may have a unique solution or may have in nitely many solutions or may have no solution depending on the boundary conditions This result is stated in a precise way belowi Theorem 63 BVP Fix real constants a ag and let ri be the roots of the characteristic polynomial pr r2 aJr ag If the roots rt 6 R then the boundary value problem given by Egs 61 and has a unique solution for all ya y E 118 G NAGY 7 ODE DECEMBER 17 2008 If the roots ri form a complex conjugate pair that is ri a i i with a B E R then the solution of the boundary value problem given by Egs 61 and belongs to only one of the following three possibilities a There exists a unique solution b There exists in nitely many solutions c There exists no solution Before presenting the proof of Theorem 6 concerning boundary value problem let us review part of the proof of Theorem 24 concerning initial value problems using matrix notation to highlight the crucial part of the calculations For the simplicity of this review we only consider the case r f r In this case the general solution of Eq 61 can be expressed as follows yt 01 eT t52 er 0102 6 R The initial conditions in Eq 6 determine the values of the constants 01 and 02 as follows 90 M750 01 5T to 02 6 to e7 0 e 0 01 yo yl yto Cir eT 0 02m e to 71 Eric T em to 02 91 The linear system above has a unique solution 01 and 02 for every constants yo and y1 iff the determinant of the coef cient matrix Z is nonzero where an to em to Z T7 er to m er to A simple calculation shows detZ ri 7 n eTT O 0 42gt r r Since r r the matrix Z is invertible and so the initial value problem above a unique solution for every choice of the constants yo and y1 The proof of Theorem 63 for the boundary value problem follows the same steps we described above First nd the general solution to the differential equation second nd whether the matrix Z above is invertible or not Proof of Theorem 63 First consider the case where ri are real numbers We have two cases r r and r r In the former case the general solution to Eq 61 is given by yt 01 e t 02 e 1 cl 02 E R 64 The boundary conditions in Eq 63 determine the values of the constants 01 and 02 as follows yo yltto 6167a to J 5257 to gt em to em to Cl yo 6 5 n r z yl yt1 clef E102entl 5 1 5 l 1 C2 91 The linear system above has a unique solution 01 and 02 for every constants yo and y1 iff the determinant of the coef cient matrix Z is nonzero where 5N to 66 r7 1 en 1 A simple calculation shows detz er tier to 7 en Loer 1 em Loer to en 2170 7 er 11720 67 So it is simple to see that detZ 0 42gt eT 1quot0 eT m m 42gt r r 68 G NAGY r ODE DECEMBER 17 2008 119 Therefore in the case 7 f 7 the matrix Z is invertible and so the boundary value problem above has a unique solution for every choice of the constants yo and y1 In the case that T 7 7 7 0 then we have to start over since the general solution of Eq 61 is not given by Eq 64 but by the following expression yt 01 02 t eTO 0102 6 R Again the boundary conditions in Eq 63 determine the values of the constants 01 and 02 as follows 90 900 516010 527506010 eTO O toeTO O 01 7 yo 91 901 Cierot1 l Czti iTm1 6T0 tieTOtl C 7 yl The linear system above has a unique solution 01 and 02 for every constants yo and y1 iff the determinant of the coef cient matrix Z is nonzero where Z on toergtg erotl tierotl A simple calculation shows detZ tleTO 1 O 7tOeToO1togt t1 7toeTot1to f 0 4 751 f to Therefore in the case 7 7 7 7 0 the matrix Z is again invertible and so the boundary value problem above has a unique solution for every choice of the constants yo and y1 This establishes part of the Theorem et us consider part ii of the Theorem and assume that the roots of the characteristic polynomial have the form Ti a i with 6 f 0 In this case the general solution to Eq 61 is still given by Eq 64 and the boundary condition of the problem are still given by Eq 65 The determinant of matrix Z introduced in Eq 66 is still given by Eq 67 However the claim given in Eq 68 is true only in the case that Ti are real numbers and it does not hold in the case that Ti are complex numbers The reason is the following calculation Let us start with Eq 67 and then introduce that Ti a i i detZ 60344410 67 Milo 7 E7 11 O 620410 ea11to 4130140 7 e i 1 30 2i6 1 0 sin 501 7 7 We then conclude that detZ 0 42gt sin t17to 0 42gt 69 where n l2 and we are using that t1 to This last equation in 69 is the key to obtain all three cases in part ii of this Theorem as can be seen from the following argument If