7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 281

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1 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 28 and i 2 Show how Euler formula (page 33) can then be ued to deduce the reult a ( a) 2 b 2 {e at co bt} {e at in bt} b ( a) 2 b 2 5 Under what condition i a linear function f (x) mx b, m, a linear tranform? 52 Explain why the function t, t 2 f(t) 4, 2 t 5 >(t 5), t 5 i not piecewie continuou on [, ) 53 Show that the function f(t) >t 2 doe not poe a Laplace tranform [Hint: Write a two improper integral: e t e dt t 2 t t dt I 2 I 2 {>t 2 } Show that I diverge] {>t 2 } 54 Show that the Laplace tranform exit [Hint: Start with integration by part] 55 If and a i a contant, how that Thi reult i known a the change of cale theor em 56 Ue the given Laplace tranform and the reult in Problem 55 to find the indicated Laplace tranform Aume that a and k are poitive contant (a) (b) (c) {f(t)} F() {e t } ; {in t} { co t} {f(at)} a F a {e at } 2 ; ( 2 ) ; (d) {in t inh t} ; {in kt} {2te t2 coe t2 } { co kt} {in kt inh kt} 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES REVIEW MATERIAL Partial fraction decompoition See the Student Reource Manual INTRODUCTION In thi ection we take a few mall tep into an invetigation of how the Laplace tranform can be ued to olve certain type of equation for an unknown function We begin the dicuion with the concept of the invere Laplace tranform or, more preciely, the invere of a Laplace tranform F() After ome important preliminary background material on the Laplace tranform of derivative f(t), f(t),, we then illutrate how both the Laplace tranform and the invere Laplace tranform come into play in olving ome imple ordinary differential equation 72 INVERSE TRANSFORMS The Invere Problem If F() repreent the Laplace tranform of a function f(t), that i,, we then ay f (t) i the invere Laplace tranform of F() and write f(t) {F()} For example, from Example, 2, and 3 of Section 7 we have, repectively, {f(t)} F() Tranform {} {t} 2 {e 3t } 3 Invere Tranform t 2 e 3t 3 Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

2 282 CHAPTER 7 THE LAPLACE TRANSFORM We hall ee hortly that in the application of the Laplace tranform to equation we are not able to determine an unknown function f(t) directly; rather, we are able to olve for the Laplace tranform F() of f(t); but from that knowledge we acertain f by computing f (t) {F()} The idea i imply thi: Suppoe 2 6 F() i a Laplace tranform; find a function f(t) uch that {f(t)} F() 2 4 We hall how how to olve thi problem in Example 2 For future reference the analogue of Theorem 7 for the invere tranform i preented a our next theorem THEOREM 72 Some Invere Tranform (a) (b) (d) t n n! n, n, 2, 3, k in kt 2 2 k (c) (e) e at a co kt 2 2 k k (f) inh kt 2 (g) coh kt 2 2 k 2 k In evaluating invere tranform, it often happen that a function of under conideration doe not match exactly the form of a Laplace tranform F() given in a table It may be neceary to fix up the function of by multiplying and dividing by an appropriate contant EXAMPLE Applying Theorem 72 Evaluate (a) (b) SOLUTION (a) To match the form given in part (b) of Theorem 72, we identify n 5 or n 4 and then multiply and divide by 4!: 5 4! 4! 5 24 t4 (b) To match the form given in part (d) of Theorem 72, we identify k 2 7, o k 7 We fix up the expreion by multiplying and dividing b 7: in7t i a Linear T ranform The invere Laplace tranform i alo a linear tranform; that i, for contant a and b {F() G()} {F()} {G()}, () where F and G are the tranform of ome function f and g Like (3) of Section 7, () extend to any finite linear combination of Laplace tranform Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

