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# DIFF EQ & TRANSFMS MATH 267

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GPA 3.98

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This 271 page Class Notes was uploaded by Ms. Helen Sipes on Sunday September 27, 2015. The Class Notes belongs to MATH 267 at Iowa State University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/214502/math-267-iowa-state-university in Mathematics (M) at Iowa State University.

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gimp 54 5 SH JMHL 39a I quot 5 Lamb 9 A L m l we a i Loamp 541 1 54 1 m Ffbm fist an Page 3quot le I39n rerf 39 3 0qu r NQM ntv quot 39 L 076 h 395 CdSoLEquot i39i 73se I Sn 6 6y 0 Canard Action kegmnl 39 a 5 n are i 17 quot3 lt t t e quotequot 4 S iJ m r 6 GI WM 3 S Cof ecl qg 33912 aquot Jot 1239 t y it Examele ya 5quot 4 306 Wow m v ca W zo 0 U N ii 3 51 y I Huffy 1f3 gt a 5 1 y 3750 nf y m 5 Y 0263 flu S s y may 7M 4 q y 36931 in if I m G 55quot uiy 1 5 3 a aw g iiy 145 4 53 squotssquotquot 4 2 iigm 4 5345 CS HXS H SEWY SW39 W a m w W 5140 5 4quotquot 51quot 39 3 3 55 83quot393 CCYWW W A534 35 As6 39 05305 c5 390 rd ACl 303 AHc5 84quot 020 z lA aw 4040 2 S a 3An 3 39 6393 WA 33 I3 39D 7 Dmo 2 300 3 8 2 0 at f aresqftl 2 I x 1 I F O I L 3 L1 LikQMMg 1 i 5quotquot Squot l 51 3 quot 39 A534 BSZASBI r3os wcwom a A g Moga Aqcgti 8qop v Wags M 15 Jag 1 72 v iiscgtquot 39 gtf gt mm is 3 vm53amp a tncm cm a u w Smut Sbs e Example Use an integration factor to nd the general solution of the equa tion ty 2ysint tgt 0 Vrite this in the form considered at the end of Tuesday s notes dt PU I y ty t That is 2 sint W 2 W 7 Therefore 2 2111i Int9 2 Mt exp t It exp 2 dt e e t Finally e w 1 9 y ut11tgt dtut t2t t tt2 1 C Etsmt alt l2 1 C t3 tcostsmt t3 Remark Before solving a di erential equation using the method of integrat ing factors you should write the equation in the form 7 r t z t dt m m Then you can either apply the formulas from class ie memorize the ap propriate formulas or you can apply the method to derive these formulas in your speci c ease It s better to understand the method and memorize less l Separable Equations De nition If a rst order ODE is of the form lac then that ODE is separable if 1013 iMll39xLM y and NW NI Remark The separate in separable equations means that we can separate llIry and Nxy into their factors each of which depends on only one variable Dividing equation by Af1l l12y using 2 and setting AMI v Ng1 M ac Y7 and 1V 391 z l MW J MM that a separable equation can be written as Mm Nani 0 3 Equation is sometimes written as Mllr lac dy 0 This is an abuse of notation but writing the equation in this way is often helpful in solving the problem Example 13 rHJ dx e We rst express this as 39 y 7 0 4 e e dx This equation is separable since referring back to equation 1 Mxy e ey NW z 1 We recast as e ey 0 and then through implicit differentiation e ey 0 Finally integrate both sides with respect to x e e l C C E R Then e l C e 2 y 1D C 9 provided that C e gt 0 or equivalently C lt e Remark We could have abused notation at equation to write e lac e l dy 0 Then e lac e1 13 e whiz Ky dyC e e l C so that and cons equentli which is equivalent to equation Example Solve the initial value problem 3 3522 90 2 We rst rewrite the equation as 3312 91 13 cosrc dag Integrating derive 313 91 sinac C From the initial condition 23e2zsin0C 2 Cz8e2 Therefore from equation 7 have 313 91 sinac 8 92 y is given implicitly as a function of a and this is the best can do 30 Example Find the general solution in the ODE 33123 1 cosac First write the equation as 3312 13 1yg zcosx ya l So nd 3 2 i 13 cosrc lac 133 and then by antidifferentiating 1n 1y3i sinrcC 2 1y3 KeSi Kgt0 2 1 3 Kesi K 0 2 y Kesinzn1l3i We extend K to also be zero because it can be easily veri ed that y 1 1 is a solution to the differential equation Therefore the general solution to the differential equation is y KeSiM 1 3 K e R Example Solve the di crcntial equatwn fry gal2 First rewrite the equation as 13 1 1 r21277 r 31 y do 0 y D so that 2 12 1 1 31 39 dyzgdrc 060 By integrating this expression nd arcsiny ln C 1 lt 3 lt1 Taking the sine of both sides yzsinln x C 1ltylt1 You can also Check that y le are also solutions to the differential equation SCConJ OFJLF ia nu ESuab OAS RQCa LA a Sunni orcler Mau OE A4 39Hu Form Pmy acuy Raw at ULer La OFJMQS Jho tt Jiffe nfia39eidn wi L rufect to t Bayw Mn An ebua bbn of lm alcove form 5 and homogeneouS F Gearm D39Uwruo39fe 39ZLe gauntion i 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fist ebua on artl rcol Him ICH Caquot39quot die Rt a a Solu39E an 0 a Jifferm al ewe ayquotLy39 Cy o for any CLmZe a 39ELe anstmfl CUCL 3522 Set we 39239 C a gem 1 11M y cH canequot andquot rye yl l39e Grizer Ctr 2 I E ayquotlo cy r at c r19 after 3 B ct ue t r1 F I I C C Ef t jczer t So a7 Ey39 C cequotar krq C eqtarbl ge wo clefquot a C it fluGare 139 439 Czar IS 0 Salu39 bA 0 ay by Cy o a we Call we camp 7 616 He general Saintfan 0 a lquotl Y CY 0 Examele Solarg ELQ IVP Yquot l39 3yoj vcoI y39la f rzquot r3 0 G gt Crncr33 o 3 F r3 59 ycgt c Q 4 e39 Suffhci 39tLo 0E Far My be It 00 rennm ns o CLooSc 4 CI 50 lm6 445 c39nf39te al conxezzmx Ya I 5a f5 feJ 7 1 mean I yto CC CLEO 7 C 1 at We c 64618 C 3C7 ak 55 c 61 1 C 2 l c39 3C1 CL 0 TitreExt ll39H Iequott 06 8quot K Le um cbue Shiatr39on to lg Illfgt I I Y Iy393yo Coll yCo 39 In bene t Iquot F F 4N or z nct red foot 11 02 ar LFco Men M sum0 tum 2w Y ltv Y ogty Vinny s nyc e ntczeQt MAN I I C Y0 varL er39tb C1 Var YD CW Systems of Differential Equations April 7 2008 Systems oi Diiieremial Equations Systems of Differential Equations In many applications two or more interrelated rates of change derivatives are at work in a process that we want to model In these situations we obtain two or more simultaneous differential equations that must be satisfied at the same time Such systems can be studied effectively using linear algebra Systems of Differential Equations The PredatorPrey Model Suppose we want to model the populations in a given predator prey ecosystem Let Ft denote the population of foxes in an ecosystem at time t and let RU denote the number of Rabbits at time t The predator population is dependent upon the prey population foxes need to eat and the prey population is dependent upon the predator population rabbits need to live A mathematical model for this system is dF dH E7a1Fb1FH Ea2H7b2FH Systems at Dilleremial Equations The PredatorPrey model For positive values of a1 a2 b1 b2 why does the system 7a1Fb1FH a2F7QFH dt describe the predatorprey model gt In the absence of rabbits RU 0 the population of foxes would decay exponentially we would have if 7511 F gt A higher rabbit population is good for the fox population so a larger RU leads to larger growth rate for the foxes The likelihood of interaction between the species is proportional to the size of each population gt The differential equation for EU can be similarly motivated Systems at Dilleremial Equations Other Differential Equations Systems of differential equations arise in many applications in engineering physics and pure mathematics gt In electrical engineering to describe complicated circuits gt In physics to describe situations where multiple forces are at work gt In number theory and other areas of mathematics to describe interrelated functions gt Many numerical algorithms for solving differential equations require that the input be a first order system Systems of Diff mial Equations Writing Higher order DEs as Systems of DEs Higher order differential equations can always be written as a system of first order differential equations Consider the springmass system we analyzed in Chapter 3 mu yu ku Ft Set X1l UTt X2l UT The differential equation can be written mxz yx2 kx1 Ft All together we have 7k y F 1sz X2WXquotE2E39 Systems of Differential Equations Writing Higher order DEs as Systems of DEs In general an arbitrary nth order differential equation yiquotJ FUN4y W can be written via n71 X1y X2y X3y Xny J as X17X2 X X3 Xn71Xnv XFIX1X2X3Xn Systems at Dmeremial Equations Solutions of Systems of Differential Equations A general system of n differential equations can be written X F1TX1X2 Xn X2 F2t X1 X2 Xn X Fntx1X2 X a solution is a set of functions X1 X2 X3 i an that each satisfy the differential equations If we also require thto X1 X2To X2 Xsfto X3 i anto an we arrive at an initial value problem for the system Systems of Differential Equations Solutions to systems viewed as Parametric Equations Consider the differential equation y y0 Olttlt2n We first convert this to a system via X1 yr X2 y Then the differential equation can be written X2 X1 0 So all together we have the system Systems at Dillerential Equations Solutions to systems viewed as Parametric Equations How do we find functions X1T and X2t that are interrelated by X X2 The pair X1T sin 1 X2 cos t works What do we get if we plot all the points X1TX2T on the xyplane for O lt tlt 27 50 The pair X1T sin 1 X2t cos t form a parametric representation for the unit circle Systems of Differential Equations The Method of Undetermined Coe icients for Higher Order Differential Equations The method of undetermined coefficients works for nding particular solutions to higher order equations The key difference is that if our guess is a multiple of the solution to the corresponding homogeneous equation then we may need to multiply that guess by t t2 t3 t4 and so on up the order of the differential equation in order to make things work Otherwise the same procedure applies We make a guess for the form of the solution based upon the form of gt in 2 p1 02101 1 pn1ty MW 90 The general solution to a nonhomogeneous equation is found by adding a particular solution of the nonhomogeneous equation to an arbitrary linear combination of the 71 linear independent solutions to the homogeneous equation yd W 1y1t C2y2t 39 Cnynlttz P V S Gen solution to homogeneous Example Find a particular solution to the equation 214 2y y 3sint 5cost The corresponding homogeneous equation is 214 2g y 0 We solved this equation previously by showing that the characteristic equation r4 27quot2 1 7quot ir a 0 so that the general solution to the homogeneous equation is yt c1 cost Cg sint C3t cos t C4t sin t From the form of gt 3 sint 5 cost we would normally guess that a solution to the nonhornogeneous equation would be of the form Yt A cost B sint However each term here is a solution to the homogeneous equation Our next guess Yt At cost Bt sint also includes solutions to the homogeneous equation Therefore our guess should be Yt At2 cos t Bt2 sin t Now we need to take all four derivatives of our guess and plug the resulting expressions into the differential equation to nd A and B After computing these plugging them into the differential equation and simplifying we nd that 8Acost 8B sint 3sint 5cost so that A B3 8 8 Example Find the solution to the tntz al value problem 3y 23 t 6t yltogt1 yam i y 0 We rst solve the corresponding hornogeneous equation ym 3y 2y 0 The characteristic equation is r3 3r22rrr2 3r2 rr 1r 2 0 Thus the general solution to the homogeneous equation is yt c1 eget c3e2t To nd the general solution to the nonhornogeneous equation need a particular solution to the nonhornogeneous equation Since gt t et is the sum of two terms containing functions of different forms we consider the two equations 3y 23 t 1 3y 23 et 2 For equation 1 we guess Y t At B Then we think again and notice that B is a multiple of a solution to the homogeneous equation so that our guess needs to be multiplied by t That is we guess that a solution is of the form Y t At2 Bt When we plug in Yt the DE 1 reduces to 6A2B 4At t so that 6A 2B 0 and 4A 1 Therefore 1 3 A B 4 4 In making the guess for a solution to equation 2 our rst impulse is to use Yt Det but this function satis es the homogeneous equation Hence we use Yt Dtet For this Yt the differential equation reduces to Det ct gt D 1 So our particular solution to the nonhomogeneous