the coef cients a1 a0 in Eq 61 and the numbers t1 to in the boundary conditions in 63 are such that Eq 69 does not hold that is mr 5 t1 7 to then detZ 0 and so the boundary value problem for Eq 61 has a unique solution for all constants yo and y1 This establishes part iia If the coef cients a1 a0 in Eq 61 and the numbers t1 to in the boundary conditions in 63 are such that Eq 69 holds then detZ 0 and so the system of linear equation given in 65 may or may not have solutions depending on the values of the constants yo and y1 In the case that there exists a solution then there are in nitely many solutions 120 G NAGY 7 ODE DECEMBER 17 2008 since detZ 01 This establishes part iibi The remaining case when yo and y1 are such that Eq 65 has no solution is the case in part iici This establishes the Theoremi Our rst example is a boundary value problem with a unique solution This corresponds to case iia in Theorem 63 The matrix Z de ned in the proof of that Theorem is invertible in this case and the boundary value problem has a unique solution for every yo an y EXAMPLE 61111 Find the solution to the boundary value problem y 4y 0 ND 17 WW4 1 We rst nd the general solution to the differential equation above We know that we have to look for solutions of the form em with the constant 7 being solutions of the characteristic equation T240 gt Tidal We know that in this case we can express the general solution to the differential equation above as follows 01 cos21 02 sin21i The boundary conditions imply the following system of linear equation for 01 and C2 Jig233 e a 9 2 131 The linear system above has the unique solution 01 1 and 02 711 Hence the boundary value problem above has the unique solution cos21 7 sin21 i lt1 The following example is a small variation of the previous one we change the value of the constant t1 which is the place where we impose the second boundary condition from 7r 4 to 7r21 This is enough to have a boundary value problem with in nitely many solutions corresponding to case iib in Theorem 63 The matrix Z in the proof of this Theorem 63 is not invertible in this case and the values of the constants to t1 yo yl a1 and a0 are such that there are in nitely many solutions EXAMPLE 61112 Find the solution to the boundary value problem y 4y 0 ND 1 WrZ 1 The general solution is the same as in Example 611 above that is 01 cos21 02 sin21i The boundary conditions imply the following system of linear equation for 01 and C2 1 90 51 1 0 01 7 1 71 y7r2 701 1 0 02 7 1 i The linear system above has in nitely many solution as can be seen from the following 1011H1011 5117 71 0 l 71 0 0 l 0 02 free Hence the boundary value problem above has in nitely many solutions given by cos21 c2 sin21 c2 6 R i G NAGY r ODE DECEMBER 17 2008 121 The following example again is a small variation of the previous one this time we change the value of the constant yl from 71 to 11 This is enough to have a boundary value problem with no solutions corresponding to case iic in Theorem 63 The matrix Z in the proof of this Theorem 63 is still not invertible and the values of the constants to t1 yo yl a1 and a0 are such that there is not solutionl EXAMPLE 61113 Find the solution to the boundary value problem y 4y07 ND 17 WrZ 1 The general solution is the same as in Examples 611 and 612 above that is 01 cos21 02 sin21l The boundary conditions imply the following system of linear equation for 01 and C2 1 y0 01 1 y7r2 751 From the equations above we see that there is no solution for cl hence there is no solution for the boundary value problem above We now use matrix notation in order to follow the same steps we did in the proof of Theorem 63 1 0 51 7 1 71 0 52 7 1 The linear system above has in nitely many solutions as can be seen from the following Gauss elimination operations 1011H1011H101 71011 00l2 00H Hence there are no solutions to the linear system abovel lt1 6121 Eigenvalueeigenfunction problems A particular type of boundary value prob lems are called eigenvalueeigenfunction problems he main example we study in this Section is the following Find all the numbers A and the nonzero functions with values solutions of the homogeneous boundary value problem way yltogto yltzgt0 be 610 This problem is analogous to the eigenvalueeigenvector problem studied in Sect 511 that is given an n X n matrix A nd all numbers A and nonzero vectors 1 solution of the algebraic linear system Av Au The role of matrix A is played by d2d12 the role of the vector space R is