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# Engineering Mathematics MATH 3321

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CHAPTER 3 Second Order Linear Differential Equations 31 Introduction Basic Terminology and Results Any second order differential equation can be written as Fm 247 27 24 0 This chapter is concerned with special yet very important second order equations namely linear equations Recall that a rst order linear differential equation is an equation which can be written in the form 1 2905 6196 where p and q are continuous functions on some interval I A second order linear differential equation has an analogous form DEFINITION 1 A second order linear di erential equation is an equation which can be written in the form 24 py qy 1W 1 where p q and f are continuous functions on some interval I The functions p and q are called the coe cients of the equation the function f on the right hand side is called the forcing function or the nonhomogeneous term The term forcing function77 comes from applications of second order linear equations the description nonhomogeneous is given below A second order equation which is not linear is said to be nonlinear Examples a y 7 53 6y 3 cos 2x Here pz 75 qz 6 x 3 cos 2x are continuous functions on foo 00 b z2 y 7 2x 3 2y 0 This equation is linear because it can be written in the form 1 as 2 2 yH77yi2y0 1 1 where pz 2m qz 2m2 x 0 are continuous on any interval that does not contain z O For example we could take I 0 oo 63 c y zyzy 7 y3 e is a nonlinear equation this equation cannot be written in the form I Remarks on Linear Intuitively a second order differential equation is linear if y appears in the equation with exponent 1 only and if either or both of y and 3 appear in the equation then they do so with exponent 1 only Also there are no so called cross product77 terms yy y y y y In this sense it is easy to see that the equations in a and b are linear and the equation in c is nonlinear Set My y pzy If we View L as an operator that transforms a twice differentiable function y into the continuous function Llyml 24 Nell90 610596 then for any two twice differentiable functions y1z and y2z Lly1 y2xl 24196 y2xl ply1 y2xl qly1 y2l 241 24305 plyl y Wl 6196 mm y2l We pyl qy1 24390 py 6100 we Lly1ml Lly2l and for any constant 0 WWW l0yxl plcyxl qlcyl cy plcy wl 6610596 ellTm Nell90 qyl cLlyl Therefore as introduced in Section 21 L is a linear dz erentz39al operator This is the real reason that equation 1 is said to be a linear differential equation I The rst thing we need to know is that an initial value problem has a solution and that it is unique THEOREM 1 Existence and Uniqueness Theorem Given the second order linear equation Let a be any point on the interval I and let Oz and B be any two real numbers Then the initial value problem 24 2496 2 6100 y f7 Ma a7 Ma B has a unique solution As before a proof of this theorem is beyond the scope of this course Remark We can solve any first order linear differential equation Section 2 1 gives a method for nding the general solution of any first order linear equation In contrast there is no general method for solving second or higher order linear di erential equations There are however methods for solving certain special types of second order linear equations and we shall study these in this chapter Extensions of these methods to higher order linear equations will be given later I DEFINITION 2 The linear differential equation 1 is homogeneous 1 if the function f on the right side of the equation is 0 for all z E I In this case equation 1 becomes 24 2496 2 6196 y 0 2 Equation 1 is nonhomogeneous if f is not the zero function on I ie 1 is nonhomo geneous if fz 7 0 for some z E I As you will see in the work which follows almost all of our attention will be focused on homogeneous equations 1This use of the term homogeneous is completely different from its use to categorize the rst order equation y fxy in Exercises 22 CHAPTER 1 Introduction to Differential Equations 11 Basic Terminology Most of the phenomena studied in the sciences and engineering involve processes that change with time For example7 it is well known that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t In mathematical terms this says that d dig ky k a negative constant 1 where y yt is the amount of material present at time t If an object7 suspended by a spring7 is oscillating up and down7 then Newton7s Second Law of Motion F ma combined with Hooke7s Law the restoring force of a spring is proportional to the displacement of the object results in the equation d2y 2 g k y 07 k a pos1t1ve constant 2 where y yt denotes the position of the object at time t The basic equation governing the diffusion of heat in a uniform rod of nite length L is given by Bu 7 3 at 812 where u uzt is the temperature of the rod at time t at position I on the rod Each of these equations is an example of what is known as a differential equation DIFFERENTIAL EQUATION A di ferential equation is an equation that contains an unknown function together with one or more of its derivatives Here are some additional examples of differential equations Example 1 2 I y i y a y y 1 b 12 7 235 2y 413 c g 27 0 Laplace7s equation d g 7 4 4 3671 TYPE As suggested by these examples7 a differential equation can be classi ed into one of two general categories determined by the type of unknown function appearing in the equation If the unknown function depends on a single independent variable then the equation is an ordinary di er ential equation if the unknown function depends on more than one independent variable then the equation is a partial di erential equationi According to this classi cation the differential equations 1 and 2 are ordinary differential equations and 3 is a partial differential equation In Exam ple 1 equations a b and d are ordinary differential equations and equation c is a partial differential equation Differential equations both ordinary and partial are also classi ed according to the highest ordered derivative of the unknown function ORDER The order of a differential equation is the order of the highest derivative of the unknown function appearing in the equation Equation 1 is a rst order equation and equations 2 and 3 are second order equations In Example 1 equation a is a rst order equation b and c are second order equations and equation d is a third order equation In general the higher the order the more complicated the equation In Chapter 2 we will consider some rst order equations and in Chapter 3 we will study certain kinds of second order equationsi Higher order equations and systems of equations will be considered in Chapter 6 The obvious question that we want to consider is that of solving a given differential equation SOLUTION A solution of a di erential equation is a function de ned on some interval I in the case of an ordinary differential equation or on some domain D in two or higher dimensional space in the case of a partial differential equation with the property that the equation reduces to an identity when the function is substituted into the equation Example 2 Given the secondorder ordinary differential equation 12 y 7 21 y 2y 413 Example 1 show that a 12 213 is a solution b 2x2 31 is not a solution SOLUTION a The rst step is to calculate the rst two derivatives of y y 12 213 y 21 612 y 2 121 Next we substitute y and its derivatives into the differential equation 122 121 7 2121 612 212 213 77 413 Simplifying the lefthand side we get 212 1213 7 412 712a3 212 413 i 413 and 413 7 413 The equation is satis ed y 12 213 is a solution b The rst two derivatives of 2 are 2 212 31 2 4x 3 ZN 4 Substituting into the differential equation we have 124 7 2141 3 2212 31 7 413i Simplifying the lefthand side we get 4x2781276x4126z 0413 The function 2 212 31 is not a solution of the differential equation I Example 3 Show that uz y cos I sinh y sin I cosh y is a solution of Laplace s equation 8211 8211 w wwi SOLUTION The rst step is to calculate the indicated partial derivatives 7 7 sin I sinh y cos I cosh