equation is 1 3 Y t t2 t tet 4 4 Using the general solution to the homogeneous equation we see that the general solution to the nonhomogeneous equation is l 3 yt c1 Cget C3th t2 t t t 4 e We need to nd the constants C1C2 C3 such that the initial conditions 1 3 01 0 quot0 y 7 y lt 4 lt 2 are satis ed We compute 3 t y t Z ct i tet Cget 2C3th and 1 y t i Zet tet Cget 46362t Thus 1ylt0C1C2C3 l 1 Z YO C2 263 3 3 i 4 2 0 2CQ Cg Solving this system we nd that 611 620 630 Therefore the solution to the initial value problem is l 3 1 t2 t t t yt 4 4 6 Example Determine the suitable form of a guess to a particular solution to the following di erential equation 214 y y y t2 4tsint We need to determine the solutions to the homogeneous equation so that our guess does not include a multiple of one of these solutions The characteristic equation for y4 ylll yll y 0 r4 r3 r2rrr3 r2 r10 By Newton s divisibility criterion the possible rational roots of above cubic are 1 and in fact r ii are both roots Therefore by long division characteristic equation is rr 1r1rr4 r3 r2r0 By long division of polynomials we see that r r 1 Therefore the characteristic equation is rr 12r 1 0 So the solutions to the homogeneous equation are 1 et tet et We can write the particular solution to 344 y y y t2 4 tsint as the sum of solutions to the differential equations 344 y y y t2 4 and 4 III II I i y y y y itSIHt Our rst guess for the particular solution to the nonhornogeneous equation y4 ylll yI y t2 4 Yt Alt Agt A3 We must ask ourselves are any of the terms above multiples of the solutions 1 ct tet ft to the homogeneous equation Yes the constant term involving only A3 is a multiple of the solution MW 1 to the homogeneous equation Therefore we need to multiply our guess by t before we start taking derivatives Ylttgt Alt3 14th Agt Our guess for the particular solution to y4 ylll y yI tsint Ylttgt ltB1t B2 COSt Clt 02 sint None of the terms in our guess is a multiple of any of the solutions 1 ct tet e to the homogeneous equation Therefore our guess for the particular solution to y4 ylll yll y t2 4tsint is Yt A1t3 A2t2 A3t B1tB2 cost 01t02 sint For a complicated function of the above form involving many unknown parameters a computer algebra system can be extremely helpful in executing the tedious algebraic calculations needed to apply the method of undetermined coefficients impulse Fun ctfan SHWoSe act is a Stanctian at fake cm ulargequot IJ39atMS i in tine interwal vegquott lttltto t and TS gra war at 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vy1y an a Fanimweqt 5H a swimma h yA P YtnHm PIP39YI 1y 1 Q HMx L LE schema S lh39bb in Y m lw quotquot quot quotPn39quot 3Z I39S SMI Qquot Ly ale CIYI 39CAA Y wLm You is a Par rcutar saga5w 395 ngMOSENeauJ 32H 39bbquot t39 m anfis 01 OKQoLaL39EIbn 396 aI JPr e wJeEermfnel 44 c emf ovarl f roh 9F quQME CFf A all work a leLVb JQP Jl fe q fam hr39 we em km a M QFSILA 0F Aloe rm W YIYzW VA 1 Solu 39an Ym ycnm Rwy 3 0 flea Mmme ace WM 2004 I v 61 ieafy OF I VsLer orgPf Ia39 eaquot ejua 39 ns is 5511 Smph el L493 ma nke 1 754g Cafceseoulc t Leary of near algelgrc Autonomous Equations and Population Dynamics In this section we examine equations of the form y y called autonomous equations7 where the independent variable does not appear explicitly The main purpose of this section is to learn how geometric methods can be used to obtain qualitative information directly from a differential equation without solving it Example The dl erentlal equation 2 W is an autonomous dl erentlal equation Recall that the general solution is yt Keri An exponential model is unrealistic in many situations eg7 modeling the number of organisms in a population because of limits to growth We can make this model more realistic by replacing the model by y haw where hy is some function which will control7 the size of the derivative We want 0 hy to be approximately 7 when y is small7 o hy to decrease as y gets larger7 and o hy lt 0 when y is suf ciently large The function hy r 7 ay satis es our requirements l Inserting My 7 7 ay into the model 3 hyy7 we obtain 2 T 7 Wy This is known as the the logistic equation The constant a is usually chosen to be for some K7 so that the model becomes By looking at this model we see that I For small Values of y the factor 1 7 is Very close to 17 so that the di erential equation is Very close to the exponential equation 9 T9 for such Values of y When y is Very close to K7 the derivative gets Very close to zero7 so solutions should level 03 around y i K The direction eld for illustrates this behavior 2Ir hv2 39 4 2 1 5 The logistic model has two equilibrium solutions7 found by setting 73 0 0r17gty gt K and y0 De nition The critical points of the dz erehtz39al equation 73 y are the solutions to the equation 0fy Instead of graphing yt as a function of t7 let us graph7 for the logistic rnodel7 y as a function of y That is7 we plot fyr1iy K2 rK4 gt gtlt l K2 K Y Here the arrows at the bottom indicate the sign of y We sometimes represent the sign of y by plotting these arows parallel to the y axis7 as in the following plot v The plot with the blue arrows on the left is called the phase plot Note that at y it appears that the concavity of yt changes How can we prove this Start by differentiating the differential equation with respect to t dy dzy df dy i gt 7 if dt fly dtz dy dt by the chain rule dzy Thus7 the sign of the second derivative W is d 0 Positive if if d and l have the same sign dy t d 7y have opposite signs df 0 Negative if Ty and dt Now7 rltligtyT y d So 1 changes from positive to negative at y Meanwhile from the differential equation itself T 1 A y dt K 7 dy weseethat Egt0for0ltyltK Therefore the sign of the second derivative 2 is 0 Positive for 0 lt y lt K2 0 Negative for K2 lt y lt K So the solutions yt are 0 Concave up for t such that 0 lt yt lt K2 o Concave down for t such that K2 lt y lt K And this is consistent with what we observe in the plots above That is there is an in ection point for the function yt corresponding to y K2 Let us now attempt to solve the logisitic differential equation Since iv T 1 E y dt K is separable we may write dy 7 dt 1 lt1 7 way gt Using the partial fraction expansion 1 1 1K 7 77 MyKM y legK we nd that 1 can be written as 1 1K 77 dyrdt y legK We then integrate both sides with respect to t to derive lnlyl ilnll iyKl TtC If the initial value y0 yo is positive then 0 lt y lt K so that y 0 d 177 0 ygt an Kgt Thus with this assumption lny7ln17yKrt0 gt 1 TtC UL liyK y TtC 55 y gt 6760 7 erteci y K K 6760 gt emec 39 1 7 gt Introducing an initial condition 110 107 implies that c 7 yo 6 7 7 LO K Inserting this value of 60 into the computations above and doing a fair bit of algebra we nd 110K yo K 7 yo W Mechanical Vibrations Linear secondrorder di erential equations model the oscillatory motion of a spring The same principles We apply here can be used to model other systems involving vibrations or periodic oscillations Examples 0 Acoustics o The vibration of a bridge under external forces such as traffic and Wind 0 Electrical current l l i De nitions 0 l is the length of the spring 0 L is the additional elongation caused by the mass 0 ut is the displacement from the original position at time it Key facts from physics Newton s Second Law of Motion F ma Force Mass gtlt Acceleration Hooke s Law The force exerted by the spring F3 on the mass directed up ward is proportional to L the elongation due to the weight of the mass Fs ltL The spring force acts upwards in the negative direction If ut represents a position function at time t then u t represents the corresponding velocity function and u t is the corresponding acceleration function If we denote the magnitude of external forces in the spring system by f t then Newton s Second Law of motion implies that a model for the system is t m X MW v v force mass gtlt acceleration To make the model precise we need to account for each force acting on the system And the forces are If m is the mass of the weight then by Newton s second law the weight exerts a downward positive magnitude force of wmg Where g is the acceleration due to gravity Using Hooke s law the force exerted by the spring at times t is F305 le 1105 Where k is some positive constant of proportionality Note that When L ut lt 0 the spring is compressed the resulting force Fst is positive and so exerted i The damping or resistive force Fdt caused say by air resistance or friction is assumed to be proportional to the velocity of the mass at time t Fold Vu t Where 7 is some positive constant of proportionality Note that if the velocity is in the negative T direction then the resistive force is positive so directed i We also account for the possibility of other variable external forces by So our model is mu t mg F5t Fdt Ft my 7 ML 7 39yut Since the force of the spring with the weight attached at rest is equal and opposite to the force exerted by the weight F 7kL w mg we have my 7 EL 0 so the model simpli es to mu t 7ut kut Ft where o m is the mass of the suspended object o 39y is the damping constant 0 k is the spring constant 0 Ft represents all other external forces Let us rst consider a simple case where there are no other external forces and no darnping so Ft 0 and 7 0 The model above simpli es to mu t kut 0 The corresponding characteristic equation is mr2k0 The roots of the characteristic equation are rj i m Therefore the general solution to the differential equation is ulttACOS gtBSm gt A cos wot B sin wot where we denote the constant Mi by um and call it the circular frequency We will now rewrite the solution ut using new pararneters R and 6 de ned in terms of the numbers above by ARcos6 and BRsin6 In other words we nd 6 and R from A and B Via the formulas B sm itan6 and RA2B2 Aicos i We can then re write ut as ut R cos 6 cos wot R sin 6 sin wot R cos 6 cos wot sin 6 sin wot Rcosw0t 6 1 Where to get the last equality we applied the trigonometric identity cosx y cos 10 cos y sinm sin y In the solution ut Rcosw0t 6 2 the parameter R is called the amplitude of the motion and measures the magnitude of oscillation The parameter 6 is called the phase angle This parameter measures the displacement of the wave from its normal position corresponding to 6 0 Notice that ut Rcosw0t 5 is just a shifted cosine curve With a perioda T given by 271 271 m 12 271 010 3 Notice that if the mass is large then T is large so that the spring Will Vibrate more slowly If k is large then T Will decrease so a more rigid spring Will cause the system to Vibrate faster aT is the amount of time it takes for the system to undergo a complete cycle Example A mass weighing 5 pounds stretches a spring 3 inches Suppose that the mass is forced an additional 6 inches down and then released with an downward velocity of 1ftsec Determine the position of the mass at all later times Also determine the period T amplitude R and phase 6 of the motion Since the mass stretches the spring 3 inches i foot and the force exerted by the mass is 5 pounds we have FltL 5i k201bft Therefore the model for this system is With g 32 ftsec gt m 332a 20a 0 The roots of the characteristic equation are r im i The general solution to the differential equation is ut Acos Bsin The initial conditions in feet are u0 u 0 1 1 o 1 gt A and B This means B 271 6 arctan 0174969 R V A2 B2 0507752 T A V1 8 Using these approximations for 6 and R we have ut 0507752 cos V128 t 0174969 With a nonzero darnping constant the model for the system becomes mu t Vu t kut 0 The corresponding characteristic equation is 77173 77quot k 0 and the roots are T177n2 7 j V72 4km 2m 7iyV21 4 ngt 2m Viv 1 j T nT 4 l 1 i 1 2m 7 Notice that the kind of solution we get depends on the sign of 72 4km For nonnegative values of VQ 4km we have ut Aer Ber if 72 4km gt 0 ut A Btelt2m if 72 4km 0 and if y2 4km lt 0 4k 2 12 ut e ltQm A cos Lt B sin at u Notice how the nature of the solution changes as y2 4km changes sign The point Where this happens 7 2 km is called the critical damping Note that 4km gt 0 74km lt 