3 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 283 EXAMPLE 2 Termwie Diviion and Linearity 2 6 Evaluate 2 4 SOLUTION We firt rewrite the given function of a two expreion by mean of termwie diviion and then ue (): { } { termwie diviion co 2t 3 in 2t linearity and fixing up contant } 2 { } 2 { 2 4 } part (e) and (d) of Theorem 72 with k 2 (2) Partial Fraction Partial fraction play an important role in finding invere Laplace tranform The decompoition of a rational expreion into component fraction can be done quickly by mean of a ingle command on mot computer algebra ytem Indeed, ome CAS have package that implement Laplace tranform and invere Laplace tranform command But for thoe of you without acce to uch oftware, we will review in thi and ubequent ection ome of the baic algebra in the important cae in which the denominator of a Laplace tranform F() contain ditinct linear factor, repeated linear factor, and quadratic polynomial with no real factor Although we hall examine each of thee cae a thi chapter develop, it till might be a good idea for you to conult either a calculu text or a current precalculu text for a more comprehenive review of thi theory The following example illutrate partial fraction decompoition in the cae when the denominator of F() i factorable into ditinct linear factor EXAMPLE 3 Partial Fraction: Ditinct Linear Factor Evaluate ( )( 2)( 4) SOLUTION There exit unique real contant A, B, and C o that ( )( 2)( 4) A B 2 C 4 A( 2)( 4) B( )( 4) C( )( 2) ( )( 2)( 4) Since the denominator are identical, the numerator are identical: A( 2)( 4) B( )( 4) C( )( 2) (3) By comparing coefficient of power of on both ide of the equality, we know that (3) i equivalent to a ytem of three equation in the three unknown A, B, and C However, there i a hortcut for determining thee unknown If we et, 2, and 4 in (3), we obtain, repectively, 6 A()(5), 25 B()(6), and C(5)(6), and o A 6, B 25, and C Hence the partial fraction decompoition i >5 25>6 >3, (4) ( )( 2)( 4) 2 4 Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

4 284 CHAPTER 7 THE LAPLACE TRANSFORM and thu, from the linearity of and part (c) of Theorem 72, ( )( 2)( 4) TRANSFORMS OF DERIVATIVES 2 3 (5) 4 Tranform a Derivative A wa pointed out in the introduction to thi chapter, our immediate goal i to ue the Laplace tranform to olve differential equation To that end we need to evaluate quantitie uch a {dy>dt} and {d 2 y>dt 2 } For example, if f i continuou for t, then integration by part give { f(t)} e t f (t) dt e t f (t) e t f (t) dt f () { f (t)} or { f(t)} F() f () (6) Here we have aumed that e t f(t) : a t : Similarly, with the aid of (6), { f (t)} e t f (t) dt e t f (t) e t f (t) dt f() { f(t)} [F() f ()] f () or { f (t)} 2 F() f() f() (7) In like manner it can be hown that 6 5 et 25 6 e2t 3 e4t ; from (6) { f (t)} 3 F() 2 f () f() f () The recurive nature of the Laplace tranform of the derivative of a function f hould be apparent from the reult in (6), (7), and (8) The next theorem give the Laplace tranform of the nth derivative of f The proof i omitted (8) THEOREM 722 Tranform of a Derivative If f, f,, f (n) are continuou on [, ) and are of exponential order and if f (n) (t) i piecewie continuou on [, ), then { f (n) (t)} n F() n f() n2 f() where F() { f(t)} f (n) (), Solving Linear ODE It i apparent from the general reult given in Theorem 722 that depend on Y() {y(t)} and the n derivative of y(t) evaluated at t Thi property make the Laplace tranform ideally uited for olving linear initial-value problem in which the differential equation ha contant coefficient Such a differential equation i imply a linear combination of term y, y, y,, y (n) : {d n y>dt n } a n d n y dt n a n d n y dt n a y g(t), y() y, y() y,, y (n) () y n, Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

5 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 285 where the a i, i,,, n and y, y,, y n are contant By the linearity property the Laplace tranform of thi linear combination i a linear combination of Laplace tranform: a n d n y n dt a n d n y dt a {y} {g(t)} From Theorem 722, (9) become a n [ n Y() n y() y (n) ()] a n [ n Y() n2 y() y (n2) ()] a Y() G(), where {y(t)} Y() and {g(t)} G() In other word, The Laplace tranform of a linear differential equation with contant coefficient become an algebraic equation in Y() If we olve the general tranformed equation () for the ymbol Y(), we firt obtain P()Y() Q() G() and then write where P() a n n a n n a, Q() i a polynomial in of degree le than or equal to n coniting of the variou product of the coefficient a i, i,, n and the precribed initial condition y, y,, y n, and G() i the Laplace tranform of g(t) * Typically, we put the two term in () over the leat common denominator and then decompoe the expreion into two or more partial fraction Finally, the olution y(t) of the original initial-value problem i y(t) {Y()}, where the invere tranform i done term by term The procedure i ummarized in the diagram in Figure 72 (9) () Y() Q() G(), () P() P() Find unknown y(t) that atifie DE and initial condition Apply Laplace tranform Tranformed DE become an algebraic equation in Y() Solution y(t) of original IVP Apply invere Laplace tranform Solve tranformed equation for Y() FIGURE 72 Step in olving an IVP by the Laplace tranform The next example illutrate the foregoing method of olving DE, a well a partial fraction decompoition in the cae when the denominator of Y() contain a quadratic polynomial with no real factor EXAMPLE 4 Solving a Firt-Order IVP Ue the Laplace tranform to olve the initial-value problem SOLUTION equation: dy 3y 3 in 2t, y() 6 dt We firt take the tranform of each member of the differential dy (2) dt 3{y} 3{in 2t} * The polynomial P() i the ame a the nth-degree auxiliary polynomial in (2) in Section 43 with the uual ymbol m replaced by Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