y g 7 cos I sinh y 7 sin I cosh y 7 cos I cosh y sin I sinh y gig cos I sinh y sin I cosh yr Substituting into the differential equation we nd that 7 cos I sinh y 7 sin I cosh y cos I sinh y sin I cosh y 0 and the equation is satis ed uz y cos I sinh ysin z cosh y is a solution of Laplace s equationi Exercises 1 1 1i Classify the following differential equations With respect to type iiel7 ordinary or partial and order a b V y 2 zyy sin 1 y em tan 1 C 92 7 0 812 Bray 8y2 7 d2 3 cl zy2 If e f g y 751y ye 71 V BuBz MailBy dig dz2 dy d3 ZyE If i e 7 g 721 For each differential equation determine Whether or not the given functions are solutions 2 y 4y 0 sin 31 cos 21 2sin 21 dsy dy x 1 x 1 1 Big E76 yz7ls1nz e7 2z72cosz el 4 my y 0 y1r1n1I7 MI 12 5 z 1y my 7 y z 12 e 12 1 I2 1l d3 d2 d 6 E 7 ST 6 0 61621 02637 cl Cg constants7 262 3631 4 2 2 7 a a 0 11117yln12y27 u2zy 1373zy2l 812 By2 Sly 7y27z yze 172 21sinhz172l 8 82 9i 7 1626712 11117 67th cos I 11217 7th sin 27ml Find the set of all solutions of each of the following differential equations y 21lnz i y 32 i y 6zcos 21 dy i3l dz y dy xy70i Determine values of 7 if possible7 so that the given differential equation has a solution of the form y em yll74y0i iy2y 78y0i iyll76yl9y0i y 7 4y By 7 2y 0i iy 72y 5y0i Determine values of 7 if possible7 so that the given differential equation has a solution of the form y I7 d2y dy 2772 i 2 0 I dz2 zdzJr y i z2yzy 79y0i i 12y 7 31y 4y 0i 33 Homogeneous Equations with Constant Coef cients We emphasized in Sections 31 and 32 that there are no general methods for solving second or higher order linear differential equations However there are some special cases for which solution methods do exist In this and the following sections we consider such a case linear equations with constant coefficients In this section we treat homogeneous equations nonhomogeneous equations will be treated in the next two sections A second order linear homogeneous di erential equation with constant coe icients is an equation which can be written in the form y ay by0 1 where a and b are real numbers You have seen that the function y 5quot is a solution of the rst order linear equation 3 ay 0 the model for exponential growth and decay This suggests the possibility that equation 1 may also have an exponential function y em as a solution If y 5 then 3 re and y r2 e Substitution into 1 gives r2 5m a r e 136 em r2 ar b 0 Since em 7 0 for all x we conclude that y em is a solution of 1 if and only if r2arb0 2 Thus if r is a root of the quadratic equation 2 then y em is a solution of equation 1 we can nd solutions of 1 by nding the roots of the quadratic equation DEFINITION 1 Given the differential equation The corresponding quadratic equa tion r2 1 ar 1 b 0 is called the characteristic equation of 1 the quadratic polynomial r2 1 ar 1 b is called the characteristic polynomial The roots of the characteristic equation are called the characteristic roots The nature of the solutions of the differential equation 1 depends on the nature of the roots of its characteristic equation There are three cases to consider 1 Equation 2 has two distinct real roots r1 4 r2 2 Equation 2 has only one real root r Oz 76 3 Equation 2 has complex conjugate roots7 r1 04 2 3 r2 Oz 7 i B B 7 0 Case I The characteristic equation has two7 distinct real roots7 r1 a r2 B In this case7 are solutions of Since 04 7 yl and y2 are not constant multiples of each other7 the pair y17y2 forms a fundamental set of solutions of equation 1 and y Cl em 025 is the general solution Note We can use the Wronskian to verify the independence of y1 and y2 60m e m a 604m B e m 50m 519m 751 a 50m 0473 ea z 7g 039 I Example 1 Find the general solution of the differential equation yH2y78y 0 SOLUTION The characteristic equation is r2 2r 7 8 0 74772 0 The characteristic roots are 71 74 r2 2 The functions y1z 5 y2z 52m form a fundamental set of solutions of the differential equation and y Cl 574m CZ 521 is the general solution of the equation I Example 2 Find a linearly independent pair of solutions of y 33 0 and give the general solution of the equation SOLUTION The characteristic equation is r2 3r rr 3 O7 and the characteristic roots are 71 O7 r2 73 Therefore the functions y1z 50m E 1 and y2z 5 31 are linearly independent solutions of the differential equation The general solution is y 011 025 01 025 l 77 Case II The characteristic equation has only one real root r 041 Then yl e and y2m m 5 are linearly independent solutions of equation 1 and y 01 e 0260m is the general solution Proof We know that y1z e is one solution of the differential equa tion we need to find another solution which is independent of y1 Since the characteristic equation has only one real root 4 the equation must be T2aTbT7042T27204T0420 and the differential equation 1 must have the form yH7204y042y 0 Now 2 Ce C any constant is also a solution of but 2 is not independent of y1 since it is simply a multiple of y1 We replace C by a function it which is to be determined if possible so that y us is a solution of Calculating the derivatives of y we have y ueu m y Que u 5 H 2 H y auew2auewu em Substitution into gives H azueam2au em u 50m7204 auemu 5M a2uem 0 This reduces to uHem 0 which implies u 0 since 5 7 0 Now u 0 is the simplest second order linear differential equation with constant coefficients the general solution is u 01 ng 01 1 Cg z and u1z 1 and u2z z form a fundamental set of solutions Since y u e we conclude that y1m 1 5 em and 242 me 1In this case a is said to be a double root of the Characteristic equation 2This is an application of a general method called mutation of parameters We will use the method several times in the work that follows are solutions of In particular7 yg me is a solution of which is independent of y1 e That is7 M and yg form a fundamental set of solutions of This can also be checked by using the Wronskian 60m eam aem 5M Imam e 6 ameml 7 axe 52 7 0 Finally7 the general solution of is y 01 em I 02x5 Note The solution y2z me can also be obtained by using Problem 15 in Exercises 32 I Example 3 Find the general solution of the differential equation yH76y9y0 SOLUTION The characteristic equation is r2 7 ST 9 0 r 7 32 0 There is only one characteristic root r1 r2 3 The functions y1z 53m y2z x 53m are linearly independent solutions of the differential equation and y0153m02m53m is the general solution I Case III The characteristic equation has complex conjugate roots Tlai 72047i Bio In this case y1z 5 cos x and y2z 5 sin Bx are linearly independent solutions of equation 1 and y 01 5 cos x I 02 5 sin Bx e 01 cos x I 02 sin 3m is the general solution Proof It is true that the functions 21z 40 erm and 22z awhile are linearly independent solutions of 17 but these are complex valued functions 79 and we are not equipped to handle such functions in this course We want real valued solutions of The characteristic equation in this case is rzarb T7 ai ri Oti 7 272047 04232 0 and the differential equation 1 has the form yH72aya232y039 We ll proceed in a manner similar to Case ll Set y us where u is to be determined if possible so that y is a solution of Calculating the derivatives of y we have y u e y 04 u em u cm H H y 04214 cm 204 u em u em Substitution into gives 04214 e 204150 iiHeD m 7 204 auew ueu m 042 32 lieu 0 This reduces to Mew ozuew 0 which implies u 5 0 since 5 7g 0 Now7 u 32 u 0 is the equation of simple harmonic motion for example7 it models the oscillatory motion of a weight suspended on a spring The functions u1z cos x and u2z sin Bx form a fundamental set of solutions Verify this Since y u e we conclude that y1z 5 cos x and 242 em sin Bx are solutions of It s easy to see that M and yg form a fundamental set of solutions This can also be checked by using the Wronskian Finally7 we conclude that the general solution of equation 1 is y 01 5 cos Bx 025 sin Bx e 01 cos Bx 02 sin Bx I Example 4 Find the general solution of the differential equation y 7 43 13y 0 80 SOLUTION The characteristic equation is r2 7 4r 13 By the quadratic formula7 the roots are