0 727407 lt 72 So o If 11142 are real then the numerators of the Values iyidyz 74km 2m are negatiVe so that 141142 lt O o If 141142 are complex Aim then A 75 lt 0 Therefore in all cases We haVe exponential decay dominating eanh solution This means that the damping gradually dissipates the energy in the system For example if m and r2 are complex then the solution is ut tilt2m A cos Mt B sin at Which can be Written Via the relations ARcos5 and BRsin5 ut Re ltzm 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3 MuSt kg cm eiynw tof a A CoNQSPOAJI39CJ b flu Macaw ALEvi we QM write rst ebua tian 46 J 63 em a at I w ABC await 2 a gewem 2 WEB z mm G W 61 0 6 3 I I 31 39 v I n I a 55 quotI 43 G 0 Ogtwt39w1quot Mam 14 nce a 34ch Znearly o39nae oenel n Se39u 39bn 57 lm JigFeren tial ejaa ien f g A 5 3 1 t 5 Q 59135ng 393 z i t f Seder S lu b g 5 L 613 4 h tq mg j Ch 73 Sys Lin Eq Lin lnd Eigenvalues A system of n linear equations in n variables aux1 amx2 alanxn 91 aux1 aux2 azanxn 92 analx1am2x2a x 19 mm 71 n7 can be expressed as a matrix equation Ax b a11 a12 a1n x1 1 a21 a22 a2n x2 b2 aml an2 o armz xn bn To see that the matrix equation is equivalent to the system of equations multiply the matrix and vector on the left side and equate components If b 0 the system is said to be homogeneous Nonsingular ie Invertible Case An invertible matrix A is called nonsinqular lfA is invertible we can always solve Ax b as follows Axb ZgtA1AXA1b gtIXA 1b gtXA 1b This solution is unique lfA is invertible the only solution to Ax 0 is the trivial solution xA3910 0 Example 1 Nonsingular Case 1 of 3 Here is an invertible matrix A and its inverse 0 1 2 92 7 32 A o 3 A4 2 4 1 4 3 8 32 2 12 By multiplying by the inverse ofA on both sides of the equation Ax 0 we obtain 92 7 32 0 0 A1AXA10 2 4 1 0 32 2 12 0 0 Therefore the only solution to Ax 0 is x 0 Example 1 Nonsingular Case 2 of 2 Now let s solve the nonhomogeneous linear system Ax b below using A391 Ox1x22x3 2 1x1 Ox2 3x3 2 4x1 3x2 8x3 O This system of equations can be written as Ax b where 0 12 x1 2 4 3 8 x3 0 The 92 7 32 2 23 XA1b 2 4 32 2 12 0 7 Singular Case If the coefficient matrix A is singular not invertible then either there is no solution to Ax b or there are Innnltely many s0IutIons to Ax b We can find these solutions by row reducing A b Linear Dependence and Independence A set of vectors xi x 2 x0 is linearly dependent if there exists scalars c1 02 cn not all zero such that 01X 02X cnxo 0 If the only solution of is 01 02 on 0 then x717 x727 x0 is linearly independent Example near n 1 of2 Determine whether the followind Wch ae linear dependent or linearly independent 0 1 2 X0 1 X0 0 7 X6 3 4 3 8 We need to determine for what coefficients 01X 02X C3X3 0 or Will c l 3 Ol I OOUJN Example near n 2 of2 We reduce the augmented matrix Ab 0 1 2 0 1 0 3 0 Ab 030 gt0120 4 3 8 0 0 0 1 0 01 303 0 0 gt 02 203 0 gtc 0 0 0 C3 Thus the only solution is 01 C2 on O and therefore the original vectors are linearly independent Example Linear Dependence 1 of 2 Determine whether the followind Wch ae linear dependent or linearly independent 1 2 1 X0 2 1 7 Xe 5 7 X3 6 5 4 5 We need to solve 01X 02X C3X3 0 or lilrlfilrlillEl lt3 l3 3 Ell illEl Example Linear Dependence 2 of 2 We thus reduce the augmented matrix Ab as before 1 2 1 0 1 2 1 o Ab 1 5 60 0 3 50 5 4 5 0 0 0 0 0 c1 202 lc3 0 7c33 7 gt 302 503 20 c5033 gt ck 5 003 0 03 3 Thus the original vectors are linearly dependent WILII lilsl llilll Linear Independence and Invertibility Consider the previous two examples The first matrix was known to be nonsingular and its column vectors were linearl inde endent The second matrix was known to be singular and its column vectors were linearly dependent This is true in general the columns or rows of A are linearly independent if and only ifA is nonsingular if and only if A391 exists Also A is nonsingular if and only if det A at 0 hence columns or rows ofA are linearly independent if and only if det A at O Linear Dependence amp Vector Functions Now consider vector functions x 1t x 2t xquott 163 X000 k12n teIa Xfft As before x1t x2t xquott is linearly dependent on lif there exists scalars c1 02 0 not all zero such that clxlt1gtltrgt ozxmm cnxltquotgtltrgt 0 for all re 1 OthenNise x1t x2t xquott i linearly independent on I Eigenvalues and Eigenvet y Vull LIV VIV39VVU MU u Illurl VI LI IV vector x into a new vector y Nonzero vectors x that A maps to multiples of themselves are important in applications We will be interested in finding numbers xi and corresponding vectors x such that Ax xix or equivalently Axilx 0 This equation has a nonzero solution if we choose xi such that det Axil 0 Such values of xi are called eigenvalues of A and the corresponding nonzero solutions x are called eigenvectors Example 5 Eigenvalues 1 of 3 all VVVLVIO VI LI Iv IIIuLI II A A42 3l K3 6 Solution Choose xl such that detAll 0 as follows detA det 3 2 2 3 det 3 6 l 2 2X 6 2 33 2242 212 327 gt x123 22 7 Example 5 First Eigenvector 2 of 3 To find the eigenvectors V he maum A we need to solve Atx 0 for A 3 and A 7 Eiqenvector for A 3 Solve Auxof2 3 WWJOWQF 1 t 3 6 3x2j to 3 9 x2 0 by row reducing the augmented matrix 1 3 0L 3 0L 3 0L1x1 3x2 0 t3 9 0 t3 9 0 to o 0 0x2 0 3 3x 3 gt X 2 of I C arbitrary gt choose X0 1 x2 Kl Example 5 Second Eia o 3 of 3 Eiqenvector for 1 7 Solve A11xo 27 3YXIHOKCgtP 3mm K 3 67KX2 K0 K3 1de K0 by row reducing the augmented matrix 9 3 o 1 13 0 1 13 0 1x 13x 0 3 1 o3 1 019 0 OJ9 1 Ox 0 13 13 1 gt 3 x2 of c arbitrary gt choose X2 W X2 K 1 J K 3 Normalized Eigenvectors From the previous example we see that eigenvectors are determined up to a nonzero multiplicative constant If this constant is specified in some particular way then the eigenvector is said to be normalized For example eigenvectors are sometimes normalized by choosing the constant so that x x x 2 1 Algebraic and Geometric Multiplicity In finding the eigenvalues A of an n x n matrix A solve detAll 0 Since this involves finding the determinant of an n x n matrix the problem reduces to finding roots of an nth degree polynomial Denote these roots or eigenvalues by A1 22 in If an eigenvalue is repeated m times then its algebraic multiplicity is m Each eige h Iquot e g eigenvalue of algebraic multiplicity m may have q linearly independent eigenvectors 1 s q s m where q is called the geometric multiplicity of the eigenvalue Eigenvecto i pe If an eigenvalue xi has algebraic multiplicity hen it is said to be simple and the geometric multiplicity is 1 also If each eigenvalue of an n x n matrix A is simple then A has n distinct eigenvalues It can be shown that the n eigenvectors corresponding to these eig are linearly independent If an eigenvalue has one or more repeated eigenvalues then there may be fewer than n linearly independent eigenvectors This may lead to com lications in solvin39 s stems of differential equa ons Example 6 Eigenvalues 1 of 5 Find the eigenvalues aw eigenvwww V me A A 0 1 1 A21 0 1 1 1 0 Solution Choose xl such that detAll 0 as follows 1 1 1 detA lIdet 1 1 1 1 1 1 Z33l2 1 2x2 12 gt 111214 1 Example 6 First Eigenvector 2 of 5 Eigenvector for A 2 Solve AAlx 0 as follows 2 1 1 0 1 1 2 0 1 1 2 0 1 210 gt1 210 gt0 3 30 11 20 2110 03 30 11 20 10 10 1x1 1x3O gt01 10 gt01 10 gt 1x2 1x320 000 0000 0x320 Example 6 2nol and 3rd Eigenvectors 3 of 5 Eigenvector for A 1 Solve AAlx 0 as follows 1110 1110 1x11x21x3O 1 1 1 O gt O O O O gt 0x2 0 1 1 1 O O O O 0 0x3 0 x2 x3 1 1 gtx1 x2 2x2 1 x3 O where x2x3 arbitrary x3 0 1 1 1 gt choose X2 I X O O Example 6 Eigenvectors of A 4 of 5 Thus three eigenvectors of A are 1 1 1 X0 1 X2 1 X6 O 1 O 1 where x2 x3 correspond to the double eigenvalue xi 1 It can be shown that x x2 x3 are linearly independent Hence A is a 3 x 3 symmetric matrix A AT with 3 real ei39envalues and 3 linearl inde endent 0 1 1 eigenvectors A21 0 1 1 1 0 Example 6 Eigenvectors of A 5 of 5 The eigenvectors in the last example are orthogonal since X1DX2 O X1DX3 O X2DX3 O Thus A is a 3 x 3 symmetric matrix with 3 real eigenvalues and 3 linearly independent orthogonal eigenvectors We could have also chosen Hermitian Matrices A selfadjoint or Hermitian matrix satisfies A A Note that if A has real entries and is symmetric see last example then A lsrlermltlan An n x n Hermitian matrix A hasjhe following properties All eigenvalues of A are real There exists a full set of n linearly independent eigenvectors of A If x and x2 are eigenvectors that correspond to different eigenvalues of A then x and x2 are orthogonal Corresponding to an eigenvalue of algebraic multipllcny m it is possible to choose m mutually orthogonal eigenvectors and hence A has a full set of n linearly independent orthogonal eigenvectors Existence and Uniqueness of Solutions to Linear and Nonlinear Differential Equations The following theorem tells us when we should expect to nd a unique solution to an lVP involving a rst order linear equation Theorem Existence Kc Uniqueness for First Order Linear Equations The initial value problem 1 pty M We yo 1 has a unique solution in the interval I t l 04 lt t lt B 046 provided The functions pt and gt are continuous on I The interval I contains the point t to corresponding to the initial value l Idea of Proof If pt is continuous we can always de ne W exp 198 ds and multiply both sides of 1 by ut Recognizing as before that for this choice of ut 1 dt we integrate both sides with respect to t to obtain may mow dt 0 WW Mtpty WOW 5 M011 00900 We can then solve for yt to obtain in general 1 C at 7 ilttgtglttgt dt 7 W W What remains is to choose 0 so that the condition yt0 yo is satis ed and to de ne y in terms of a de nite integral It turns out that the function 7 i t S S S 270 at 7 W a M gt d W satis es the initial condition D Example Find an interval in which the VP W 2y 4t 211 2 has a unique solution We rst recast the equation in the form given in the theorem 2 y Ey 4t Here 2 t gt 4t We need to nd an interval in t 1 On which p and g are continuous7 2 Containing t 1 This is accomplished by choosing Itltgt00oo By the theorern7 this lVP has a unique solution on I In fact7 we can solve this lVP using the method of integrating factors to nd that 1 9tt2t37 tgt07 Example What does the theorem say about the initial value problem ty 2y 4t2 y0 0 Nothing The theorem does not apply Why When we rewrite the equation in standard forrn7 2 y 7y 4t t the function pt g is discontinuous at t 0 There is no interval that can be chosen about t 0 so that pt is con tinuous The hypotheses of the theorem are not satis edl This does not mean that a solution does not exist y t2 satis es the differential equation and the initial condition The theorem we stated above applies only to linear differential equations that can be written in the form 1 pty 90 What can we say about more general equations7 namely nonlinear equa tions that can be written in the form dy i t dt f y for some function f Theorem General Existence and Uniqueness Theorem for First Order lVPs 61 a Let the functions fty and a y be continuous in some rectangle a lt t lt 6 v lt y lt 6 containing the point t07y0 Then there erists a nontrivial interval t07hlttltt0h7 contained in 04 lt t lt B such that the initial value problem 1 ts7 We 90 2 has a unique solution within the interval to 7 h lt t lt to h Proof We defer the proof to a more advanced course in differential equations Example Is there an interval in which the VP dy 3242 7 0 71 dx Mil 7 9 has a unique solution First note that this ODE cannot be written in the form 1 pty 90 That is7 this ODE is not linear So we cannot apply the existence and unique ness theorem for rst order linear equations We can apply the more general theorern above 7 it works for linear and nonlinear equations Both 3x242 3f 3x242 and 7 7 2 29 1 51 29 1 are continuous everywhere except the line y 1 may Luckily7 our initial value y0 71 corresponding to the ordered pair 07 71 is not on the line y 1 Thus7 the theorem says that there is a rectangle about 07 71 in which diy7324x2 dz W7 90 1 has a unique solution In