Math 2233 Homework Set 9

Transcription

1 Math Homewk Set 9. Compute the Laplace transfm of the following functions. a ft t Let ft t. L[f] te t dt Integrating y parts, with u t du dt dv e st dt v s e st we get te t dt vdu uv vdu t s e st s e st s e st dt s ft t n Now let ft t n. L[f] t n e t dt Integrating y parts, with u t n du nt n dt dv e st dt v s e st

2 we get t n e t dt vdu uv vdu t n s e st + n s n s L [ t n ] t n e st dt n n L [ t n ] s s nn n s L [ t n ]. nn L [t] n! s n+ s n s e st nt n dt. Use the fmula sint eit e it i to compute the Laplace transfm of sint. Let Then L [sint] ft sint i e it e it. e it e it e st dt i e s+it e s+it dt i i s + i e st e it + i s i s + i s + i s i s + i e st e it i s +. Invert the following Laplace transfms. a L [f] s + 4 s +

3 Note that the denominat is a sum of squares: s + 4 s + s + [ ] L [sin x] L sin x fx sin x L [f] s + s 4 s + s 4 s + 4 s Since the denominat is easily factized, we ll try a partial fractions expansion. s + 4 s A s B s A s + B s + 4 s A + B 5 B 5 s 4 A 5 + B A 5 So now L [f] s + s 4 s + 4 s 5 s L [ e 4x] + 5 L [ex ] L [ 5 e 4x + 5 ] ex + 5 s f x 5 e 4x + 5 ex c L [f] s + s + s + 5 In this prolem, the denominat does not factize easily, ut it can e written as a sum of squares: s + s + s + 5 s + s + s + + s + s + + and so we ll try to realize it as a linear comination of the Laplace transfms of the fm We have L [f] d L [f] s + s s + L [ e at cos t ] s + s + + s a s a +, L [ e at sin t ] f x e x cos x s + s + + s a + L [ e x cos x ] L [ e x cos x ]

4 4 Again the denominat does not factize easily, so we ll try to express the denominat as a sum difference of squares. s + L [f] s s + s + s s + + s + s + s + + s + s + s + s s + + s + L [e x cos x] + L [e x sin x] e L [f] L [e x cos x + e x sin x] s s + 4s + 5 f x e x cos x + e x sin x The denominat does not factize easily, so we ll try to first try to express it a sum difference of squares. L [f] s s + 4s s + s s s + 4s s s + + s s + + s + s + + s + + 5L [ e x sin x ] L [ e x cos x ] L [ 5e x sin x e x cos x ] f x 5e x sin x e x cos x 4. Use the Laplace transfm to solve the given initial value prolems. a y y 6y ; y, y Taking the Laplace transfm of oth sides of the differential equation yields L[y ] L[y ] L[6y] s L[y] sy y sl[y] y 6L[y] s L[y] s sl[y] + 6L[y] s s 6 L[y] s + L[y] s s s 6 s s + s the differential equation f y ecomes an algeraic equation f L[y]. To undo this Laplace transfm we first carry out a partial fractions expansion of the right hand side of the equation f L[y]. s s + s A s + + B s As + Bs + s This expansion must e valid f all values of s; in particular when s and when s. In the fmer case we have so we must have A 4 5. s 4 A + B + 5A In the latter case, we have s A + B + 5B

5 5 so B 5. We then have s L[y] s + s 4 5 s s 4 5 L[e x ] + 5 L[ex ] [ 4 L 5 e x + ] 5 ex Hence, taking inverse Laplace transfm of oth sides y 4 5 e x + 5 ex y y + y ; y, y Taking the Laplace transfm of oth sides of the differential equation yields s L[y] sy y sl[y] y + L[y] s s + L[y] L[y] s s + s s + + s + We now consult a tale of Laplace transfm and spot the following identity L [ e at sint ] s a + which looks just like the right hand side of our expression f L[y] once we thake a and. We conclude L[y] L[e x sinx] y y y ; y, y yx e x sinx Taking the Laplace transfm of the differential equation we get Thus, s L[y] sy y sl[y] y L[y] s s L[y] s + 4 s s L[y] s + 4 L[y] s 4 s s Note that the denominat is easily written as a difference of squares; viz., L[y] s 4 s s s s s + s s

6 6 so to invert this Laplace transfm, the following Laplace transfms from the tale of Laplace transfms might e useful. L[e at s a cosht] s a We have L[e at sinht] s a s L [y] s s s s s s s s s So taking a and and using the two Laplace transfms aove, we have L[y] L[e x cosh x] [ L e x sinh ] x [ L e x cosh x e x sinh ] x so yx e x cosh x e x sinh x d y + y + y 4e t ; y, y Taking the Laplace transfm of oth sides of the differential equation we get s L[y] sy y + sl[y] y + L[y] L[4e t ] s + s + L[y] s s + s + L[y] 4 s + + s s + s + s + 5s + 7 s + s + L[y] s + 5s + 7 s + We now determine the partial fractions expansion of the right hand side. The general ansatz is P x s + a A s + a + B s + a + C s + a and so we will try to find constants A, B, C such that Multiplying oth sides y s + we get Plugging in s we find s + 5s + 7 s + A s + + B s + + C s +. s + 5s + 7 A s + + B s + + C C C 4.

7 7 Plugging in s yields Plugging in s yields We now solve 7 A + B + C A + B + 4 A + B. 4 4A + B + C 4A + B + 4 4A + B A + B 5. A + B A + B 5 f A and B. Sutracting the first equation from the second we otain Now the first equation yields A + A. + B B. Thus, A, B, and C 4. Applying this partial fractions expansion to the equation f L[y] now yields L[y] s + s + 7 s + s + + s s + Now from a Tale of Laplace transfms we find L [ t n e at] n! s + a n+ Hence so s + L[e t ] s + L[te t ] s + L[t e t ] L[y] L[e t ] + L[te t ] + 4 L[t e t ] L [ e t + te t + t e t] Taking the inverse Laplace transfm of oth sides we finally get yt t + t + e t.