T17 T2 774 74L4113 i Alix16752 7 Alix736 7 4i6i 72H 2 2 T 239 2 The characteristic roots are the complex numbers 71 2 32 r2 2 7 32 The functions y1z 52m cos 3m y2z 52m sin 3x are linearly independent solutions of the differential equation and y 01 52 cos 3x Cg 521 sin 3x 52 01 cos 3x Cg sin 3m is the general solution I Example 5 Find two linearly independent solutions of y 16y 0 SOLUTION The characteristic equation is r2 16 0 and the complex numbers 71 0 4i 4i r2 0 7 4i 742 are the characteristic roots The functions 0 0 y1z e 1 cos 4x cos 4m y2z 5 1sin 4x sin 4x are linearly independent solutions of the differential equation I In our next example we nd the solution of an initial value problem Example 6 Find the solution of the initial value problem 24 217152 0 MW 27 MO 6 SOLUTION The characteristic equation is r 27 7 15 0 r 5r 7 3 0 The characteristic roots are 71 757 r2 3 The functions y1z e Sm y2z 53m are linearly independent solutions of the differential equation and y 01 6751 CZ 531 is the general solution Before applying the initial conditions we need to calculate y ym 7501 575m 302 53m Now7 the conditions y0 27 y 0 76 are satisfied if and only if 0102 2 75013CZ 76 81 The solution of this pair of equations is 01 g 02 and the solution of the initial Value problem is ye 5m53m I Recovering a Differential Equation from Solutions You can also work backwards using the results above That is we can determine a second order linear homogeneous differential equation with constant coefficients that has given functions u and v as solutions Here are some examples Example 7 Find a second order linear homogeneous differential equation with constant coefficients that has the functions 52m vz 5 31 as solutions SOLUTION Since 52m is a solution 2 must be a root of the characteristic equation and r 7 2 must be a factor of the characteristic polynomial Similarly 5 31 a solution means that 73 is a root and r 7 73 r 3 is a factor of the characteristic polynomial Thus the characteristic equation must be r 7 2r 3 0 which expands to r2 r 7 6 0 Therefore the differential equation is y y76y0 I Example 8 Find a second order linear homogeneous differential equation with constant coefficients that has y 01 e741 02 z e741 as its general solution SOLUTION Since 5 and ze are solutions 74 must be a double root of the characteristic equation Therefore the characteristic equation is r 7 742 r 42 0 which expands to r2 7 8r 7 16 0 and the differential equation is y 8yf16y 0 I Example 9 Find a second order linear homogeneous differential equation with constant coefficients that has em cos 2x as a solution SOLUTION Since 51 cos 2x is a solution the characteristic equation must have the complex numbers 1 2i and 1 7 2i as roots Although we didn t state it explicitly 51 sin 2x must also be a solution The characteristic equation must be r 7 1 2ir7 17 21 0 which expands to r2 7 2H 5 0 and the differential equation is yH72y5y0 I 82 Exercises 33 Find the general solution of the given differential equation H 34H2378y 0 to 34H7133 4234 0 03 34H7103 2534 0 q 34 235340 U 34H431334 0 6 34 0 T 34 23 0 00 234 53 7 3y 0 to 34H71234 0 10 34 1234 0 11 34 7 23 1 2y 0 12 373334 0 13 3 7 3 7 3034 0 14 234 33 0 15 234 23 34 0 16 34 23 1 3y 0 17 34 7 83 1634 0 18 534 3 7 2y 0 Find the solution of the initial value problem 19 34 7 53 634 0 340 17 30 1 20 34 43 334 0 340 27 30 71 21 34 23 34 0 340 737 30 1 22 34 g 0 347r17 37r 1 23 y 7 22 2y 0 240 17 MO 1 83 2 25 26 27 28 29 30 31 32 33 3 q 3 U 38 F 24 42 4y 0 214 2 y 11 Find a differential equation y Ly 1 by 0 that is satis ed by the given functions 1 M7 24295 575m 5 363m y2m 2x55 5 1 y y y1z cos 2x y2z 2 sin 2m y1z Zm cos 4m y2z 5 21 sin 4m Find a differential equation y ay 1 by 0 Whose general solution is the given expression y Clem2 0252 y 0153m l 025 y 015 m cos 3x 025 m sin 3m y Clem2 026m2 y 01 cos 4x Cg sin 4m Find the solution y of the initial value problem y 7 y 7 2y O y0 a y 0 2 Then nd 04 such that 7 0 as x 7 00 Find the solution y ofthe initial value problem 4y 7y O y0 27 y 0 3 Then nd 3 such that 7 0 as x 7 oo Given the differential equation y 7 2a 7 1y aa 7 1y O a Determine the values of a if any for which all solutions have limit 0 as 1 H 00 b Determine the values of a if any for which all solutions are unbounded as 1 H 00 Exercises 37 39 are concerned with the differential equation 1 y ay 1 by 0 Where a and b are constants Give a condition on a and b which will imply that a 1 has solutions of the form yl 5W yg em a 3 distinct real numbers b 1 has solutions of the form yl e yg me 04 a real number 5 sin 3m 04 B c 1 has solutions of the form yl 5 cos 3m yg real numbers Prove that if a and b are both positive7 then all solutions have limit 0 as x 7 oo 84 39 4 to 43 44 45 46 Prove a If a 0 and b gt 07 then all solutions of the equation are bounded b If a gt 0 and b 07 and y is a solution7 then lim k for some constant k 700 Determine k for the solution that satis es the initial conditions a7y401 y0 Show that the general solution of the differential equation y 7 Lazy 07 w a positive constant7 can be written y 01 cosh um 02 sinh Luz Suppose that the roots r1 r2 of the characteristic equation 2 are real and distinct Then they can be written as T1 13 r2 04 7B where 04 and B are real Show that the general solution of equation 1 in this case can be expressed in the form y e 01 cosh Bx 02 sinh 3m Euler Equations A second order linear homogeneous equation of the form dzy dy 2 7 7 0 E mdm2adx y where 04 and B are constants7 is called an Euler equation Prove that the Euler equation can be transformed into the second order equation with constant coefficients 2 d y dy 7 7 b 0 122 adz y Where a and b are constants7 by means of the change of independent variable 2 ln m Find the general solution of the Euler equations xzy 7 my 7 8y 0 xzy 7 me 2y 0 xzy 7 3zy 4y 0 xzy 7 my 5y 0 62 Systems of Linear Differential Equations Introduction Up to this point the entries in a vector or matrix have been real numbers In this section and in the following sections we will be dealing with vectors and matrices whose entries are functions A vector whose components are functions is called a vectorvalued function or vector functionl Similarly a matrix whose entries are functions is called a matrix function The operations of vector and matrix addition multiplication by a number and matrix multipli cation for vector and matrix functions are exactly as de ned in Chapter 5 so there is nothing new in terms of arithmetic However there are operations on functions other than arithmetic operations elgl limits differentiation and integration that we have to de ne for vector and matrix functions These operation from calculus are de ned in a natural way Let Vt f1 t f2 t l l l fn be a vector function whose components are de ned on an interval l Limit Let c E It If lim fit 1 exists for i 12 l l in then 17 11mm lt11mf1t limf2t limfntgt a1 a2 an t7gtc t7gtc t7gtc t7gtc Limits of vector functions are calculated componentwiser Derivative lf f1 f2 Hi fn are differentiable on I then V is differentiable on I and V UN f2 t7 film That is V is the vector function whose components are the derivatives of the compo nents of V Integral Since differentiation of vector functions is done componentwise integration must also be componentwiser That is vtdtltf1tdtf2tdt fntdtgtl Limits differentiation and integration of matrix functions are done in exactly the same way 7 componentwise Systems of Linear differential Equations Consider the thirdorder linear differential equation y WM WW WM 7 f0 where p q 7 f are continuous functions on some interval It Solving the equation for y we get u 7Tty 7 409 7 WM W 237 Introduce new dependent variables 11 12 13 as follows 11 y 12 1 1 y rs 1 2 y Then H y l 13 Tt11 4t12 PWIS f and the thirdorder equation can be written equivalently as the system of three rstorder equations 11 12 2 Is 13 TtI1 4t12 Pt13ft Note This system is just a very special case of the general system of three7 rstorder differential equations 11 a11tzl a12tzg a13tzgt b1t 12 a21tzl a22tzg a23tzgt 122t 13 