fact7 this equation is separable7 and we can show explicitly that the solution 11173222x47 zgt72 is the unique solution in the rectangle satisfying the initial condition y0 71 is 96711 W gt 7271 lt1 An important corollary follows from these existence and uniqueness the orerns Corollary Suppose y fty and the functions f and are continuous in some rectangle R Then two solutions with di erent initial conditions cannot intersect in R Proof Suppose yl and y2 are two different solutions to the ODE y f t y lf they intersect at a point7 say 71l0 92t07 then yl and y2 are two di erent solutions to the same lVP7 namely 11 E 107 910 112 This contradicts the uniqueness part of the theorem on lVPs D Example The VP 1 1137 110 0 Has many different solutions7 includling7 for t 2 07 2 32 2 32 it 7 it 0 91 3 7 12 3 7 93 The nonlinear ODE appearing here does not satisfy the hypotheses of the general theorem for existence and uniqueness because of1 7 7 23 3y 3 y is not continuous in any rectangle that includes the point Ly 00 appearing in the initial condition We summarize some of the important differences between lVPs involving linear and nonlinear ODEs in the following table Linear Equations y7 py g Nonlinear Equations A formula exists for the solution perhaps involving integrals Solutions may or may not be ex pressible explicitly They may be given implicitly We can identify all points of discon tinuity of the solution by determin ing where the coef cient functions p and g are discontinuous The points of discontinuity of solu tions are more dif cult to discern A general solution containing an arbitrary constant can be written down Particular solutions can be found by choosing this constant ap propriately It is not always possible to write down an expression involving an arbitrary constant which encom passes all solutions Applications Mixing Problems 0 Involve the concentration of a substance in some medium 7 Flow rate and concentration of drugs in the bloodstream Concentration of chemicals in a body of water 0 This type of problem is common in many disciplines 0 These problems correspond to interesting differential equations To answer applied problems involving differential equations think about the following H What is the question is asking D What information are you given What variable assignments should you make that are consistent with the question and given information 00 What are the relationships between the given rates derivatives and the other parameters 4 Cf Can you use your model to solve the problem Example Suppose that a tank contains 20 gallons of solution of a certain chemical and that 5 lb ofthe chemical are dissolved in the solution A solution containing the same chemical with a concentration of2 lbgallon is allowed to flow into the tank at a rate of3 galmin The miccture is drained o at the same rate At what time after the process starts does the tank contain 25 lb of the chemical Variables t time7 t amount of chemical in tank at time t7 With these assignments7 d d7 rate of change of the chemical in the tank at time t Key Relationship Between Parameters lt Rate Of Charge Of lt Rate owing in gt 7 lt Rate owing out gt amnt of chem1ca1 Rate of chemical owing in concentration of in ow gtlt rate of in ow 2 gtlt 3 lbmin Rate of chemical owing out W concentration of out ow gtlt rate of out ow W gtlt 3 lbmin So we have the differential equation dx 3 7 7 is dt 20 We can solve this using integrating factors First rewrite as di zi6 dt 20 quot The integrating factor is a est2039 The general solution is t 40 Ce StZO Using the given information about the initial amount of Chemical in the tank 0 5 lb we see that 540on 0735 Therefore the solution the the lVP is zt 40 i 35e 3t20 We want to nd the value of t when t is 25 zt 25 s 40 i 3553 0 25 gt 15 35e 3t20 3 73t20 gt 7 7 e gt ln 3 i 7 3t 7 7 2039 Hence the tank contains 25 lbs of the Chemical after 20 3 20 3 1 20 7 i t 77 ln 7 7 ln ln 7 minutes 3 7 3 7 3 3 Remark 0 How much chemical will accumulate in the tank in the long run Will the amount of chemical increase without bound Will the amount of chemical in the tank approach some limiting value To answer these questions we use the form of the solution zt 40 i 355W to observe that as t a 00 the amount of chemical in the tank approaches 11310 zt 11310 40 i 35e 3t20 3310 40 i 35 40 o How much time will it take for the amount of chemical in the tank to reach 3 times its initial amount To answer this question we need to determine the t for which This means that 3553 0 25 St20 57 gt 73t20 ln57 20 gt ti ln57 Does this answer makes sense Is this last CCEPTCSSlOTZ negatiue Why or why not Example A tank holds 100 gallons of a solution that contains 40 lb of a chemical A solution containing 2 lbgal of the chemical runs into the tank at a rate of2 galmin The mimture runs out at a rate of3 galmin How much chemical is in the tank after 50 min Variables t time yt amount of Chemical in tank at time t The model is d 3 3 7y2gtlt27 y477y dt 10073t2t 100 That is7 dy 3 7 7 4 dt 100 i t y Using the method of integrating factors7 we derive t 7 ex Lott 7 ex 1 73111100 7 tdt M 7 p 100 it 7 p dt explt73 ln100 7 tgt expltln100 7 0 3 100 7 0 3 Multiplying boths sides of the DE dyjL 3 4 dt 100 y by the integration factor7 we nd that 3 dy 7 737 7 73 7 100 t dt 100 t 100 it y 4100 i 0 3 We now simplify this expression to 73d 74 73 100 i t a 3100 i t y 4100 i t and express the left side as the derivative of the product My 100 7 t 3y7 100 i t 3ygt 4100 i 0 3 s 100 i 0 31 4100 i t 3dt O s 100 i 0 31 2100 7 t 0 Solving for y we nd that m 200 7 21 C100 i t3 The initial condition means that 110 407 and therefore 200720O10070340 2001003040 so we derive 7 160 100339 Therefore the amount of chemical in the tank at time t is 160 t 20072t771007t3 ylt gt 1005 gt We now use this expression to compute the amount of chemical in the tank after 50 minutes This is what we were asked to do originally 160 50 200 i 100 i 7503 M 1003 Example Assume that the rate of change of the supply St ofa commodity is proportional to the di epenee between the demand Dt and the supply Find afopmula for St if the demand is constant ie Dt D0 We should be careful to distinguish functions of t in our model from constants which do not depend upon t The given assumptions imply dS a rD 7 S TD0 7 S where r is some constant of proportionality This differential equation can be recast in standard form as dS E TS rDO The integration factor is W eXp r dt e 6 Therefore7 the differential equation is equivalent to d a eTtS rDoe eTtS rDOe dt 0 d5 erti ertrS eTtrDO gt dt gt gt eTtS D06 C We conclude that the supply function in terms of the dernand7 D07 is St D0 05 F 1 39 The r r quot of q quot in a certain area increases at a rate proportional to the current population and in the absence of other factors the population doubles each week There are 200000 mosquitoes in the area initially and predators birds bats and so forth eat 20000 mosquitoes each day Determine the population of mosquitoes in the area at time t Let Pt be the population of mosquitoes in thousands at time t in days In the absence of predators the rate of change of the population is dP 7 P dt T Solving the differential equation for P and using the initial condition P0 200 we nd that Pt 2005 Remember this Pt represents the population without predation I Since without predation the population of mosquitoes doubles each week we know that l 137 2130 gt 200677 400eOT gt r 1n 239 In the presence of predators the rate of change of the population is dP 7 P720 dt T Therefore using the value for r above we see that 1n72pi20 gt 7EPt720 The Pt in this model represents the population with predation I The integration factor is Mt exp 71H dt exp imTQt exp 7 ln 2 expln 247 247 Now multiply both sides of the DE dP ln2 dt 7 Pt 720 by the integration factor and recognize the left side as the derivative of My 2 t7y to derive d 7 24713 72024 dt Therefore 24713 720247dt C To evaluate the integral on the right note that 71n2t 2 7dte lv tdt 6 7 7ie l tii2quot7 711 ln 2 ln 2 Therefore t7 27t7 2 Pt 140 O m 7 or equivalently 140 P t 2t7o 1n Since P0 200 we know that 140 O 200 i 7 ln 2 Therefore 1 140 140 t7 2 7 20077 ln2 ln2 M 4444 V m NAM JH beaver5gan 0 S Asaiu o an s Nam am GQIMM Y Qai 39 OUr ALSCUQSM 0 S r 3 ampxPanS39ionS a Jigbib Elua o onf MS be Pura y Formal 72 5 UQMNquot Mt CanScclerJ ilxe in erwll a anWQFSQACQ 5Q 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LLJHFLtvc 3 iquot ch MM WM L m Final quotHe ver5Q Laplace fra f wm a Rs 2 stwq 6y fie EL aNM GivePAS 1 i X uzs H g 393 QILt353nL1 1 o 7 0 34 Ha axf azua i39b Na 59 en n 7 in Le 412 an 93 35 a He E CK Loq o i a 1 ALCQ39E 5141 39 EXamelU F AJ La LaplaCe trms arm 0F f z OI 475 tary fquot 2 ml V Nome wt 2 dwit my SY Pre Q Ov tamem 1amp6 HUTclaitnaY We M Exa le39 FM EL Laelacg e tfce 1265 Estef t eiE m 2 So e quot ct4t Pas43 ULQM F45 1 iifaf x JFaNeS gurem IF F63321i tg e X K39t 5 SgtQZQJ Ctquot iamp 7 t F swc SgtaC CanarerrehJ i fif2 zuFCS quot lxm 3 quotin 21 FLYd 1quot 3 we amplebeg tLZ 51am v4 39a x bx C siz ML39l 6XC 39 I 2quot 12quot M39H c1 6 305 e enmn r 2Qltg c i s w 3 uw foeI um FLt f T e 1u cbcoskt quot039 Solving Differential Equations with The Laplace Transform The usefulness of the Laplace transform in solving initial value problems comes from the fact that f is related t0 5U 15 Theorem Suppose that f is continuous and f is piecewise continuous on any interval 0 g t g A Suppose futher that there exist constants Ka and M such that ft Keat for t Z M Then ft exists for s gt a and moreover f t 8 ft f0 Proof Suppose that the points of discontinuity of f on 0 g t g A are 0 t1ltt2ltlttn A Then integrating over each subinterval A estf t dt f1 estf t dt 2 e stf t dt 0 0 t1 A e Stf a dt tn We evaluate each integral using integration by parts u e do f t dt v ft du sest dt A 6 Stf t dt 0 tt1 tt2 tA fa fowl0 ea f ttt1 6 Stflttttn 8t1 3tflttdt8t2 3tflttdt A s e St t dt tn ea sti e W e st2fltt2 estlm A faWm est ftn sO nest dt A 6 3AfA f0 3 0 ftest dt 0 The terms in parentheses above telescoped or cancelled each other out successively leaving us With only the rst and last terms o To get the integral A 80 fte dt the corresponding integrals over subintervals were re cornbined to form a single integral over the entire interval So all together we have A A 0 6 Stf t dt 6 3AfA f0 s0 ftest dt Taking the limit as A gt 00 on each side we obtain A00 e Stf d dt 0 f0 3000 ftest dt Equivalently f ltt flt0 8 fltt Since the second derivative is the derivative of the derivative the same formula says that f t 8 f t f 0 8l8 ft f 0 f 0 82 ft 8 0 f 0 In general by iterating the relation for higher order derivatives we get flt gtlttgt 8 fltt s 1flt0 s 2f lt0 8fn 2lt0 JAR Um This last identity is only valid under certain assumptions Theorem Suppose that the functions ft f t f 1t are continuous and that f t is piecewise continuous on any interval 0 g t g A Suppose further that there exist constants 0 g t g A such that fort Z M ft S K6 f t S K6 Then ft exists and flt gtlttgt Siam s 1flt0 s 2f lt0 8fltn 2gtlt0 fut Um 397 f 1t Keat Example Use the Laplace transform to solve the initial value problem 34 634 7y 0 340 1 30 0 Clearly we could solve this by nding the roots of the characteristic equation but let us see how the Laplace transform can lead us to the same solution Since the Laplace transform is linear the differential equation irnplies y 65W 75w 0 1 Using the recursion formula from the theorem above we may write the left side of 1 in terms of f Applying the above theorem we get my 8210 y lt0 6l8 y 210 7 My 0 Combining terms with y and combining terms with y0 gives us 32 68 7 y 6 sy0 y 0 0 Plugging in the initial conditions we get 68 2 s 68 7 y 6 s 0 gt y 82 68 7 Factoring the denominator we expand by partial fractions 68 A B Liymmtm To nd the numbers A and B multiply both sides of the last equality by s ls 7 to get 63As7Bs l ABs7A B Therefore equating coef cients on both sides we nd that ABl and 7A B6 Solving this system results in 14 31 8 8 This means that 78 