6 286 CHAPTER 7 THE LAPLACE TRANSFORM From (6), {dy>dt} Y() y() Y() 6, and from part (d) of Theorem 7, {in 2t} 2>( 2 4), o (2) i the ame a Y() 6 3Y() Solving the lat equation for Y(), we get 26 or ( 3)Y() Y() 6 (3) 3 26 ( 3)( 2 4) 62 5 ( 3)( 2 4) Since the quadratic polynomial 2 4 doe not factor uing real number, it aumed numerator in the partial fraction decompoition i a linear polynomial in : ( 3)( 2 4) A B C Putting the right-hand ide of the equality over a common denominator and equating numerator give A( 2 4) (B C)( 3) Setting 3 then immediately yield A 8 Since the denominator ha no more real zero, we equate the coefficient of 2 and : 6 A B and 3B C Uing the value of A in the firt equation give B 2, and then uing thi lat value in the econd equation give C 6 Thu Y() 62 5 ( 3)( 2 4) We are not quite finihed becaue the lat rational expreion till ha to be written a two fraction Thi wa done by termwie diviion in Example 2 From (2) of that example, y(t) It follow from part (c), (d), and (e) of Theorem 72 that the olution of the initialvalue problem i y(t) 8e 3t 2 co 2t 3 in 2t EXAMPLE 5 Solving a Second-Order IVP Solve y3y2y e 4t, y(), y() 5 SOLUTION Proceeding a in Example 4, we tranform the DE We take the um of the tranform of each term, ue (6) and (7), ue the given initial condition, ue (c) of Theorem 7, and then olve for Y(): d2 y dy dt2 3dt 2{y} {e 4t } 2 Y() y() y() 3[Y() y()] 2Y() 4 ( 2 3 2)Y() 2 4 Y() 2 (4) ( 2 3 2)( 4) ( )( 2)( 4) The detail of the partial fraction decompoition of Y() have already been carried out in Example 3 In view of the reult in (4) and (5) we have the olution of the initial-value problem y(t) {Y()} 6 5 et 25 6 e2t 3 e4t Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

7 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 287 Example 4 and 5 illutrate the baic procedure for uing the Laplace tranform to olve a linear initial-value problem, but thee example may appear to demontrate a method that i not much better than the approach to uch problem outlined in Section 23 and Don t draw any negative concluion from only two example Ye, there i a lot of algebra inherent in the ue of the Laplace tranform, but oberve that we do not have to ue variation of parameter or worry about the cae and algebra in the method of undetermined coefficient Moreover, ince the method incorporate the precribed initial condition directly into the olution, there i no need for the eparate operation of applying the initial condition to the general olution y c y c 2 y 2 of the DE to find pecifi contant in a particular olution of the IVP The Laplace tranform ha many operational propertie In the ection that follow we will examine ome of thee propertie and ee how they enable u to olve problem of greater complexity c n y n y p REMARKS (i) The invere Laplace tranform of a function F() may not be unique; in other word, it i poible that { f (t)} { f 2 (t)} and yet f f 2 For our purpoe thi i not anything to be concerned about If f and f 2 are piecewie continuou on [, ) and of exponential order, then f and f 2 are eentially the ame See Problem 44 in Exercie 72 However, if f and f 2 are continuou on [, ) and { f (t)} { f 2 (t)}, then f f 2 on the interval (ii) Thi remark i for thoe of you who will be required to do partial fraction decompoition by hand There i another way of determining the coefficient in a partial fraction decompoition in the pecial cae when { f(t)} F() i a rational function of and the denominator of F i a product of ditinct linear factor Let u illutrate by reexamining Example 3 Suppoe we multiply both ide of the aumed decompoition ( )( 2)( 4) A B 2 C 4 (5) by, ay,, implify, and then et Since the coefficient of B and C on the right-hand ide of the equality are zero, we get Written another way, ( 2)( 4) A or A , ( ) ( 2)( 4) 6 5 A where we have haded, or covered up, the factor that canceled when the lefthand ide wa multiplied by Now to obtain B and C, we imply evaluate the left-hand ide of (5) while covering up, in turn, 2 and 4: ( )( 2)( 4) B and ( )( 2)( 4) C 4 3 Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