a31t11 a32t12 a33tIst 113t Example 1 a Consider the thirdorder nonhomgeneous equation y 7 y 7 8y 12y 26 Solving the equation for y we have ym 712y 8y y 2amp3 Let 11 y 1 1 12 y 7 1 2 zg y Then m y 1g 71211812132 t and the equation converts to the equivalent system 11 12 2 13 13 71211812zg26t b Consider the secondorder homogeneous equation t2yH 7 ty 7 3y 0 Solving this equation for y we get 3 1 H77 7 y 7t2yty To convert this equation to an equivalent system7 we let 11 y 11 12 y Then we have 1 I2 1 7 31 11 2 t2 1 t 2 238 Which is just a special case of the general system of two rstorder differential equations 11 a11tzl a12tzg b1t 12 a21tzl a22tzg 122 t General Theory Let a11t7 a12t7 m a1nt7 a21t7 m annt7 121t7 122t7 m bnt be continuous functions on some interval If The system of n rstorder differential equations 11 a110311 a12tI2 a1nt1nt 5175 12 a210311 a22tI2 a2nt1nt 5275 S N mm m4mm m is called a rstorder linear di ereritial systeml The system S is homogeneous if b1t E b2t E E bnt E 0 on If S is nonhomogeneous if the functions bit are not all identically zero on I that is7 if there is at least one point a E I and at least one function bit such that 12 a f 0 Let At be the n gtltn matrix a11t a12t a1nt Ag a21t a22t a2nt an1t an t am t and let x and b be the vectors 11 b1 12 122 X 7 b i M M Then S can be Written in the vectormatrix form A xh S The matrix At is called the matrix of coe cients or the coe cient matrix Example 2 The vectormatrix form of the system in Example 1 a is 0 l 0 0 x 0 0 l X 0 712 8 l 2et a nonhomogeneous systeml The vectormatrix form of the system in Example 1b is x 7 0 1 11 0 7 0 1 11 7 32 1 2 0 7 32 1 2 a homogeneous systeml A solution of the linear differential system S is a differentiable vector function 11t 12 Ina that satis es S on the interval 11 3 Example 3 Verify that V lt 372 gt is a solution of the homogeneous system 7 0 1 11 X7 32 1 12 of Example 1 SOLUTION VL 3 7 32 l 0 1 3 7 32 7 6 7 321 32 32 6 V is a solution 62 1 e 2 Example 4 Verify that V 262 e is a solution of the nonhomogeneous system 462 l e 2 0 1 0 0 X 0 0 1 X 0 712 8 1 26 of Example 1 a 240 SOL UTION rt gtgtm mm mm e mmm e o rt to m rt 862t to m 0 ct mmm as 862t 6 E v 2amp2t e 4 2t e 262 e 462 e 862t e 0 1 0 e2 get 0 i 0 0 1 262 get 0 712 8 l 462 e 26 0 1 0 e2t 0 1 0 e 0 0012 2t 001 et 0 712 8 1 Mt 712 8 1 get 2a 1 2 1 2 g 2 g m 0 ct Nlb Nlb NlH V is a solution THEOREM 1 Existence and Uniqueness Theorem Let a be any point on the interval 1 and let a1 a2 11 1 an be any n real numbers Then the initialvalue problem a1 x Atx 131 xa W has a unique solution Exercises 62 Convert the differential equation into a system of rstorder equations 11 y 7 tyl By sin 2t 21 y y 2e 2t1 3y 7 y y eti 41 my 6y ky cos At m7 5 k 1 are constants ln Exercises 5 8 a matrix A and a vector b are given Write the system of equations corresponding to X AtX b 241 51Atlt C gibwa t3 6 At on 2 71 2 et TAO 72 0 1 bt 2w 1 2 3 0 eZt 0 0 1 5 blttgt7 lt 2 3 t 37 71 81At 72 72 t bt 2t 3 7 Write the system in vectormatrix formi 91 11 7211 12 sin t 12 1173127200st 101 11 girl 7 62t12 12 e tzl 7 36 111 11 211123133 2t 12 1173127200st 13 2117I24zgt 121 11 t211zg7tzg3 12 7362 213 7 2672i 13 211 7212 413 r1 131 Verlfy that u 772 gt 1s a solution of the system 1n Example 1 1 67 3t 5 5t 14 Verify that u 3 1 et is a solution of the system in Example 1 7367 t 967 2 16 a 1 762 151 Verify that W e2 2te2 is a solution of the homogeneous system associated With 462 4te2t the system in Example 1 a 242 7 39 t 16 Verlfy that V Sn 1s a solutlon of the system 7 cos t 7 2 sm t x 7 2 1 x 0 7 73 2 2sint 726 2 17 Verify that V 0 is a solution of the system 3672 1 73 2 x 0 71 0 x 0 71 72 243 36 Vibrating Mechanical Systems Undamped Vibrations A spring of length lo units is suspended from a support When an object of mass m is attached to the spring the spring stretches to a length 1 units If the object is then pulled down or pushed up an additional yo units at time t 0 and then released what is the resulting motion of the object That is what is the position yt of the object at time t gt 0 Assume that time is measured in seconds We begin by analyzing the forces acting on the object at time t gt 0 First there is the weight of the object gravity F1 mg This is a downward force We choose our coordinate system so that the positive direction is down Next there is the restoring force of the spring By Hooke s Law this force is proportional to the total displacement 1 yt and acts in the direction opposite to the displacement F2 7km with k gt 0 The constant of proportionality k is called the spn39ng constant If we assume that the spring is frictionless and that there is no resistance due to the surrounding medium for example air resistance then these are the only forces acting on the object Under these conditions the total force is F F1 F2 mg klll 2405 mg kll 905 Before the object was displaced the system was in equilibrium so the force of gravity mg plus the force of the spring ikll must have been 0 mg 7 kll 0 Therefore the total force F reduces to F 7kyt By Newton s Second Law of Motion F ma force mass x acceleration we have k 7k t d if t ma an a my Therefore at any time t we have k k a y t RW 0r Mt t 0 9 m When the acceleration is a constant negative multiple of the displacement the object is said to be in simple harmonic motion 106 Since km gt 07 we can set w wkm and write this equation as 2 y t w yt 07 1 a second order linear homogeneous equation with constant coefficients The characteristic equation is 72 w2 0 and the characteristic roots are iwi The general solution of 1 is y 01 cos wt Cg sin wt ln Exercises 36 Problem 5 you are asked to show that the general solution can be written as y Asinwt 07 2 where A and 0 are constants with A gt 0 and 0 6 027139 For our purposes here7 this is the preferred form The motion is periodic with period T given by 727139 T 7 LA a complete oscillation takes 27rw seconds The reciprocal of the period gives the number of oscillations per second This is called the frequency denoted by f 1 727139 f Since sin wt 0 oscillates between 71 and 17 yt Asin wt 0 oscillates between 7A and A The number A is called the amplitude of the motion The number 0 is called the phase constant or the phase shift The gure gives a typical graph of Figure 1 107 Example 1 Find an equation for the oscillatory motion of an object given that the period is 27r3 and at time t 0 y 1 y 3 SOLUTION In general the period is 27rw so that here 2 2 l l and therefore an 3 w 3 The equation of motion takes the form yt Asin 3w 0 Differentiating the equation of motion gives y t 3A cos3t 0 Applying the initial conditions we have y0 1 Asin 0 y 0 3 3A cos 0 and therefore Asin 01 Acos 01 Adding the squares of these equations we have 2 A2 sin2 0 A2 cos2 0 A2 Since A gt O A Finally to nd 0 note that 2sin 01 and 2cos 01 These equations imply that 0 7r4 Thus the equation of motion is W sin3t in l Damped Vibrations If the spring is not frictionless or if there the surrounding medium resists the motion of the object for example air resistance then the resistance tends to dampen the oscillations Experiments show that such a resistant force R is approximately proportional to the velocity v y and acts in a direction opposite to the motion R icy with c gt 0 Taking this force into account the force equation reads F 7km 7 cm 108 Newton s Second Law F ma my then gives my t 7W0 7 Clot which can be written as H C k 7 y 7y 7y 7 0 37 k m all constant 3 m m This is the equation of motion in the presence of a damping factor The characteristic equation k 72 7 7 0 m m has roots i 7ci V62 7 4km 7 2m 39 There are three cases to consider 0274kmlt0 0274kmgt0 0274km0 Case 1 02 7 4km lt 0 In this case the characteristic equation has complex roots 0 V 4km 7 02 T1 2m C 2 zw7 r27277zw Wherew m m The general solution is y eke27quot 01 cos wt Cg sin wt which can also be written as yt Amie27quot sin wt 0 4 Where7 as before7 A and 0 are constants7 A gt 07 0 6 027139 This is called the underdamped case The motion is similar to simple harmonic motion except that the