7 18 ys 1 87 Thus to nd yt we need to nd the functions Whose Laplace transforms are 78 18 3 1 s 7 We showed yesterday that Ct 1c ifclt0 6 dt 0 The integral diverges if c 2 0 Therefore and OO 00 1 fate 7 5 dt es7t dt s gt 7 0 0 8 7 We conclude that 1 1 t i 7t eis1 e 87 From these calculations and by the linearity of the Laplace transform 78 18 7 t 1 7t t 5 8187 gt M 8 86 In general we often want to know how to undo a Laplace transform That is given a Laplace transform Fs what function ft is such that ft Fs If we denote the inverse Laplace transform of Fs by 1Fs then we look for an t so that 1F8 t If we can t easily compute 1Fs by hand there are many references for common inverse Laplace transforms See p 319 of our textbook Computer algebra systems can also compute many inverse Laplace transforms Just as the Laplace transform is a linear operator the inverse of the Laplace transform is also linear If Fs c1F1s C2F2lt8gt anns then 1F8 01 1F18 C2 1F28 39 39 39 1Fn8 This means that you can compute the inverse Laplace transform term by term as we did in the last example Example Find the inverse Laplace transform of 832 4s 12 882 4 Notice that the denominator is factored completely over the reals The partial fraction expansion is given by 882 4812 iA B3C This implies that 832 4s12 A824B82CS AB32 Cs4A By equating coefficients of each power of s we nd AB8 C 4 4A12 Therefore A 3 B 5 C 4 832 431273 53 4 3 5s 4 3324 is 324324 324 Hence using the table on page 319 of our textbook 1832 4s 12 Slt824 35cos2t2sin2t Under appropriate conditions we can differentiate the Laplace transform by differentiating under the integral sign diims Ame87a dt 0 463715 dt tft Differentiating again and again we nd that al2 2 d n 2F8 t 1 F8 t ft d8 d8 Example Find the Laplace transform of tea sin bt From the above calculations and the linearity of the Laplace transform teat sin bf i eat sin bf i eat sin bt d8 d8 39 Entry 9 in our table p 319 implies that d t d b a M ds e 8m d8 s a2b2 b 8 a2 b22 23 2a Therefore 2bs a teat sin bt 2 3 a2 2 The Laplace transform of a function involving it can often be computed by taking an iterated derivative With respect to s of the Laplace transform corresponding to a simpler function More on the Differential Equation II I ay by cy 0 Repeated Roots We now know how to handle the differential equation II I ay by cy 0 in the cases Where b2 4cm y 0 Recall that we considered separately the cases b2 4acgt0 or b2 4aclt0 What do we do When b2 4ac0 In this case the roots of the Characteristic equation ar2 Iquot C 0 are b b2 4ac 7 b 67 2a 2a 2a How do we solve the differential equation II I ay by cy 0 when the two roots r1 m of the Characteristic equation ar2 Iquot C 0 are the same number That is what do we do when I T1T2 In the case where b2 4cm 0 the zero b g gives us only one solution mot e btQa But the general solution to the differential equation ay by cy0 1 is of the form yltt 01y1t C2y2ta 2 where y2t is a solution to 1 linearly independent to y1 How do we nd a general solution of the form 2 Answer We guess that yt is the product of y1t and another function That is we guess that W vty1t De 21 where vt is a function that we need to gure out If we assume that yt has this form then by the product rule a b a y t v te btQ vte btQ 3 Applying the product rule to 3 and simplifying we get 9 b2 yltt Ullltte bt2a Evlltte bt2a 4a2vltte bt2a 4 Then substituting 3 and 4 into our original DE ay by cy 0 we get I 2 Ulltte bt2a 12 II t bt2a a v e a Ultte bt2a b l bt2a bt2a b v tc 2avte c vtebt2a 0 We now rnultiply both sides of this equation by emf2 1 Next we collect the terms involving vtv t and 11t b av t b bv t c vt 0 5 Equation 5 in turn implies that b2 2192 4ac 6 4 W o lt gt Since 92 4ac 0 we know that b2 4ac 0 so that av t b2 2b24ac i b24ac iii 4a 7 4a i4ai From equation 6 we obtain and hence Integrating once we get 1t c1 sorne c1 E R Integrating again we see that 11t clt Cg sorne C1C2 E R Therefore since y1t TM2a yt Ulttgty1t c1t c2y1t C1tebtlt2a gym2 0 Therefore yt is a linear combination of the functions y1t TM200 and 312 t trim2 1 In this case we can compute the Wronskian of these two functions bt2a bt2a 6 te Wlty17y2gtlttgt b bt2agt bt bt2agt 6 bt2a e 1 6 Since the Wronskian is never zero we know that y1t e WQa and MW te btQa are linearly independent on the real numbers and therefore form a fundamental set of solutions to the original differential equation We have therefore proven the following useful theorem Theorem In the case b2 4ae 0 the general solution to the di erential equation ay by ey 0 is given by wt 616 bt2a CQte btQa Example Find the general solution to the di erential equation y 6y 9y 0 Here a 1 b i 6 e 9 so 92 4oe 62 419 36 36 0 The Characteristic equation r2 67quot 9 0 has the repeated roots b f3 2 By the theorem on the last page the general solution to the differential equation is yt 01 e3t egte3t We can now combine the three cases on b2 4ac in the following theorem Theorem Consider the di erential equation ay by cy 0 8 Let r1 and r2 be the roots of the correspondong characteristic equation ar2brc0 Then one of the three following cases occurs 1 If h2 4ac gt 0 then r1 and r2 are real and the general solution to the di erential equation 8 is yt clerlt Cg mt If 92 4ac lt 0 then r1 and r2 are complex numbers say A iui In this case the general solution to 8 is yt c1eAt cos at c2eAt sin ut If 92 4ac 0 then r1 r2 and the general solution to 8 is got 616 bt2a CQte btQa Reduction of Order Or how to get a new solution from an old one To solve ay by Cy 0 in the case Where b2 4cm 0 we made a lucky guess based upon a known solution We can extend this to a general procedure that Will give us a solution to the equation 34 pty qty 0 from a single known solution Reduction or Order Algorithm If y1t is a known solution to the differential equation 34 pty qty 0 then you might be able to nd another solution by trying W vty1t and then using the differential equation to determine What the function 11t is 1 Example Given that in t t is a solution to the di erential equation 2t2y 3ty y 0 t gt 0 nd the second linearly independent solution Let s try the reduction of order algorithm setting yt utt1 Then by the product rule y39t WOW 1 WW2 y t Wan 1 Wan 2 2utt3 Let s now plug this back into the original differential equation 2t2 Wan 1 2u tt2 wan 3 3tu tt1 utt2gt utt1 0 This simpli es to 2m t u t 0 9 We now think of 9 as a rst order equation in the function wt u t We can then write 9 as 2tw t wt 0 or ZtiZ l w We can solve this separable equation by writing the last equation in the form dw i dt w72 t39 Integrating both sides gives lnw lntKlnt12K w gt 0 K en Thus by exponentiating we nd that wt celntl2 ctlQ c 6K Our assumption that wt ctl2 gt 0 is satis ed since t gt 0 by assumption Therefore v t wt ctlQ Integrating this equation we nd that 2 W gee2 k2 Since y1t F1 2 ya vltty1ltt Citl2 0271 Ci g0 C2 k We can drop the last term since it contains the solution we started with We see that y2t t12 is an independent solution since F1 tlQ 1 5 i i 2 32 Wy1y2t t2 t32 2t H In particular Wy17y21 2 7 039 Nonhomogeneous Equations and the Method of Undetermined Coe icients We now connect our study of homogeneous equations 34 pty qty 0 to the nonhornogeneous case What do we do With more general differential equations like u pltty qltty 9t Where gt is a function not identically equal to zero It turns out that solutions to the nonhornogeneous equation are connected to solutions of the homogeneous equation Theorem If Y1 and Y2 are two solutions to the nonhomoge neous equation y pltty qltty 905 then their di erence Y1 Y2 is a solution to the corresponding homogeneous equation 34 pty qty 0 1 If in addition we have two linearly independent solutions yl and y2 to equation 1 then Y1t Y2t C1y1ltt C2y2t for some appropriate choice of constants c1 and Cg Proof Suppose that Y1 and Y2 are two solutions to the nonhornogeneous equation y pltty qltty 90 Then since Y1 ptY1 qtY1 W and Y2 WW5 qtY2 905 We see that Y1 Y2 PtY1 Y2l QtY1 Y2 Y1 Y2 ptY1 WW5 qtY1 qtY2 Yr mm mm Yr my gm 9t 9t 0 This proves the rst part of the theorem Since Y1t Y2t is a solution to the homogeneous equation we may express it as a linear combination of two linearly independent solutions y1t and y2t to the corresponding homogeneous equation for some appropriate choice of constants In other words Y1t Y2t 61111 t C2y2t for some constants c1 and Cg D This brings us to the following important theorem Theorem The general solution to the nonhomogeneous equa tion y pltty qltty W 2 can be written in the form yltt 01y1t C2y2t Y where yl and y2 are a fundamental set of solutions to the ho mogeneous equation 34 pty qty 0 3 01 and Cg are arbitrary constants and Yt is any solution to the nonhomogeneous equation Proof If yl and y2 form a fundamental set of solutions for the homogeneous equation 3 and Yt is a solution to the nonhomogeneous equation 2 then W C1y1t C2y2t Wt is a solution of 2 for any Choice of constants since CU PWZU My Cly1ltt C2y2t YW W 61111 t 6221200 Yltt qlttlt61y1ltt 62112 t Yltt 61yl plttyl qltty1 m 0 62 my qltty2 m 0 Y plttY qlttY 90 0 0 gt gt We now know that W 61y1t C2y2t Wt is a solution Why is it the general solution Let Mt be some arbitrary solution to the nonhomogeneous equation With yl and y2 a fundamental set of solutions for the homogeneous equation We need to show that Mt has a representation of the form C1y1ltt C2y2t YW Since Yt and Mt are both solutions of the nonhomogeneous equation their difference is a solution to the homogeneous equation The theorem on page one implies that for some appropriate choice of c1 and Cg W Yt 6134105 6274205 Therefore qb is representable as a linear combination of y1y2 and Yt so we are justi ed in calling the form W Cry1t C2y2t W the general solution to the nonhomogeneous equation y pltty qltty 90 How do we apply this theorem to solve a nonhomogeneous differential equation y pltty qltty W 7 We need to do three things 1 Find the general solution Ciyi t C2y2t to the corresponding homogeneous equation 34 pty qty 0 2 Find some particular solution Yt to the nonhornogeneous equation y pltty qltty 90 3 Add together the functions from steps 1 and 2 The dif culty lies in step 2 How do we nd some particular solution to y pltty qltty W 7 One technique we can use is called the method of undetermined coe icients The Method of Undetermined Coe icients Idea We guess at the form of the particular solution Yt to the nonhomogeneous differential equation 11 pltty qltty 905 but without specifying the coefficients Example Find a particular solution to the di erential equation y3y4y362t lt4 and use this to nd the general solution We know that the derivative of an exponential function is some number times the original function It would be reasonable to assume that we could nd a solution of the form In this case Y t 2146275 Y t 41462 Plugging these calulations into 4 we get 31453 chaff 4amp2 36 Y t Y t Ya In other words 414th 614th 414th 3e2t or equivalently 6Ae2t 3th gt A Therefore 1 Y t 2t lt gt 2 is a solution to the nonhornogeneous differential equation y 3y 4y 36 Note that the general solution to the corresponding homogeneous equation y 3y 4y 0 C1 4t CQ t Therefore by the theorem above the general solution to the nonhornogeneous equation yll 3y 4y 3627 is given by l yt C1 4t 026quotj gem Example Find a particular solution to y By 4y 2 t 5 If we assume that Yt Ara t is a solution and proceed as in the last example we obtain Ari t 3Aet 4Aet 2 t But here the left side reduces to zero That is we have 0 2 t This happened because as we saw on the last page is a solution to the homogeneous differential equation In other words it makes the left side of 5 zero that is y 3y 4y 0 We therefore need to be more industrious in nding the form of our particular solution Yt to the nonhomogeneous equation Since et is a solution to the corresponding homogeneous equation let us suppose that Yt vtet and use the method of reduction of order to nd a function 11t that works From the assumption Wt 11006 3 6 we use the product