played by the vector space of all in nitely differentiable functions f with domain 0 C R satisfying 01 We mentioned in Sect 511 that given any n X n matrix A there exist at most n eigenvalues and eigenvectorsl In the case of the boundary value problem in Ed 6110 there exist in nitely many solutions A and yz as can be seen in the following result Theorem 64 39 39 quot The L boundary value problem in Eq 610 has in nitely many solutions labeled with a subindex n E N and given by n27r2 mrz An77 sin7l 122 G NAGY 7 ODE DECEMBER 17 2008 Proof of Theorem 64 We rst look for solutions having eigenvalue A 0 In such a case the general solution to the differential equation in 610 is given by 51 5217 51752 6 RA The boundary conditions imply the following conditions on 01 and 02 0 y0 cl 0 ya 0102Z 01520i Since the only solution in this case is y 0 there are no nonzero solutions We now look for solutions having eigenvalue A gt 0 In this case we rede ne the eigenvalue as A M2 with M gt 0 The general solution to the differential equation in 610 is given by 5157 025 where we used that the roots of the characteristic polynomial 7 2 7 M2 0 are given by Ti iMi The boundary conditions imply the following conditions on 01 and C2 e l1 1i M 0 ya 6167M 026M 67M el j 02 0 l 1 Z e t j 6M detZeMie M7 0 42gt M7 0i Hence the matrix Z is invertible and then we conclude that the linear system above for cl Denoting by we see that 02 has a unique solution given by 01 02 0 and so y 0 Therefore there are no nonzero solutions y in the case that A gt 0 e now study the last case when the eigenvalue A lt 0 In this case we rede ne the eigenvalue as A 7M2 with M gt 0 and the general solution to the differential equation in 610 is given by 516717 52617 where we used that the roots of the characteristic polynomial 7 2 M2 0 are given by Ti iiMi In a case like this one when the roots of the characteristic polynomial are complex it is convenient to express the general solution above as a linear combination of realvalued functions 01 cosMz 02 sinMzi The boundary conditions imply the following conditions on 01 and C2 0 y0 cl 0 ya 01 cosMZ 02 sinM Since we are interested in nonzero solutions y we look for solutions with 02 f 0 This implies that M cannot be arbitrary but must satisfy the equation 02 sinMZ 0i sinM 0 42gt MnZ 7m n E N We therefore conclude that the eigenvalues and eigenfunctions are given by 2 2 n 7r mrz n i 2 7 en 51117 Choosing the free constants an 1 we establish the Theoremi D G NAGY r ODE DECEMBER 17 2008 123 EXAMPLE 614 Find the numbers A and the nonzero functions with values solutions of the following homogeneous boundary value problem y Ay 90 07 WW 0 This is also an eigenvalueeigenfunction problem the only difference with the case studied in Theorem 64 is that the second boundary condition here involves the derivative of the unknown function y The solution is obtained following exactly the same steps performed in the proof of Theorem Gilli We rst look for solutions having eigenvalue A 0 In such a case the general solution to the differential equation is given y 01 021 0102 6 R The boundary conditions imply the following conditions on 01 and C2 0y0617 0y 7r62 Since the only solution in this case is y 0 there are no nonzero solutions We now look for solutions having eigenvalue A gt 0 In this case we rede ne the eigenvalue as A M2 with M gt 0 The general solution to the differential equation is given by 5157 025 where we used that the roots of the characteristic polynomial 7 2 7 M2 0 are given by Ti iMi The boundary conditions imply the following conditions on 01 and C2 a l 1 WM 0 yw iMcle M M526 iMe M Mel r 52 0 Denoting by Z l 1 1 Me m Mew we see that detZ Me r 67 0 Hence the matrix Z is invertible and then we conclude that the linear system above for cl 02 has a unique solution given by 01 02 0 and so y 0 Therefore there are no nonzero solutions y in the case that A gt 0 We now study the last case when the eigenvalue A lt 0 In this case we rede ne the eigenvalue as A 7M2 with M gt 0 and the general solution to the differential equation in 610 is given by 5167quot 5261 where we used that the roots of the characteristic polynomial 7 2 M2 0 are given by Ti iiMi As we did in the proof of Theorem 64 it is convenient to express the general solution above as a linear combination of realvalued functions 01 cosMz 02 sinMzi The boundary conditions imply the following conditions on 01 and C2 0 y0 cl 5 cosM7r 0i 0 yTr 7M01s1nM7r M02 cosM7r 2 Since we