damping factor e c2mt causes yt 7 0 as t 7 00 The oscillations continue inde nitely with constant frequency f w27r but diminishing amplitude 144727quot This motion is illustrated in Figure 2 I Figure 2 109 Case 2 02 7 4km gt 0 In this case the characteristic equation has two distinct 7c V 02 7 4km 7c 7 V62 7 4km 7 T2 39 2m 2m real roots T1 The general solution is yt y Clem Cgent 5 This is called the ouerdamped case The motion is nonoscillatory Since m lt xF c 71 and r2 are both negative and yt 7 0 as t7 00 I Case 3 02 7 4km 0 In this case the characteristic equation has only one real root 3 T1 7 2m7 and the general solution is 240 y 016702mt 02te7c2mt39 This is called the critically damped case Once again7 the motion is nonoscillatory and yt70 as 25700 I In both the overdamped and critically damped cases7 the object moves back to the equilibrium position 7 0 as t 7 00 The object may move through the equilibrium position once7 but only once Two typical examples of the motion are shown in Figure 3 t Figure 3 Forced Vibrations The vibrations that we have considered thus far result from the interplay of three forces gravity7 the restoring force of the spring7 and the retarding force of friction or the surround ing medium Such vibrations are called free vibrations 110 The application of an external force to a freely vibrating system modi es the vibrations and produces what are called forced uibrations As an example we ll investigate the effect of a periodic external force F0 cos 39yt where F0 and 39y are positive constants In an undamped system the force equation is F 7km F0 cos 39yt and the equation of motion takes the form k F0 H i 7cos t y mu m 7 We set w km and write the equation of motion as F y Lazy 70 cos 39yt 7 in As we ll see the nature of the motion depends on the relation between the applied frequency 39y27r and the natural frequency of the system w27r Case 1 39y 7 Ad In this case the method of undetermined coefficients gives the particular solution F 2t cos 39yt Y and the general equation of motion is i Fom y As1nwt 0 W cos 39yt 8 If wy is rational the vibrations are periodic lf wy is not rational then the vibrations are not periodic and can be highly irregular In either case the vibrations are bounded by Fom 1141 L724 Case 2 39y Ad In this case the method of undetermined coefficients gives F0 1 t it t z mum sin in and the general solution has the form F yAsinwt 0 tsin wt 9 The system is said to be in resonance The motion is oscillatory but because of the t factor in the second term it is not periodic As t a co the amplitude of the vibrations increases without bound 111 A typical illustration of the motion is given in Figure 4 I Figure 4 Exercises 36 H to 03 F U a T 00 to An object is in simple harmonic motion Find an equation for the motion given that the period is in39 and7 at time t 07 y 17 y 0 What is the amplitude What is the frequency An object is in simple harmonic motion Find an equation for the motion given that the frequency is 17r and7 at time t 07 y 07 y 72 What is the amplitude What is the period An object is in simple harmonic motion with period T and amplitude A What is the velocity at the equilibrium point y 0 An object in simple harmonic motion passes through the equilibrium point y 0 at time t 0 and every three seconds thereafter Find the equation of motion given that y0 5 Show that simple harmonic motion yt 01 cos wt 02 sin wt can be written as a yt Asinwt 0 b yt Acoswt 110 What is the effect of an increase in the resistance constant c on the amplitude and frequency of the vibrations given by 4 Show that the motion given by 5 can pass through the equilibrium point at most once How many times can the motion change directions Show that the motion given by 6 can pass through the equilibrium point at most once How many times can the motion change directions Show that if 39y 7 w then the method of undetermined coefficients applied to 7 gives Fom z m cos ME 112 10 Show that if w7 is rational then the Vibrations given by 8 are periodic 11 Show that if y M then the method of undetermined coe icients applied to 7 gives F0 2wm z t sin wt 113 43 Inverse Laplace Transforms and InitialValue Problems In Section 42 we saw that the Laplace transform of the solution y ofthe initial value problem v ay by WE y0 a7 MO is given by Mimi Yltsgt where Fs fz is the Laplace transform of 1 Now that we know yz the obvious question is What is The general problem of nding a function with a given Laplace transform is called the inversion problem The inversion problem and its application to solving initial value problems is the topic of this section If f is continuous on 0 00 and if the Laplace transform fm Fs exists for s gt A then the function F is uniquely determined by 1 that is the operator E is itself a function Our rst result states that E is a oneto one function A proof of this result is beyond the scope of this introductory treatment THEOREM 1 If f and g are continuous functions on 0 00 and if fm gz then f E 9 that is x gz for all z E 0 00 The following de nition gives the terminology and notation used in treating the inversion problem DEFINITION If Fs is a given transform and if the function 1 continuous on 0 00 has the property that fz Fs then f is called the inverse Laplace transform of Fs and is denoted by 1 00 71F8 The operator E 1 is called the inverse operator of E There is a general formula for the inverse operator E l corresponding to 1 Section 41 but use of the formula requires a knowledge of complex valued functions a topic which is treated in more advanced courses The relationship between E and E 1 is given by the following equations r1 Ei z M r E 1Fs Fs for all functions 1 continuous on 0 00 such that fz For convenience here we reproduce the table of Laplace transforms given at the end of Section 41 129 Table of Laplace Transforms 1W F8 Elfl 1 1 7 sgt0 s 1 50m sgta 3704 s 0 cos Bx 8232 sgt i B 0 sin Bx 8232 sgt M 3704 e cos z Siagt2 z sgta emsin Bx m sgta l m 71172 sgt0 711 me 71172 W sgtr 82732 x cos Bx sgt0 82322 l 233 msin z sgt0 82522 A simple way to interpret Theorem 1 is that the table can be read either from left to right or from right to left That is7 the table is simultaneously a table of Laplace transforms and of inverse Laplace transforms Example 1 a If fz Fs 4 then x 5 b If mm Fs 829 then m cos3z 7 7 32 7 32 7 3772 C H mm FS 324s13 3229 377229 5 21 cos 3m then x The properties of the Laplace transform operator E can be used to derive corresponding properties of its inverse operator 71 For our purposes7 the most important property is that of linearity THEOREM 2 The operator E 1 is linear that is 1Fs 13 E 1Fs 1Gs17 and 130 71CFS c 1Fs c any constant The proof is left as an exercise Example 2 Find E l if F 777 8 32 324 SOLUTION Since E 1 is a linear operator LL 33 324 1 2 7K1 7 73 1 7 33 94 Now reading the table from right to left we see that 1Fs 1 71 7 731 71 7 A 83 75 and E 824 78111 2m Therefore K1 i 7 i 753 73sin 2m l s 3 32 4 39 The translation property of E Theorem 5 Section 42 is also useful in nding inverse transforms The analog of Theorem 5 for inverse transforms is THEOREM 3 If f is continuous on 0 00 and if fm Fs exists for s gt A then for any real number r 71Fs 7 7 em x The following examples illustrate the kinds of manipulations that typically occur in calculating inverse Laplace transforms The basic strategy is to try to re Write a given expression Fs as sum of terms which appear in the table Example 3 Find 1Fs if SOLUTION From the table To put in a form in the table we complete the square in the denominator 32 7 23 10 and adjust the numerator 1 1 1 1 3 3272310327231937129g3712939 From the table 1 1 3 71 7 71 i z A E 8272810 3E 87129 35 s1n3m Putting the two results together we have 4 1 3272371710 4x531 1 em sin 3m I Example 4 Find 1F3 if 23 1 F S 32 7 23 7 8 SOLUTION By factoring the denominator we can write 231 231 F 8 3272378 32374 Now by partial fraction decomposition A 3 1i 32374 374 32 Therefore 231 3 7 1 1 7 1 1 7 1 7E1 4z 172 I 3272374 2 374l2 32 25 25 Example 5 Find 1F3 if 2 4 F3 8 3 7 232 7 