rule to nd Y t Y t and plugging these values back into the original differential equation 34 33 4y 26 3 7 We nd omitting a lot of work that v 511 2 Leting wt v t this equation reduces to w 5w 2 8 Using the method of integration factors we nd that the general solution to 8 is 2 wt gcc5t CER In other words Integrating we nd that 2 c 5 t t k v 5 56 Finally using assumption 6 at the top of the page we see that 2 Y t etvt Etet 6 Ice t keep throw out We throw out the last two terms in Yt since we know that they are solutions to the homogeneous equation Therefore 2 Yt tt ltgt 56 is a particular solution to the nonornogeneous differential equation Notice that the solution is a function of the form Ate t 7 Which is t times What our initial guess was A Rule of Thumb If you nd yourself in the situation above Where your guess turns out to be a multiple of a solution to the homogeneous equation try multiplying by your guess for Yt by t If this does not work try multiplying by t2 Let Pnt dot a1t 1 an be a polynomial of degree 71 Here are some guidelines for choosing Yt If gt has the form you should try for Yt t tSA0t Amt 1 An tSA0t Altn l Anth Pm Pm teo t Pm teo t sin t or P teo t cos t tSA0t Alt 1 14 th cos t tSB0t Ban 1 B th sin t Here 8 is the lowest power of t that Will assure that the resulting guess for Yt is not a solution to the corresponding hornogeneous equation Second Order Linear Differential Equations We7ve looked at special cases of the differential equation ail by co 07 where a7 b7 and c are constants If we consider the more general homogeneous equations a WW My 07 where pt and qt are functions of t7 when can we be assured that a solution exists Theorem Let I be an interval that contains the point to and suppose the functions pt qt and gt are continuous on I Then there is emactly one solution to the initial value problem a pty my 907 We 107 y to 16 valid in the interval I Example Find the largest interval in which the solution to the initial value problem cos ty ty By 17 y0 2 yO 3 is certain to eccist Rewrite the differential equation in the form appearing in the theorem t 5 i 1 y Ev y The coef cient functions t 5 1 E E are continuous in the interval 7717277172 This interval contains t 0 Therefore7 there is a unique solution to the lVP on the interval 771727172 Anytime we have two solutions of a second order homogeneous linear dif ferential equation7 their linear combination is also a solution This result is sometimes called the principle of superposition Theorem If yl and y2 are solutions to the di erential equation a WW My 0 then the linear combination clyl 02y2 is also a solution for any choice of constants 0102 The superposition principle results from the basic differentiation rules gum m gm gm goat eggs 0 e R Key Questions If we have two solutions yl and y2 to the equation a pty my 07 1 ls there a choice of constants for which yt 01y1t 02y2t also satis es some arbitrary initial conditions We 107 4 26 7 2 Can we get all the solutions through some linear combination C191 02927 C1702 E R 7 To answer these questions7 we need to de ne a few facts from linear algebra The determinant of the matrix a b c d 7 b d adicb is denoted de ned by a C How do determinants and matrices show up in differential equa tions In order for a linear combination Mt 01910 027120 to satisfy the initial conditions yt0 yo y t0 ye we need 90 030 01711030 0292t0 16 yto 0111 to UziWe lf we solve this system Via substitution after some elementary algebra we nd 7 10 1 to 7 16 120 C 710 130 16 110 1 91t0y t0 11 t0yzt07 2 91t0y t0 yitoyzt039 Using determinants this can be written as 90 712030 11to 1 617 16 9W0 CZ 91 1 11to 1 to 7 9 to 1 The denominator in each case is the same Webmo l 3183 wmgw eggtomato We call Wy1y2 to the Wronskian of y1 and y2 evaluated at to Our calculations above show that so long as the denominators in the expressions for c1c2 above are nonzero we can always satisfy the initial conditions Theorem Suppose that yl and y2 are two solutions to y WW My 0 and that the Wronshian Wyl7 12 lily2 13112 is not zero at the point to where the initial conditions 1W0 107 yto 16 are assigned Then there is a choice of the constants c1cz for which 11 Gil t 021209 satis es the di erential equation and the initial condiditions This means that we can always nd a solution to an lVP y pty 1002 07 We 107 y to 26 with a linear combination of two particular solutions y1t and y2t to the differential equation provided Wy1y2 to 31 0 In fact as shown on the last page the solution is Mt 01110 02120 where c1 and c2 are de ned by W 382 l lit 3823 l 4 Theorem If yl and y2 are two solutions of the di erential equation a pty 1001 07 1 and if there is a point to where the Wronshian ofyL and y2 is nonzero then the family of solutions 9 01910 02120 includes each solution to 1 for some choice of constants 0102 Proof Suppose you have an arbitrary function 6t satisfying the differential equation a pty 1001 07 we need to see that 0t h1y1t h2y2t for some constants 1162 lf yl and y2 are two particular solutions to 1 and to is a point where the Wronskian of y1 and y2 is nonzero the theorem on the last page implies that we can choose 01 and 01 so that a solution to the initial value problem i pty qty 07 We 9t07 alto 9 to 2 is given by 9 Gil10 02220 Therefore the two functions yt 01y1t ngg t and 0t are both solutions to the lVP By the uniqueness of solutions to the lVP 2 from the rst theorem in today7s notes 6 and y are really the same function That is 90 y 01110 0212 Example Find the Wronshidn of the functions 7215 t672t 6 7 dtt0 We form matrix with the two functions in the rst row and their respective derivatives in the second row 672 teat 72e 2t e Zt 7 2te 2t 39 To compute the Wronskian at t 07 we evaluate at the determinant at t 0 6720 0 6720 W572t7 WW0 726720 6720 7 2 0 6720 i 10 721 1474i0r Why is the Wronskian important It tells us when two particular solutions form a fundamental set of solu tions ie7 it tells us when we can expect to get every solution through a linear combination of two solutions We can also extend the notions of the determinant Wronskian to higher order differential equations Example We proved yesterday that ifrL and r2 are two di erent real roots of the quadratic equation arz br c 0 then for any choice of constants c1 and c2 yt Clem egem is a solution to the di erential equation ay by cy 0 Do the solutions em and em form a fundamental set of solutions for the di erential equation on the interval foo oo That is are there any solutions that are not of the form yt Clem egem for some choice of constants c1 and c2 The Wronskian of y1t em and y2t em equals erlt ergt Mal17112 Tlerlt rzergt T26r1r2t 7 T16r1r2t r1 7 r2e t Since r1 and r2 are different r1 31 r2 so that r1 7 r2 31 0 No matter which t we Choose in the interval foo oo We t em 31 0 By the last theorem above this means that y1t em and y2t em form a fundamental set of solutions In other words every solution yt to the differential equation ay by cy 0 is of the form yt Clem egem for some Choice of constants c1 and c2 The Laplace Transform Integral transforms are useful tools for solving linear differential equations An integral transform of a function f t is a function of the form B 1 Kltstgtflttgt dt where Ks t is a given function called the kernel of the trans formation The limits a and 6 include real numbers as well as 00 and 00 We will focus on a particular type of integral transform called the Laplace Transform To understand the Laplace transform we need to know how to compute an improper integral 0 ft dt lim um dt u gtoo If this limit exists then the integral is said to converge If the limit does not exist then the integral is said to diverge or fail to exist Example Let C be some real nonzero constant Compute OO eCt dt 0 This is an improper integral so we compute the limiting value of the corresponding nite integral 0 u eCt dt lirn eCt dt 0 u gtoo 0 1e if e lt 0 The integral diverges if e Z 0 In our 1ater discussion We will encounter a number of important applications in which We attempt to integrate functions that are not continuous A function f is said to be piecewise continuous on an interval m if the interval can be partitioned into a nite number of points at0 ltt1 lt ltta3 so that o f is continuous on each open subinterval tH ltt lt tr 0 f approaches a nite 1imit as the endpoints of each subine terval are approached from within the subinterval v In short a function is piecewise continuous ifit jumps at a nite number of points in the interval The Laplace Transform of a function f t is a function Fs de ned by the equation we Fltsgt 0 e m dt Notice that in the de nition above 0 s is a free parameter and o t is the variable of integration Therefore the resulting integral Will be a function of 3 if the corresponding limit exists The Laplace transform of some functions does not exist We need the following theorem to determine When the Laplace transform of a function represents a convergent integral Comparison Test for Integrals Suppose f is a a piecewise continuous function o If f 3 gt When t Z M for some positive constant M and if gt dt converges then ft dt also con verges a o If t 2 gt Z 0 for t Z M and if gt dt diverges M then f t dt also diverges Example Use the comparison test for integrals to determine a su icient condition on a given piecewise continuous function t so that the Laplace transform of t exists By the comparison test we need to insist that ft is eventually bounded by some nice function gt so that the integral arising from the Laplace transform of gt converges Let s take gt Keat for some constants K and a For Which such constants does gt exist The Laplace transform of gt is gt OOO Ke steat dt K eat St dt 0 K 0 t Ts ifa slt0 K it a 0 The integral diverges if a s 2 0 K a s ifsgta The integral diverges if s g a By the comparison test for integrals it suf ces to require that ft be eventually bounded by gt Keat Then the Laplace transform of f t will exist for s gt a This proves the following theorem Theorem Suppose that f is a piecewise continuous function on the interval 0 g t g A for any positive A Assume that ft Keat whentZM Then the Laplace transform of ft exists for s gt a In a more general context we allow the parameter s to be complex In this case the existence theorem above can be phrased in more general terms Example Compute the Laplace transform of the constant function ft 1 and give the interval of existence The constant function 1 satis es 1 eat for a Z 0 Thus we require that s gt 0 In this case 00 St 6 st 75 1 1 e 1 alt lim Example Show that if the Laplace transform off and 9 exists for s gt a1 and s gt a2 respectively then C1ft 62905 C1 ft C2 gt provided that 8 gt maxa1a2 By applying the de nition of the Laplace transform we see that 01fltt c2900 0 6 Stc1fltt c2glttdt c1 e st t dt 62 e stgt dt 0 0 cl ft c2 gltt The last two terms above represent convergent integrals as long as s is greater than maxa1a2 The above example shows that the Laplace transform L is a linear operator Example Find the Laplace transform of t 56 2t 3 sin4t Since the Laplace transform is linear we can compute the transform of each term in f t separately 2t The Laplace transform of e is convergent for s gt 2 since e 2tl eat for any a Z 2 Therefore e2t 6 St 2tdt es2tdt 0 0 82 The laplace transform of sin 4t is valid for s gt 0 since sin 4t 3 eat for any a Z 0 The transform itself is sin4t 6 sin4t dt 0 u lim 6 St sin 4t dt u gtoo st tu u 39 6 StCOS4t dt 20 6 cos 4t dt 1 s 6 sin4t 75 s u t S 4 4 4 4 t040 6 sin tdtgt 2 co 8 6 sin4t dt 0 32 1 6 s1n4t l l 5 We now use the equality above to solve for sin 4t That is 32 1 sin4t 1 6 sin 4t 1 implies 14 116 4 L 4t 4 8m 1 3216 3216 Putting together the facts 1 14 4 d L 4t 82 an Sin 1 8216 e2t we see that ft 5e2t 3 sir14t 5 e2t 3 sir14t 5 12 82 8216 The Laplace transform Obtained here is valid for 8 gt maX 20 0 Variation of Parameters Goal Find a formula for a solution to a general nonhomogeneous equation y pltty qltty W in terms of the functions pt qt and gt and the functions yl