8 288 CHAPTER 7 THE LAPLACE TRANSFORM The deired decompoition (5) i given in (4) Thi pecial technique for determining coefficient i naturally known a the cover-up method (iii) In thi remark we continue our introduction to the terminology of dynamical ytem Becaue of (9) and () the Laplace tranform i well adapted to linear dynamical ytem The polynomial P() a n n a n n a in () i the total coefficient of Y() in () and i imply the left-hand ide of the DE with the derivative d k ydt k replaced by power k, k,,, n It i uual practice to call the reciprocal of P() namely, W() P() the tranfer function of the ytem and write () a Y() W()Q() W()G() (6) In thi manner we have eparated, in an additive ene, the effect on the repone that are due to the initial condition (that i, W()Q()) from thoe due to the input function g (that i, W()G()) See (3) and (4) Hence the repone y(t) of the ytem i a uperpoition of two repone: y(t) {W()Q()} {W()G()} y (t) y (t) If the input i g(t), then the olution of the problem i y (t) {W()Q()} Thi olution i called the zero-input repone of the ytem On the other hand, the function y (t) {W()G()} i the output due to the input g(t) Now if the initial tate of the ytem i the zero tate (all the initial condition are zero), then Q(), and o the only olution of the initial-value problem i y (t) The latter olution i called the zero-tate repone of the ytem Both y (t) and y (t) are particular olution: y (t) i a olution of the IVP coniting of the aociated homogeneou equation with the given initial condition, and y (t) i a olution of the IVP coniting of the nonhomogeneou equation with zero initial condition In Example 5 we ee from (4) that the tranfer function i W() ( 2 3 2), the zero-input repone i and the zero-tate repone i 2 y (t), ( )( 2) 3et 4e 2t y (t) ( )( 2)( 4) 5 et 6 e2t 3 e4t Verify that the um of y (t) and y (t) i the olution y(t) in Example 5 and that y (), y () 5, wherea y (), y () EXERCISES 72 Anwer to elected odd-numbered problem begin on page ANS- 72 INVERSE TRANSFORMS In Problem 3 ue appropriate algebra and Theorem 72 to find the given invere Laplace tranform ( ) ( 2) Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it

9 73 OPERATIONAL PROPERTIES I y 3y3y2y e t, y(), y() y(), 9 4 y 2yy2y in 3t, y(), y(), y() 2 9 The invere form of the reult in Problem 5 in Exercie 7 are ( )( 2) a ( a) 2 b eat co bt b 23 ( 2)( 3)( 6) ( a) 2 b eat in bt 2 In Problem 4 and 42 ue the Laplace tranform and thee 24 invere to olve the given initial-value problem ( )( )( 2) 4 yy e 3t co 2t, y() ( 2)( 2 4) 42 y2y5y, y(), y() Dicuion Problem 28 ( 4 9 )( 2 ) 43 (a) With a light change in notation the tranform in (6) i the ame a 3 ( )( 2 4) { f(t)} { f (t)} f () 722 TRANSFORMS OF DERIVATIVES In Problem 3 4 ue the Laplace tranform to olve the given initial-value problem 3 32 dy y, y() dt 2 dy y, y() 3 dt 33 y6y e 4t, y() 2 34 yy 2 co 5t, y() 35 y5y4y, y(), y() 36 y4y6e 3t 3e t, y(), y() 37 y y 22 in 22t, y(), y() 38 y9y e t, y(), y() With f(t) te at, dicu how thi reult in conjunction with (c) of Theorem 7 can be ued to evaluate {te at } (b) Proceed a in part (a), but thi time dicu how to ue (7) with f (t) t in kt in conjunction with (d) and (e) of Theorem 7 to evaluate {t in kt} 44 Make up two function f and f 2 that have the ame Laplace tranform Do not think profound thought 45 Reread (iii) in the Remark on page 288 Find the zero-input and the zero-tate repone for the IVP in Problem Suppoe f(t) i a function for which f (t) i piecewie continuou and of exponential order c Ue reult in thi ection and Section 7 to jutify f () lim F(), : where F() { f (t)} Verify thi reult with f(t) co kt 73 OPERATIONAL PROPERTIES I REVIEW MATERIAL Keep practicing partial fraction decompoition Completion of the quare INTRODUCTION It i not convenient to ue Definition 7 each time we wih to find the Laplace tranform of a function f(t) For example, the integration by part involved in evaluating, ay, {e t t 2 in 3t} i formidable, to ay the leat In thi ection and the next we preent everal laboraving operational propertie of the Laplace tranform that enable u to build up a more extenive lit of tranform (ee the table in Appendix III) without having to reort to the baic definition and integration Copyright 22 Cengage Learning All Right Reerved May not be copied, canned, or duplicated, in whole or in part Due to electronic right, ome third party content may be uppreed from the ebook and/or echapter() Editorial review ha deemed that any uppreed content doe not materially affect the overall learning experience Cengage Learning reerve the right to remove additional content at any time if ubequent right retriction require it