43 8 SOL UTION The quadratic factor in the denominator cannot be factored into linear factors By partial fraction decomposition 234 2 7237176 F 7 S 372327438 372327438 Next we complete the square in the denominator of the second term 2 7236 7 2 7236 7 2 7236 372 3274371787372 32743447372 372217439 132 Finally7 we adjust the numerator of the second term so that we can use the Table 2 72s6 7 2 7237227 2 72372 2 372 37224isi2 37224 7372 37224 3722439 Now 7 2s4 7 1 372 2 1 1 2 72 s723274s8l 72 s7224s7224 1 372 2 7 71 7 7 71 71 7 2E 372 2E 37224l 3722 t4l 2 52 7 2 522 cos 2m 7 522 sin 2m l Solution of InitialValue Problems Here we complete the application of Laplace transforms to the solution of initial value problems For our rst example7 we finish Example 27 Section 42 Example 6 Find the solution of the initial value problem 1 7 2y 253 y0 72 SOLUTION From the Example7 if y is the solution7 then Ellz 22M mew mam s i 3 Ell90 7 2 ym S i 3 sryzl 7 y0 7 Mam 8 i 3 8 DEB1 2 7 S i 3 s e 2gt iyltzgti 7 S j 3 7 2 Therefore7 lyz Ys 7 7 2 372s3 37239 Now7 by partial fraction decomposition 2 25 E 372s3 372 33 Therefore7 7 25 25 2 7785 25 YltSgtisi2is3isi2i 372is3 and 8 5 2 5 E71 7 7 J 7 62m 573 I 372 33 133 Next we nish Example 3 of Section 42 Example 7 Find the solution of the initial value problem 24 4y 36 y0 57 MW 2 SOLUTION From the Example if y is the solution then EllKm 4yz men men S i 2 EllMl 4Llyltmgtl S f 2 52 l9l 3907y04 yz 72 lt324gt iyltzgtiesselte2gt Sf 32 4 ym S f 2 53 2 Therefore 3 5S 7 2 372324 32439 Now by partial fraction decomposition 3 3 3 3 8 85Z 372324 372 32439 Therefore W8 3 7 s 5372 1g s in 2 372 324 324 8372 8324 8324 and 3 1 37 s 11 2 E lY pl ifl fifl W l 8 8 372l 8 94 8 94 gezmcos2zisin2z I Example 8 Find the solution of the initial value problem 24 7 52 6y z 71 y0 07 y 0 1 134 SOLUTION If y is the solution then We 7 5m 6yltzgti Elm 71 7 1247 an 2 7 S EllMl 7 sayen mm 1 ew 7 sylt0gt 7 Mo 7 5 sagen 7 240 6mm 7 1833 32 7 55 6 ym 71 18 23 32 7 55 6 ym 18723 1 Therefore YS 17s 1 17s 1 3262753176 32753176 32372s73 s72s7339 Now by partial fraction decomposition W h l hilsbl l and 1 71 1 i s72s73 372 373 Therefore 1 1 1 1 3 1 7 1 y if 7 7 7 S 36 9169 4372gt9lt373gt and 1 1 3 7 ym75m7452m 53m I In many applications of differential equations it is not required to determine the solutions explicitly lnstead What is needed is information about the solutions Often such information can be obtained by analyzing their Laplace transforms The next example illustrates this type of application Example 9 Consider the differential equation y 7 y 7 6y 25 on 0 00 together with the condition y0 71 From Chapter 3 we know that the general solution of the differential equation has the form 016 2m 025 Ae m Where Cl 02 are arbitrary constants and A is a constant which can be determined 135 The question we want to examine is Can we choose a value for 04 y 0 so that solution of the resulting initial valueproblem y i 2 i 6y 26 240 17 MO a has limit 0 as x a 00 Since 5 2m a 0 and 54 a 0 as x a 00 we want to choose 04 so that the coefficient of the 53m term is 0 If y is the solution of the initial value problem then Ell106 we 7 6yltzgti aw 1 new 7 Ell06 7 mm 8 i1 MM 7 sylt0gt 7 Mo 7 s lyMl 7 240 7 new 8 i1 3236 yz3a1 Sil 3236 yz Si11a3 Therefore swam Yltsgt 2 1 6 3 313236 3236 2 1 04 3 313233 323339 Now by partial fraction decomposition ii B C 717 7 313233T31 32 33 31 32 33 1043 7 D E 770 3233T32 33T32 33 Combining these results we have 04 1 1 204 3 1 1 1 Y 5 32 10 lt3BgtT2lt31gt39 Clearly if 204 3 0 that is if 04 32 then the 53w term in is eliminated The resulting solution is e 2m e This solution has initial values y0 1 y 0 g and lim O I mace 136 Exercises 43 Fmd 4we FS 87 mg 3 FS 3225 FS7 sill Sk m39 mg i Fs 7713 Mg g ig3 Use partial fraction decomposition to nd the inverse Laplace transform mgg3j mgg 3 mggg5 mgii mg ggi F il mg 4 323 71si 2 Find the solution of the initial value problem 137 to DJ q y 74y0 y072 y 2yem y01 y 73y52m y02 y ysinm y01 y 4y 0 y0 27 MW 2 y 2y y4e m y027 MO 1 y 3y 2y6em y027 y 071 y 7 23 5y 357 y017 y 0 1 Given the initial value problem y 7 y 7 6y 257m y0 oz yO 71 See Example 9 What value should be assigned to 04 so that the resulting solution will have limit 0 as x 7 oo What initial conditions should be assigned with the differential equation y y 5712 so that lim 0 Where y is the solution 700 Consider the differential equation y y 7 2y 3sin 2x together with the initial value y0 2 For What values of B y 0 will the resulting solutions be bounded Consider the differential equation y 33 2y z together with the initial value y 0 72 For What values of Oz y0 will the resulting solutions be bounded The Laplace transform method applies to initial value problems in which the initial values are speci ed at z O Actually7 the method can be applied when the initial 138 x dt dt idiidi idi E dm2 7 dm dm idt dm dzidt dt idtz39 35 Find the solution of the initial value problem y 7 33 1 2y m y1 O7 y 1 1 by rst solving the transformed problem 36 Use the Laplace transform method to solve the initial value problem y72y1x y271 139 55 Matrices and Vectors In the preceding section we introduced the concept of a matrix and we used matrices augmented matrices as a shorthand notation for systems of linear equations In this and the following sections we will develop this concept furtheri Recall that a matrix is a rectangular array of objects arranged in rows and columns The objects are called the entries of the matrix A matrix with m rows and n columns is called an m X n matrix The expression m X n is called the size of the matrix and the numbers m and n are its dimensions A matrix in which the number of rows equals the number of columns m n is called a square matrix of order n Until further notice we will be considering matrices whose entries are real numbers We will use capital letters to denote matrices and lower case letters to denote its entries If A is an m X n matrix then aij denotes the element in the ith row and j h column of A an a12 dis aln a21 a22 a23 a2n A a31 a32 ass 39 39 39 an aml am2 ams 39 39 39 amn We will also use the notation A aij to represent this display There are two important special cases that we need to de ne A 1 X n matrix a1 a2 Hi an also written as a1 a2 Hi an since there is only one row we use only one subscript is called an n dimensional row vector An m X 1 matrix is called an m dimensional column vector The entries of a row or column vector are often called the components of the vector On occasion we7ll regard an m X n matrix as a set of m row vectors or as a set of n column vectors To distinguish vectors from matrices in general well use lower case boldface letters to denote vectors Thus we7ll write ua1a2iuan and v Arithmetic of Matrices Before we can de ne arithmetic operations for matrices addition subtraction etc we have to de ne what it means for two matrices to be equali 187 DEFINITION Equality for Matrices Let A aij be an m X n matrix and let B bij be a p X 4 matrix Then A B if and only if 1 m p and n L that is A and B must have exactly the same size and 2 aij bij for all i and j In short A B if and only if A and B are identical Example 1 a b 3 7 71 I ifandonlyif a7b7lc4z3i I 2 c 0 2 4 0 Matrix Addition Let A aij and B blj be matrices of the same size say m X n Then A B is the m X n matrix C cij where cij aij by for all i and j That is you add two matrices of the same size simply by adding their corresponding entries A B aij Iij Addition of matrices is not de ned for matrices of di ererit sizes a b I y a z b y E l 2 xampe a lt0 dgtltz w 62 dw 2 4 73 74 0 6 72 4 3 b lt 2 5 0 gt lt 71 2 0 gt lt 1 7 0 gt 1 3 2 4 73 c 5 73 is not de ned I 2 5 0 0 6 Since we add two matrices simply by adding their corresponding entries it follows from the properties of the real numbers that matrix addition is commutative and associative That is if A B and C are matrices of the same size then A B B A Commutative A B C A B C Associative A matrix with all entries equal to 0 is called a zero matrix For example 0 0 0 0 0 0 0 0 and 0 0 0 0 are zero