and y2 the linearly independent solutions to y pty qty 0 Idea Start With the general solution Ciyit C2y2ta 0162 E C to the corresponding homogeneous equation 34 pty qty 0 Replace the constants c1 and Cg by functions of t and determine if the requirement that the resulting function satis es the nonhomogeneous differential equation y pltty qltty 905 provides information about the functions We replace the constants With functions that change With t and so vary the parameters We Will determine conditions on U1lttgt and u t so that 305 U105 34105 U205 34205 W 77 New Cl 77 New 02 is a solution to y pty qty 90 If we dxffezenuate 1W H1 WW u2WW Wt WWW WWW WWW WWW To stmphfy ththgs we wtn Impose the zequuement thet WWW WWW 0 Th Luck was Lagxarge s lde Joseph Louis Lagrange 1736 1313 Thezefoze we have Wt u1WW WWW Takmg the deuvetwe of Y t gwes YW WWW u1W1 t WWW WWW w plug the expxesslons YtY t end Y t Into the diffexenual equetton we would hke Yt to settsfy y pW qW W We get ua mm u1lttgtyrlttgt uslttgtyslttgt u2lttgty5lttgt W u1lttyiltt u2lttgtyslttgt qlttgtu1lttgty1lttgt u2lttgty2lttgt got Let s now group all of the terms involving yl together and do the same for yg u1lttgtyrlttgt plttgtyalttgt qlttgty1 lt0 u2lttgt M plttgtyslttgt qlttgty2lttgt 1130011105 U 2ty t 90 Since yl and y2 are solution to the homogeneous equation 34 pty qty 0 we can reduce the expression above to U1lttgt 0 Ugt 0 ul 0111 t 12 ty 2t 90 Therefore we need to nd U1lttgt and u t so that U3 011105 U 2ty392t 90 If we put this last equation together with Lagrange s requirement we need U1t and Ugt to satisfy We have two equations two unknowns t and u 2t so t U 2ty2t 0y 0 t u 2ltty 2ltt 90 U 1 wyi 1 7 U3 we can solve this system to nd MW wwmw 1 Mensa mower Oi mW i m 0115 t y2tyit39 Since the denominator in each expression is equal to WwM y m w we see that W y2lttgltt 1 Wlt J17y2gtlttgty mW Wltfyiay2gtlttl To nally nd u1t and u t explicitly we integrate both sides of these equations mW LWQEWWH E 7 y1tgt U2lttgt 7 Wlty13y2lttdt02 If we can nd these integrals then our solution to U1lttgt y pltty qltty 90 Y U1ty1t U2ty2t Theorem prt and qt are continuous on the open interval I and if y1 and y2 are two linearly independent solutions to the homogeneous equation 34 pty qty 0 then a solution to the nonhomogeneous equation y pltty qltty W is given by t y2393 to Wlty17y2lt8 where to is any point in I y139lt3 als7 it y1t t lll i ll l als y2t The general solution to the nonhomogeneous equation is W 61111 t 6274205 Yd Example Find the general solution to y y tant 0 lt t lt 712 We rst solve the corresponding homogeneous equation y y 0 The characteristic equation is r2 r 0 so that 0i 4 r Therefore y1 t cost and y2t sint are the linearly independent solutions to the homogeneous equation The Wronskian of these two solutions is cos t sin t Wy1y2t cos2tsin2t 1 sin t cos t From the method of variation of parameters a particular solution to the nonhomogeneous equation is t t t t ytcost st m W63 774 1 774 1 V V Wy17y1t Wy17y1t Why did we use 7r4 as the lower limit of integration Because the number 7r4 lies in the interval 0 lt t lt 7r2 where we want to solve the equation We could have chosen any other point in this interval as the lower limit of integration How are we going to nd the integrals appearing in our expression for Yt We simplify the trigonometric quotient appearing in the integrand t t g 2 sinstans ds 8m Sds 4 714 COSS t 1 cos2 8 d8 7T4 coss t t secs ds cossds 774 774 St st sin 8 377 lnsec s tan 3 4 lnsect tant ln 1 sint 7 The second integral in the expression for Yt is much easier to compute t t st 2 cosstansds sins ds coss cost 774 774 3774 2 Therefore 2 Yt cos t lnsect tant ln 1 sint The terms in red cancel after s1nt costgt H 2 expanding each expression g ln 1gtcost gsint 1 cos t lnsect tant lt This is a solution to the homogeneous equation When we plug it into the differential equation we get 0 So it suf ces to take Yt cos t lnsect tan t as the particular solution to y y tan t The general solution to y y tant is yt c1 cost Cg sint cos t lnsect tan Systems of First Order Linear Differential Equations A general system of rst order linear differential equations has the form l 1101 P11t1 p12t2 p1ntn 91t l 1102 P21t1 P22t2 39 39 39 p2ntn 92W We represent this system in matrix form by x Ptx gt or equivalently 131 P1105 P12ltt 39quot WM 1101 9105 1102 1721t 1722t 39 39 39 P2nt 2 92W mil 17711 17712 t 39 39 39 pron t 71 gn t When the last vector gt 0 the system is said to be homogeneous In this case we have What does it mean for a vector to be a solution to a differential equation Example The vectors x1t 2t 2 3t are solutions to the equation I 1 1 X X 4 1 since d 3 3t 1 1 1 375 3 3t lt1 i i i X 7 and xi 4 dt 6 3t 4 1 4 1 2 3t 6e3t Also d 6 75 et i 1 1 ft dt 2et 2 t 4 1 2et and similarly for gt62 Suppose in general that 901105 1kltt 10 t 10 k t Mt 21 xltkgtlttgt 2 0 n1t xnkt are solutions to x Ptx We have the following superposition principle for rst order systems of linear differential equations Theorem If the vector solutions XG and Xlt2gt are solutions to then the linear combination is also a solution for any constants c1 and Cg By iterating this theorem we deduce that if x1x2 x are solutions to then so is for any constants C1C2 cn Theorem If the vector functions x1x2x are lin early independent solutions of the system XI Ptx for each point in the interval oz lt t lt 6 then each solution x of the system can be expressed as a linear combination of 2 xmpclt X say in exactly one way The linear combination of solutions in the last theorem is called the general solution to the system of linear equations The vector functions x1x2 x are a fundamental set of solutions to the system of differential equations XI Ptx The Wronskian of a set of solutions x1x2 x is the determinant of the matrix xm Xlt2gt xml The next theorem says that provided Pt consists of continuous functions the Wronskian of a set of solutions is either always zero or never zero Theorem If x1x2x are solutions of x Ptx on the interval oz lt t lt 6 and Pt is a matrix of continuous functions on oz lt t lt 6 then the Wronsliian W05 detx1t XQW X t7 is either always zero or never zero Recall that the set of vectors x1x2 x is linearly independent at a point to if detx1t0 xlt2gta0gt x t0 7r 0 If this determinant is zero then x1x2 xml are linearly dependent at to The last theorem says that provided Pt consists of continuous functions on a given interval then 0 A set of solutions x1x2 x is a fundamental set of solutions if at some point to in the interval Wltl0gt detxlt1t0 I Xlt2lltl0 I I X01 l0 75 0 o If x1x2 x are solutions to an equation x Ptx and the Wronskian equals zero at some point to but is not identically zero then at least one of the component functions of the matrix Pt is discontinuous at to Example The Wronskian of the solutions 3t t e e xlt1gtlttgt e xlt2gtlttgt 263t 2 t to the equation 1 Xl X 4 1 equals 3t t e e Wt 4 2t 263t 2 t Since Wt is not identically zero the two solutions 3t t 2lt i 10 t 7 263t 2 t e 1ltt form a fundamental set of solutions That is any solution to can be written as a linear combination c1x1t 622ltt Example Suppose are solutions to the system x Ptx for some Pt The Wronskian of these solutions is 1 5 Wt 3t 10t 7t 2t 3t Since the Wronskian is nonzero on 00 0 and 0 00 the vectors 1101 t x2t o Are Linearly independent on 00 0 and on 0 oo 0 Form a fundamental set of solutions on 00 0 and on 07 00 Moreover since the Wronskian is zero at t 0 at least one of the components of Pt is discontinuous at t 0 Mechanial Vibrations with Additional Forces So far we have considered the model for an oscillating spring mu yu kru 0 We now look at the more general case Where additional forces F3 are applied to the system The model becomes nonhomogeneous mu yu kru F3 Often the additional forces F3 are periodic so can be approximated by trigonometric functions For our purposes let s imagine that the external force at time t is given by the forcing function Fst F0 cos wt for some numbers F0 and w We can apply the method of undetermined coefficients to nd the general solution to mu yu kru F0 cos wt 0 01U1t CgU2t A cos out B sin wt Where U1 and U2 are linearly independent solutions to the corresponding homogeneous equation c1 and Cg are arbitrary constants and A B are xed numbers coming from the method of undetermined coefficients The general solution to mu 7U kru F0 cos wt ut c1u1t CgUglttgt A cos out B sinwt r homogeneous part nonhomogeneous part has two parts From our study of systems Without external forces ie the homogeneous case we know that if damping is involved in the system then the solutions U1 and U2 are dominated by functions Which decay exponentially The general solution to the homogeneous equation uct c1u1t Cqut is therefore called the transient part of the solution The particular solution to the nonhomogeneous equation Ut Acoswt B sinwt oscillates for all time so is called the steady state solution or the forced response As we did in the last section we will often want to express the steady state solution in terms of a single trigonometric function by requiring that ARcos6 and BRsin6 Then using trigonometric identities Ut Acoswt B sinw Rcoswt 6 In terms of the parameters in our model mu 7U kru F0 cos wt we have the following relationships F 2 2 RXO cos w sin67Kw where A m2w3 w22 V2w2 and tag A question of great importance is How does the amplitude R in the steady state solution depend upon the frequency w of the external force Certain values of to result in resonance and make the amplitude R large To answer these questions we need to look at the amplitude of the system F F R 0 0 A m2ltw3 we Tm as a function of w We rst consider the cases Where the frequency of the external force is small What happens as w gt 0 k limOA liqum2w2 012 waQ mwg m k 42 w gt m Therefore 7 of F0 ti illaniillnwKimwgi k3 So the smaller to is the closer the amplitude is to Fgkr What about When the frequency w is large lim A lim m2w w22 y2w2 00 Therefore lim Rf lim F0 0 w gtoo i w gtoo A i So the amplitude of the steady state solution dies out as the frequency w gt 00 Is there some to greater than 0 that Will make R greater than the two values we observed here Since we can nd Where R Rw w E 0 00 is largest by nding Where A AW m2w3 012 72012 to E 0 00 is smallest This happens at the global minimum of the function fw m2w w22 V2w2 w E 0 00 Let s expand fw 77120161 2m2w3w2 m2w4 V2012 Then the critical points for w E 0 oo occur Where 0 f w 4m2w3272 4m2w3w w4m2w2272 4m2w3 In other words the only critical point for f w on 0 00 occurs Where 4m2w2 2v2 4m2w3 0 so that 4 2 222 2 2 w w w3rwo 1r2 4m2 2m2 2m2w If 17 then v2 w 7 10 2m2w3 is complex so that there are no critical points on 0 oo Otherwise assuming V2 2 2 2 m gt lt8mw0 we nd I 2 V2 2 2 V2 2 2 2 f coo W 12m wO Z mQ 27 4mw0 8m2w3 4 gt 0 Therefore the point V2 2 w wo 2m2w0 corresponds to a global minimum for fw Since A f 7 this means that A has a global minimum on 0 00 at I V2 l w 10 2m2w3 Finally since F0 R we see that R has a global maximum on 0 00 at 72 w wmax WO 1 Notice that When the damping constant 7 is small wmax is close to w The actual maximum value for R is then Rwmax R coo 1 2m2w3 F F 2 72 7010 Smk WU 39 1 m This last approximation is only valid only for small values of 7 Two important conclusions about resonance 0 The closer the value of 4mk is to 72 the larger the value Rmax Will be This conclusion is valid for any value of 7 o For small values of y F0 Rmax V010 Recall that we de ned critical damping to refer to the situation Where 72 4km If you don t want your system to oscillate out of control you should avoid values for the damping constant 7 close to the critical darnping value If your darnping constant is small then you should assure that the values of too and F0 are such that F0 V010 is not too large Of course sometimes resonance is good For instance if your spring system is designed to detect weak oscillations In this case choosing a small 7 Will ensure that the amplitude of the exerted