are interested in nonzero solutions y we look for solutions with 02 f 0 This implies that M cannot be arbitrary but must satisfy the equation cosM7r 0 42gt Mn7r 2n 1 n E N 124 G NAGY 7 ODE DECEMBER 17 2008 We therefore conclude that the eigenvalues and eigenfunctions are given by 2 An7 ynzcnsinw1 nEN 1 lt1 EXAMPLE 61115 Find the numbers A and the nonzero functions with values solutions of the homogeneous boundary value problem 12y zy y7 9107 9Z07 gt11 This is also an eigenvalueeigenfunction problem the only difference with the case stud ied in Theorem 64 is that the differential equation is now the Euler equation studied in Sect 3 instead of a constant coefficient equationi Nevertheless the solution is obtained following exactly the same steps performed in the proof of Theorem 64 Writing the differential equation above in the standard form of an Euler equation 12ynizy7Ay07 we know that the general solution to the Euler equation is given by 01 02 lnzzr0 in the case that the constants T 7 7 0 where Ti are the solutions of the Euler charac teristic equation 7 7 7177 70 Ti 1ix1A1 1n the case that T f 7 then the general solution to the Euler equation has the form clzr cgzmi Let us start with the first case when the roots of the Euler characteristic polynomial are repeated 7 7 7 01 In our case this happens if 1 A 01 In such a case 7 0 1 and the general solution to the Euler equation is 51 02 lnz111 The boundary conditions imply the following conditions on 01 and C2 0 y1 cl 0 ya 51 02 ln hence 02 01 We conclude that the linear system above for cl 02 has a unique solution given by 01 02 0 and so y 01 Therefore there are no nonzero solutions y in the case that 1 A 01 We now look for solutions having eigenvalue A satisfying the condition 1 A gt 01 In this case we redefine the eigenvalue as 1 A 2 with M gt 01 Then Ti 1 i M and so the general solution to the differential equation is given by M1 5110710 52I1u7 The boundary conditions imply the following conditions on 01 and C2 09161C27 1 1 017 0 0 ya Clg w 52g1u E 1 Wm 52 7 0 Denoting by c ln 0 1 1 Z gum gum G NAGY r ODE DECEMBER 17 2008 125 we see that detZ Zquot 7 7 a 0 42gt Z a ill Hence the matrix Z is invertible and then we conclude that the linear system above for Cl 02 has a unique solution given by 01 c2 0 and so y 0 Therefore there are no nonzero solutions y in the case that l A gt 0 We now study the second case when the eigenvalue satis es that l A lt 0 In this case we rede ne the eigenvalue as l A 7M2 with M gt 0 Then Ti l i in and the general solution to the differential equation is given by W MW As we did in the proof of Theorem 64 it is convenient to express the general solution above as a linear combination of realvalued functions z 51 coslnz c2 sinlnzi The boundary conditions imply the following conditions on 01 and 02 0 yl cl Z 39 1 Z 0i 0 ya 0 cos M ln 0 sinln CZ SlnTM nlt Since we are interested in nonzero solutions y we look for solutions with 02 f 0 This implies that M cannot be arbitrary but must satisfy the equation sinln 0 42gt Mnlna 7m n E N We therefore conclude that the eigenvalues and eigenfunctions are given by An cnz sin n E N n7rln 1 TM 2 2 1 7277 ln Z 62 OVERVIEW OF FOURIER SERIES This Section is a brief introduction to the Fourier series expansions of periodic functions We rst recall the origins of this type of series expansions We then review basic few notions of linear algebra that are satis ed by the set of in nitely differentiable functionsi One crucial concept is the orthogonality of the functions sine and cosinei We then introduce the Fourier series of periodic functions and we end this Section with the study two particular cases The Fourier series of odd and of even functions which are called sine and cosine series respectively 621 Origins of Fourier series The study of solutions to the wave equation in one space dimension by Daniel Bernoulli in the 1750s is a possible starting point to describe the origins of the Fourier series The physical system is a vibrating elastic string with xed ends the unknown function with values ut 1 represents the vertical displacement of a point in the string at the time t and position I as can be seen in the sketch given in Fig 24 A constant c gt 0 characterizes the material that form the string The mathematical problem to solve is the following initialboundary value problem Given a function with values de ned in the interval 0 K C R satisfying 0 nd

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.