matricesi Often the symbol 0 will be used to denote the zero matrix of arbitrary size The zero matrix acts like the number zero in the sense that if A is an m X n matrix and 0 is the m X n zero matrix then A00AAi 188 A zero matrix is an additive identity The negative of a matrix A denoted by 7A is the matrix Whose entries are the negatives of the entries of Al For example if a b c 7a 7b 7c A th 7 A def en lt7d 7e 7f Subtraction Let A aij and B bij be matrices of the same size say m X n Then A 7 B A 7B To put this simply A 7 B is the m X n matrix C clj Where cij aij 7 by for all i and j you subtract two matrices of the same size by subtracting their corresponding entries A 7 B aij 7 Iij Note that A7 A 0i Subtraction of matrices is not de ned for matrices of di erent sizesi 2473 740676479 I 250 71207330 Multiplication of a Matrix by a Number Example 3 The product of a number k and a matrix A denoted by kA is the matrix formed by multiplying each element of A by the number kl That is kA kaij This product is also called multiplication by a scalar Example 4 2 71 4 76 3 712 73 l 5 72 73 715 6 I 4 0 3 712 0 79 Matrix Multiplication We now want to de ne matrix multiplication While addition and subtraction of matrices is a natural extension of addition and subtraction of real numbers matrix multiplication is a much different sort of product We7ll start by de ning the product of an ndimensional row vector and an ndimensional column vector in that speci c order 7 the row on the left and the column on the right The Product of a Row Vector and a Column Vector The product of a l X n row vector and an n X 1 column vector is the number given by a1 a2 a3 Hi an b3 a1b1 agbg agbg anbni 189 This product has several names including scalar product because the result is a number scalar dot product and inner product It is important to understand and remember that the product of a row vector and a column vector of the same dimension and in that orderl is a number The product of a row vector and a column vector of dz erent dimensions is not de ned Example 5 3 72 5 74 371727451 10 1 72 3 71 4 f3 722347173475 79 75 Matrix Multiplication If A aij is an m X p matrix and B blj is a p X 71 matrix then the matrix product of A and B in that order denoted AB is the m X n matrix C cij Where cij is the product of the ith row of A and the jth column of B an a12 quot39 a1p 511 quot39 blj quot39 bln 521 quot39 52139 quot39 527 an an quot39 aip C Cz39j bpl quot39 bpj quot39 bzm am am amp Where cij ailblj aigbgj aipbm l NOTE Let A and B be given matrices The product AB in that order is de ned if and only if the number of columns of A equals the number of rows of B If the product AB is de ned then the size of the product is no of rows of Agtltnol of columns of B That is A BC mprXn mgtltn 3 0 Alt142gt and B7l 2l 315 K172 Since A is 2 X 3 and B is 3 X 2 we can calculate the product AB Which Will be a 2 X 2 matrix 3 0 AB 7 1 4 2 71 2 1347121 1042272 31 5 1 33171513012572 33 Example 6 Let 190 We can also calculate the product BA since B is 3 X 2 and A is 2 X 31 This product will be 3 X 31 3 0 3 12 6 BA 71 2 1 4 2 5 72 6 i I 3 1 5 Example 6 illustrates a signi cant fact about matrix multiplication Matrix multiplication is not commutative AB BA in general While matrix addition and subtraction mimic addition and subtraction of real numbers matrix multiplication is distinctly different Example 7 a 1n Example 6 we calculated both AB and BA but they were not equal because the products were of different size Consider the matrices 12 7103 Clt34gt and Dlt5 72gt 147 1 You should verify this 17 28 17 Here the product CD exists and equals lt On the other hand the product DC does not exist you cannot multiply D C 1 2X3 2X2 E 4 71 and 9 75 i 0 2 3 0 In this case EF and FE both exist and have the same size 2 X 21 But EF lt 33 730 gt and FE lt 36 719 gt verify b Consider the matrices 6 12 73 so EF7 FEi I While matrix 39 quot quot is not 39 it is 39 quot Let A be an m X p matrix B a pgtlt 4 matrix and C a 4X n matrixi Then ABC ABC1 By indicating the dimensions of the products we can see that the left and right sides do have the same size namely m X n A BC and AB C 1 quotIX pgtltn mgtltq 11X A straightforward but tedious manipulation of double subscripts and sums shows that the left and right sides are in fact equali There is another signi cant departure from the real number systemi If a and b are real numbers and ab 0 then either a 0 b 0 or they are both 01 Example 8 Consider the matrices A071andB351 02 00 191 Neither A nor B is the zero matrix yet as you can verify 1900 00 07 WW As we saw above zero matrices act like the number zero with respect to matrix addition Also A00AAi Are there matrices which act like the number 1 with respect to multiplication a 1 1 a a for any real number a Since matrix multiplication is not commutative we cannot expect to nd a matrix I such that AI IA A for an arbitrary m X n matrix Al However there are matrices which do act like the number 1 with respect to multiplication Example 9 Consider the 2 X 3 matrix a b c A d e 1 Then OHO O 1 Let I2lt1j and I3 0 0 10 abc abc I2Alt01gtltd e e b 100 b A13lta C 010 a a I def 001 def Identity Matrices Let A be a square matrix of order n The entries from the upper left corner and of A to the lower right corner that is the entries an agg agg i i ann form the main diagonal of A For each positive integer n gt 1 let In denote the square matrix of order n whose entries on the main diagonal are all 1 and all other entries are 0 The matrix In is called the n X n identity matrix In particular 10 10 0 3333 I2 I3 01 0 I4 andsooni 01 0 01 0 010 0 0 01 If A is an m X n matrix then ImA A and ALL AA 192 In particular if A is a square matrix of order n then ALL InA A so In does act like the number 1 for the set of square matrices of order n Matrix addition multiplication of a matrix by a number and matrix multiplication are connected by the following properties In each case assume that the sums and products are de ned ll AB C AB AC This is called the left distributive law 2 A BC AC BC This is called the right distributive law 3 MAB kAB AkB Another way to look at systems of linear equations Now that we have de ned matrix multiplication we have another way to represent the system of m linear equations in n unknowns 1111 1212 1313 ainrn 7 1 1 2111 2212 2313 a2n1n 1 2 3111 3212 3313 asnrn 1 3 amiri am212 amsrs amnrn bm We can write this system as an L112 an am 11 b1 L121 L122 L123 a2n I2 b2 L131 L132 ass asn 13 b3 am am am am In bm or more formally in the vectormatrix form Ax b l where A is the matrix of coef cients x is the column vector of unknowns and b is the column vector whose components are the numbers on the righthand sides of the equations This looks like77 the linear equation in one unknown ax b where a and b are real numbers By writing the system in this form we are tempted to divide77 both sides of equation 1 by A provided A is not zero77 Well take these ideas up in the next section The set of vectors with n components either row vectors or column vectors is said to be a vector space of dimension n and it is denoted by Rnl Thus R2 is the vector space of vectors 193 with two components R3 is the vector space of vectors with three components and so on The zy plane can be used to give a geometric representation of R2 3dimensional space gives a geometric representation of R31 Suppose that A is an m X n matrixi If u is an n component vector an element in R then Au v is an m component vector an element of Rm Thus the matrix A can be viewed as a transfor mation a function a mapping that takes an element in R to an element in Rm A maps R into Rmi Example 10 The 2 X 3 matrix A lt 1 7 3 gt maps R3 into R21 For example lt32gt ltgtv lt32gtEltgt lt11 m g lta2gtv If we view the m X n matrix A as a transformation from R to Rm and in general then the question of solving the equation Ax b can be stated as Find the set of vectors x in R which are transformed to the given vector b in Rmi As a nal remark an m X n matrix A regarded as a transformation is a linear