force F0 is pronounced In general adjust the system to your advantage by noting the relationship between the constants m y k F0 wg Be particularly careful about When these parameters collude to to yield large oscillations in the resulting system We can Visualize the relationship between these parameters by plotting RicF0 as a function of wwo for different values of F 7i m Forced Vibrations Without Damping What happens When we add energy into the system7 but neglect clamping entirely In this case 7 07 so our model becomes mu ku F0 cos wti There are two imporatant subcases here w we and w we Where we kmi In the case Where w y am the method of undetermined coef cients and a bit of work implies that the general solution to mu kiu F0 cos out is given by ut c1 cos wot Cg sinwot coswt F0 mw3 w2 Where the constants c1 and Cg are determined by the initial conditions If those initial conditions are u0 0 and u 0 0 then the constants become F0 d 0 mw3 w an 62 C1 Thus the solution becomes F u cos wt cos wot We can use trigonometric identities for cosx j y to write this as ut WW3 012 sin 1 gw sin ltw0 m The solutions ut imwgl W2 sin 7WD gw sin 7WD wt amplitude are oscillatory but the amplitude of the oscillation also Varies in a regular way Systems with this type of motion exhibit a beat also known as amplitude modulation Example Solve the initial value problem u u 05 cos 0873 u0 0 u 0 0 In this case we haVe w0Wl w08 F005 From the formulas aboVe the solution is ut 277778sin01tsin09t A graph of the solution and bounding functions is u 27777wmon 3 wi77713mnltvngtmuuv In the case where w we called the resonant case variation of pararneters can he used to show that the systern mu ku F0 cos wot has the general solution F0 75le wot 2mm ut 1 cos wot 2 sl39nolot Because of the t in the coelhcient of the last terrn solutions of this form hecorne arbitrarily large ast grows Example Fmol the solutm t0 u u cost am 0 710 039 Here the natural ireouency we m 1 which matches the ireouency in the forcing function The tl al values is general solution with these i ut itsint MATH 267 Sections A3 C l Homework No 9 Reading Section 94 On line notes on general method to solve linear homogeneous systems of equations with constant coef cients Section 95 up to pg 497 included Section 98 Suggested Problems Section 95 Problems 1 3 25 27 Section 98 1357 Problems to be handed in class due Monday April 18th Problem 1 10 points Calculate the solution of 1 i A2 20 1 1 with 1 1 0 A 0 1 0 1 0 1 Problem 2 10 points Calculate the exponential EA where 702 Problem 3 10 points Calculate the solution of with MATH 267 Section E1 Homework No 8 Reading Sections 747 757 767 78 omit considerations on qualitative behavior of solutions See on line notes for method of solution Suggested Problems Section 74 Exercise 56 Section 75 Exercises 1517 Section 76 Exercises 17 37 9 Section 78 Exercises 1112 Problems to be handed in in class on Monday April 16 Problem 1 Find the solution of the following boundary value problem where 2 Ai where 71 0 0 0 0 0 1 0 0 0 A 0 0 71 0 1 0 0 1 71 0 0 0 0 0 71 i A2 where 0 0 0 2 0 0 72 1 A 7 1 2 0 0 72 0 0 0 Remarks 0 A comparatively small collection of differential equations can be solved in closed form using simple functions 0 In this course we will look at methods for nding solutions and when this is not feasible or possible we will examine qualitative methods to study solutions of differential equations An important qualitative analysis tool for studying a differential equation is a directionslope eld De nition A direction eld or slope eld is a collection of vectors which are tangent to the graph of solutions to a differential equation Example Direction eld for y4 y lt X XW X W1 X X X XW X W1 X X X X W W x W X X X X X X W1 X X X X X W 4 X X X rr N m X W W 4 W 4 W 5 44 f t 2 H X d 1 Example Direction eld for d7 E y Y J3939 39 W WWW A l 1 eee 2 Q 3 De nition A solution to a differential equation fyt corresponding to 0 is called an equilibrium solution Salutians may canuerge to or diverge from equilibrium solutians depending on initial values In general to nd equilibrium solutions set 0 in the differential equation and solve for y Question How do we tell if solutions asymptotically convergediverge tofrom an equilibrium solution Answer This can be determined from the sign of the derivative above below an equilibrium solution 3 Example Find the equilibrium solutions of the differential equation dy 7 a 1 dt y 39J where a and I are constants These occur at 0 7 ay I so that I y lt a Example Find all solutions to the differential equation 1 7 W 1 310 ya where 11 and ya are xed constants We rst recast the equation 1y Li 7 1 gt d 1 gt 1 dt ay ay I yba a Next integrate both sides with respect to t to derive 111 y ba at K where K is the constant of integration Now emponentiate both sides to obtain Ey ba 2 Le where L 6K y ba 2 i116 gt y ba Ce where C iL represents some constant We nally apply the initial condition y0 ya to nd that ya ba C ea390 ba C Therefore C ya ba We conclude that the solution to the VP 1 7 W 1 310 ya is y b ya 1739 a The Method of Integrating Factors In the last example we solved a differential equation by rewriting it and integrating both sides What do we do if this is not possible For rst order linear equations we can try to apply the method of in tegrating factors Key Idea Multiply both sides of the equation by a function so that one side takes the form of the product rule f939 f39g g f The dif culty lies in nding this function 0 Example Solve the initial value problem y39 2ye2t y0 2 First multiply both sides by some unknown function 110 the integration factor 110W 211099 62 1105 1 The product rule says that d E 010931 110W yu39 2 C hoose 110 so that the right side of 2 matches the left side of 1 We re set if Wt 4110 This means that Wt 110 d 1 1 gt E 111110 2 since 111110 8 110 2110 gt 2 provided 110 gt 0 gt 111110 2t C integrate gt 110 K 2t emponentiate where K 9C Since 110 is an integration factor we do not need the most general solution 110 we just need a function 110 that works We may take 110 6 Plugging this back into 1 we obtain ezty 2ye 2t 1 gt e2ty 1 rewrite the left side gt 6 2111 t C integrate gt y tea Ce solve for Now we use the initial condition y0 2 and the general solution above to conclude that 2 y0 0 62390 CB2quot C Therefore the solution to the inital value problem is y ten 262i In the last example we found 11 t by solving a differential equation How do we solve for 110 in general if this is even possible The most general rst order linear ODE can be written in the form 1 am my mag 0 3 If we set 1005 1105 t 7 l t glt gt W and m gt M then 3 can be written in the form dy 7 t t dt p y 9 Multiply both sides by an integration factor 110 to obtain dy 1105 120911099 11t9t 4 We want 110 so that the left side equals 1 a 11 This means that Mt Mt dt ptut gt a pt prouded 11t gt 0 1 gt 111110 pt derivative rule for ln f gt ln 11t ptdt C integrate gt 110 C exp ptdt exponentiate 7 Therefore we may take our integration factor to be um exp paw 5 So if we can compute this integrah we will obtain 11t in closed form We can then rewrite 4 as d 1a 1t t dt 9 l l 9 Integrating this expression we obtain y um 11tgtdt 0 Finally solving for y 1 C y m 11t9tdt m 6 Therefore we have solved the most general rst order linear differential equa tion provided we can compute the integrals appearing in 5 and 6 Unfortunately this is not always possible 72 Review of Matrices E5 For theoretical and couwwmuu masons we review resulm of matrix theory in this section and the next i A matrix A is an m X n rectangular array of elements arranged in m rows and n columns denoted an an am a a a 21 22 2 A7 717 am am awn v39vi Some examples of 2 x 2 matrices are given below 1 2 1 3 1 372139 A B c 3 4 2 4 4539 677139 Transpose I E The transpose ofA av 19 AT 41 an an am an a21 am a a a a a a 21 22 2 12 22 2 A 3 A am am awn am am quot39 awn a For example 14 12 13 123 A 3A B 3325 34 24 456 36 Conjugate I E The conjugate ofA ay 9 K EV an L712 am an L712 am a a a a a a 21 22 2 21 22 2 A 3 A am am 2 awn am am 39 39 39 awn a For example 1 23i 1 273139 A gtA 374139 4 34i 4 j Adj 0th I E The adjoint ofA is 37 and is denoted by Aquot an L712 am an L721 am a a a a a a 21 22 2 12 22 2 A 3A am am awn am am 39 39 39 awn a For example Square Matiices order n all an A 721 722 a a n1 n2 For example E A square matrix A has the same number of rows and columns That is A is n X n In this case A is said to have Vectors E A column vector x is an n X 1 matrix For example 1 x 2 3 n A row vector x is a 1 X n matrix For example y 1 2 3 Note here mat y x7 and mat in general it x is a column vector x then x7 is a row vector The Zero Matrix 3 The zero matrix is defined to be 0 0 whose dimensions I depend on the context For example 00 00 000 0 0 000 00 000 00 Matrix Equality g Two matrices A av mm B by Luv equal if ax by for I all i and j For example 1 2 l 2 A B gtAB 3 4 3 4 Matrix 7 Scalar Multiplication 3 The product of a matrix A L1 and a constant k is defined I to be kA kay For example 1 2 3 75 710 715 A 37514 4 5 6 720 725 730 Matrix Addition and Subtraction E The sum oftwo m X n matrices A av mm B by m I defined to be A B av by For example 1 2 5 6 6 8 A B gtAB 3 4 7 8 10 12 at The difference of two m X n matrices A av and B by is defined to be A B ay by For example hr 2M 2 mi 11 Matrix Multiplication E5 The product ofan m X n matrix A av mm m n X r I matrix B by is defined to be the matrix C CV where n Cy Z ricka kl l Examples note AB does not necessarily equal BA 1 2 1 3 14 38 5 11 A3 4J39Bz 4 343 916l11 25 19 212 10141 1212 41614 20 3 0 320 04r3 5 1 D 1 2 2CD j 0 I 1250 01076 17 4 BA Vector Multiplication E The dot product oftwo n A 4 WW x amp y is defined as x y Z My kl 1 The inner product of two n X l vectOIs x amp y is defined as Ky 1quot 226i k1 3 Example 1 71 x y 27 3139 gt xTy 171 227 3139 3i5 5i 712 9139 3f 5 5139 xy xTy 171 22 30 30575139 18 21139 Vector Length 3quot Note here that we have used the fact that ifx a bi then x 9 6 nbiXaibi aZ bz lxlZ v i Example 1 x 2 gt quotxquot xx 2 J11 22 3 4i3 74139 3 4139 m J5 Orthogonality g Two n A 1 4 WW x amp y are orthogonal if xy 0 I Example 1 11 XP Y4 DOW11124310 3 71 Identity Matrix E5 The multiplicative identity matrix I is an n X n matrix given by 1 0 0 1 001 41 For any square matrix A it follows that Al IA A aquot The dimensions of I depend on the context For example 1 0 0 1 2 3 1 2 3 AI1 2Y1 Ol1 zl 1B 0 1 0 4 5 6 4 5 6 K3 4A0 U K3 4 0 0 1 7 8 9 7 8 9 Inverse Matrix g 9 A square matrix A is nonsingular or invertible if there exists a matrix B such that that AB BA I Otherwise A is singular The matrix B if it exism is unique and is denoted by A391 and is called the inverse ofA It turns out that A391 exists iff detA 5 V Lulu A391 can be found using row reduction also called Gaussian elimination on the augmented matrix AlI see example on next slide The three elementary row operations 0 Interchange two rows Multiply a row by a nonzero scalar 0 Add a multiple of one row to another row Example Finding the Inverse Matrix 1 of 2 E Use row reduction to find the inverse of the matrix A below if it exism 0 1 2 0 3 4 73 8 l Solution If possible use elementary row operations to reduce AlI 0 1 AI 1 0 4 73 such that the left side is the identity matrix for then the right side will be A39l See next slide Example Finding the Identity Matn39X 2 of 2 0 1 4 J 7 732 4 72 12 792 32 732 12 792 7 7 4 32 72 A4 i Thus Matrix Functions E5 The elements u u mum can be menu u u real 4 name I In this case we Write x10 am am am x0 x29 AU any any alto mt ama ama awe 41 Such a matrix is continuous at a poing or on an interval a b if each element is continuous there Similarly WlLll differentiation and integration 2 jAzdz ag 1111 Example amp Differentiation Rules Example AU 32 sin 31A 6 cost cost 4 dt ism 0 3 Atdt7 4 1 Many of the rules from calculus apply in this setting For example dCA Cal A dt 39 Where C is a constant mattix dt dAB7 dB dt 7dr dt

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