transformation since Ax y AX Ay and Akx kAxi These two properties follow from the connections between matrix addition multiplication by a number and matrix multiplication given above Exercises 55 2 4 72 0 1 72 7 72 1iLetA 71 3 B 42 Clt72 4gtDlt3 0gti 4 72 73 1 Compute the following if possible a AB b 72B c CD d AD e 2AB f A73 71 0 72 3 76 5 2iLetA 2 B 3 72 0 Clt72 4 73gtD 3 0 4 2 3 1 4 4 Compute the following if possible 194 F 5 533 7 9 a AB b BA c CB d CA e DA f DB g AC h 32 BB 0 71 71 0 2 0 5 0 LetA 0 2 B 372 Clt 4 3gtDlt3 3 4 72 2 5 7 7 Compute the following if possible a 3A 7 2BC b AB c BA d CD 7 2D e AC 7 BD f AD 2DC Let A be a 3 X 5 matrix B a 5 X 2 matrix C a 3 X 4 matrix D a 4 X 2 matrix and E a 4 X 5 matrix Determine which of the following matrix expressions exist and give the sizes of the resulting matrix when they do exist a AB b EB c AC d AB CD e 2EB 7 3D f CD 7 CEB g EB DA 0 73 5 71 72 3 A 2 4 71 and B 5 4 73 iLet CAB and DBA1 1 0 2 0 2 5 Determine the following elements of C and D without calculating the entire matrices a 632 b 013 C d21 d d2 2 3 0 1 3 A 4 75 and B 1 4 3 iLet CAB and DBA1 71 3 Determine the following elements of C and D without calculating the entire matrices a 021 b 033 c dlg d du LetA 1 3 B 12 3 and C 2 3 5 iPutDAB201 2 4 5 0 73 7 1 0 Determine the following elements of D without calculating the entire matricesi a dgg b dlg c d23 1 73 1 2 73 2 0 72 Let A 72 4 0 B 1 73 and C 4 5 71 Put 3 71 74 71 3 2 1 0 72 D 2B 7 301 Determine the following elements of D without calculating the entire matricesi a dll b d23 c d3 Let A be a matrix whose second row is all zeros Let B be a matrix such that the product exists AB exists Prove that the second row of AB is all zeros 195 H 53 H H H to H 9 H F H 5 iLet Alt iLet Alt Let B be a matrix Whose third column is all zeros Let A be a matrix such that the product exists AB exists Prove that the third column of AB is all zeros gtvBlt gtvolt W gtA Calculate if possible b AC and CA c AD and DA a AB and BA This illustrates all the possibilities When order is reversed in matrix multiplication 21 gtBlt gto 3 4 712 Calculate ABC and ABC1 This illustrates the associative property of matrix multipli cationi LetAlt gtBlt0 2gtclt 3 5 Calculate ABC and ABC1 This illustrates the associative property of matrix multipli cation 12 710 054 213 23 61 12 713 240 7213 10 01 723 30 72 74 Let A be a 4 X 2 matrix B a2 X 6 matrix C a 3 X 4 matrix D a6 X 3 matrix Determine Which of the following matrix expressions exist and give the sizes of the resulting matrix When they do exist a ABC d DCAB b ABD e AZBDC c CAB Let A be a 3 X 2 matrix B a 2 X 1 matrix C a 1 X 3 matrix D a 3 X 1 matrix and E a 3 X 3 matrix Determine Which of the following matrix expressions exist and give the sizes of the resulting matrix When they do exist a ABC d DAB AB b ABEAB e BCDCBC c DC 7 EAC 196 25 Some Numerical Methods As indicated previously there are only a few types of rst order differential equations for which there are methods for nding exact solutions Consequently we have to rely on numerical methods to nd approximate solutions in situations where the differential equation can not be solved In this section we illustrate two elementary numerical methods Our focus here is on the initial value problem 1 ay M900 yo 1 where f and 8f8y are continuous functions on arectangle R and mo yo 6 R That is the initial value problem satis es the conditions of the existence and uniqueness theorem EULER7S METHOD Although this method is rarely used in practice we present it because it has the essential features of more advanced methods We begin by setting a step size It gt 0 Then we define m values zk 0 kh where k is a natural number The values zk are the values of z where we try to approximate the solution to Next we give some notation for our approximations to We will use the notation it 3 gm By the de nition of the derivative we know that y 90k 1 i y 9 ykz gth 7 which using our notation leads to the approximation yk1 M gm 3 i lt2 Substituting this approximation into 1 gives the approximate equations yk1iyk 7 f zk k with 0 iven h 739 7 y g Rearranging terms gives ka yk hfmk yk for k O 1 where yo is given 3 This method is known as Euler s Method with step size It Example 1 Use Euler s method with a step size of 005 to approximate the solution to the initial value problem 1 7y sin 90 MW 1 58 Before we begin7 note that we can give the exact solution to this initial value problem You might naturally ask why we are going to bother approximating the solution if we know how to solve it The answer is simple We want to illustrate how well or poorly Euler s Method works The exact solution is 7 cosz sinz 35 Now we ll use Euler s Method with a step size of h 005 to approximate this solution From 3 we have ka yk 005 iyk sin005klt for k 07 17 where yo 1 Some simple calculations give U1 ug 0904998958 us 0864740681 620 0686056580 Noting that yzo is supposed to approximate 20 y 1 0702403501 to 9 decimal places7 we can see that our approximation of y1 has an error of y1 yzo 07024035017 0686056580 0016346921 Actually7 that s not too bad I IMPROVED EULER7S METHOD Here we give an improvement to Euler s method As above7 we define h gt 0 to be the step size of the method and take zk 0 kh where k is a natural number Also7 we continue to use the notation yk z for the approximations to when x m We can use two different approximations for the derivative Namely7 N yk1 7 yk y N f k 1 k yk1gu Since wait st yltsigtgt and wait mist mist substitution gives yk1 Ms M h f 901s 60 59 and yk1 M f x mm yam Adding these equations and solving for ka gives h yk1 yk 5 100 yk fltk17 yk1gtgt Unfortunately this scheme is not easy to implement because ka occurs on both the left and right side of the equation for this reason it is called an implicit scheme We avoid the implicit nature of this equation by replacing ka on the right side by its Euler approximate using yk as a guess That is h i i yk1 yk 5 HM 24k fltk17 M Where M M WWW M and yo IS glVeH 4 This method 4 is commonly known as the Improved Euler s Method with step size It Example 2 Apply the lmproved Euler s Method with step size 005 to the initial value problem y ysinz y01 As noted in Example 1 this initial value problem has the exact solution 7 cos x sin z ge m Using the lmproved Euler s method with a step size of h 005 gives the values yl 0952499479 yg 0909747970 yg 0871504753 2420 0702609956 Again ygo is an approximation to yt20 y1 07024035012 In this case the error at z 1 is given by M1 7 ml z 206454773 x 104 This is much better than our earlier result from Euler s method I The accuracy of these methods can be predicted In general Euler s Method lymk 7 ykl leh2 and Improved Euler s Method 7 ykl szhg Where the constants L1 and L2 are dependent upon the actual solution but independent of the step size It and the number k Notice that the error estimates above imply 60 Euler s Method has an error which is a factor of h over every interval z interval there are essentially k 1h steps of size h across each z interval and the Improved Euler s Method has an error which is a factor of h2 over every m interval It is not hard to see that small values of h should give a much smaller error for Improved Euler s Method than for Euler s Method Exercises 25 1 to DJ q Use both the Euler and Improved Euler Methods with a step size of h 001 to estimate y2 Where y is the solution of the initial value problem Compare your values with those of the exact solution Use both the Euler and Improved Euler Methods with a step size of h 002 to estimate y1 Where y is the solution of the initial value problem y my y02 Compare your values with those of the exact solution Use both the Euler and Improved Euler Methods with a step size of h 005 to approximate the solution of y y4 i y y0 2 Compare your values with those of the exact solution 4 M95 W for z 120 110 320 1920 1 Use the lmproved Euler s Method with a step size of h 01 to approximate y02 Where yt is the unique solution of y sinm7y3 y0 1

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