HW 5 Solution

# HW 5 Solution - 228 4 Chapter 6 The Laplace Transform The...

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Unformatted text preview: 228 4. Chapter 6. The Laplace Transform The function f (t) has a jump discontinuity at t = 1. 7. Integration is a linear operation. It follows that A 0 cosh bt · e−st dt = = 1 2 A 0 1 2 A 0 ebt · e−st dt + 1 2 A 0 1 2 A 0 e−bt · e−st dt = e(b−s)t dt + e−(b+s)t dt . Hence 0 A cosh bt · e−st dt = 1 1 − e(b−s)A 1 1 − e−(b+s)A + . 2 s−b 2 s+b 1 1 1 s 1 + =2 . 2 s−b 2 s+b s − b2 Taking a limit, as A → ∞ , ∞ 0 cosh bt · e−st dt = Note that the above is valid for s > |b| . Sec 6.1, Pg 311 8. Proceeding as in Problem 7, A 0 sinh bt · e−st dt = 1 1 − e(b−s)A 1 1 − e−(b+s)A − . 2 s−b 2 s+b 1 1 1 b 1 − =2 . 2 s−b 2 s+b s − b2 Taking a limit, as A → ∞ , ∞ 0 sinh bt · e−st dt = The limit exists as long as s > |b| . 10. Observe that eat sinh bt = (e(a+b)t − e(a−b)t )/2 . It follows that A 0 eat sinh bt · e−st dt = 1 1 − e(a+b−s)A 1 1 − e−(b−a+s)A − . 2 s−a+b 2 s+b−a Taking a limit, as A → ∞ , ∞ 0 eat sinh bt · e−st dt = 1 1 1 1 b − = . 2 s−a+b 2 s+b−a (s − a)2 − b2 230 Taking a limit, as A → ∞ , ∞ 0 Chapter 6. The Laplace Transform teat · e−st dt = 1 . (s − a)2 Note that the limit exists as long as s > a . Sec 6.1, Pg 311 17. Observe that t cosh at = (t eat + t e−at )/2 . For any value of c , A 0 t ect · e−st dt = − t e(c−s)t s−c A A + 0 0 1 (c−s)t e dt = s−c = 1 − eA(c−s) + A(c − s)eA(c−s) . (s − c)2 ∞ 0 Taking a limit, as A → ∞ , tect · e−st dt = 1 . (s − c)2 Note that the limit exists as long as s > |c| . Therefore, ∞ 0 t cosh at · e−st dt = 1 1 1 + 2 (s − a)2 (s + a)2 = s2 + a2 . (s − a)2 (s + a)2 18. Integrating by parts, A 0 tn eat · e−st dt = − tn e(a−s)t s−a A A + 0 0 n n−1 (a−s)t t e dt = s−a =− An e−(s−a)A + s−a A 0 n n−1 (a−s)t t e dt . s−a Continuing to integrate by parts, it follows that A 0 tn eat · e−st dt = − − An e(a−s)A nAn−1 e(a−s)A − − s−a (s − a)2 n!Ae(a−s)A n!(e(a−s)A − 1) − ··· − . (n − 2)!(s − a)3 (s − a)n+1 n! , (s − a)n+1 That is, A 0 tn eat · e−st dt = pn (A) · e(a−s)A + in which pn (ξ ) is a polynomial of degree n . For any given polynomial, A→∞ lim pn (A) · e−(s−a)A = 0 , as long as s > a . Therefore, ∞ 0 tn eat · e−st dt = n! . (s − a)n+1 234 Chapter 6. The Laplace Transform Hence L−1 [Y (s)] = 3 + 5 cos 2t − 2 sin 2t . Sec 6.2, Pg 320 9. The denominator s2 + 4s + 5 is irreducible over the reals. Completing the 1 − 2s 5 − 2(s + 2) = . s2 + 4s + 5 (s + 2)2 + 1 We ﬁnd that L−1 square, s2 + 4s + 5 = (s + 2)2 + 1 . Now convert the function to a rational function of the variable ξ = s + 2 . That is, ξ2 5 2ξ −2 = 5 sin t − 2 cos t . +1 ξ +1 Using the fact that L [eat f (t)] = L [f (t)]s→s−a , L−1 1 − 2s = e−2t (5 sin t − 2 cos t) . s2 + 4 s + 5 10. Note that the denominator s2 + 2s + 10 is irreducible over the reals. Completing the square, s2 + 2s + 10 = (s + 1)2 + 9 . Now convert the function to a rational function of the variable ξ = s + 1 . That is, 2s − 3 2(s + 1) − 5 = . s2 + 2s + 10 (s + 1)2 + 9 We ﬁnd that L− 1 ξ2 2ξ 5 5 −2 = 2 cos 3t − sin 3t . +9 ξ +9 3 Using the fact that L [eat f (t)] = L [f (t)]s→s−a , L−1 2s − 3 5 = e−t (2 cos 3t − sin 3t) . s2 + 2s + 10 3 12. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y (0) − y (0) + 3 [s Y (s) − y (0)] + 2 Y (s) = 0 . Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2 Y (s) − s − 3 = 0 . Solving for Y (s), the transform of the solution is Y (s) = Using partial fractions, s+3 2 1 = − . s2 + 3 s + 2 s+1 s+2 Hence y (t) = L−1 [Y (s)] = 2 e−t − e−2t . 13. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y (0) − y (0) − 2 [s Y (s) − y (0)] + 2 Y (s) = 0 . s+3 . s2 + 3 s + 2 234 Chapter 6. The Laplace Transform Hence L−1 [Y (s)] = 3 + 5 cos 2t − 2 sin 2t . 9. The denominator s2 + 4s + 5 is irreducible over the reals. Completing the square, s2 + 4s + 5 = (s + 2)2 + 1 . Now convert the function to a rational function of the variable ξ = s + 2 . That is, 1 − 2s 5 − 2(s + 2) = . s2 + 4s + 5 (s + 2)2 + 1 We ﬁnd that L−1 ξ2 5 2ξ −2 = 5 sin t − 2 cos t . +1 ξ +1 Using the fact that L [eat f (t)] = L [f (t)]s→s−a , L−1 1 − 2s = e−2t (5 sin t − 2 cos t) . s2 + 4 s + 5 10. Note that the denominator s2 + 2s + 10 is irreducible over the reals. Completing the square, s2 + 2s + 10 = (s + 1)2 + 9 . Now convert the function to a rational function of the variable ξ = s + 1 . That is, 2s − 3 2(s + 1) − 5 = . s2 + 2s + 10 (s + 1)2 + 9 We ﬁnd that L− 1 ξ2 2ξ 5 5 −2 = 2 cos 3t − sin 3t . +9 ξ +9 3 Using the fact that L [eat f (t)] = L [f (t)]s→s−a , L−1 2s − 3 5 = e−t (2 cos 3t − sin 3t) . s2 + 2s + 10 3 Sec 6.2, Pg 320 12. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y (0) − y (0) + 3 [s Y (s) − y (0)] + 2 Y (s) = 0 . Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2 Y (s) − s − 3 = 0 . Solving for Y (s), the transform of the solution is Y (s) = Using partial fractions, s+3 2 1 = − . s2 + 3 s + 2 s+1 s+2 Hence y (t) = L−1 [Y (s)] = 2 e−t − e−2t . 13. Taking the Laplace transform of the ODE, we obtain s2 Y (s) − s y (0) − y (0) − 2 [s Y (s) − y (0)] + 2 Y (s) = 0 . s+3 . s2 + 3 s + 2 248 Hence L−1 [F (ks)] = Chapter 6. The Laplace Transform 1 f k t k . 1 t (c) From part (b), L−1 [F (as)] = a f ( a ) Note that as + b = a(s + b/a). Using the ct fact that L [e f (t)] = L [f (t)]s→s−c , L−1 [F (as + b)] = e−bt/a 1 f a t a . 26. First write F (s) = n! . s ( 2 )n+1 Let G(s) = n!/sn+1 . Based on the results in Problem 25, 1 −1 s L G 2 2 in which g (t) = tn . Hence L−1 [F (s)] = 2 (2t)n = 2n+1 tn . = g (2t), 29. First write F ( s) = Now consider e−4(s−1/2) . 2(s − 1/2) e−2s . s 1 g 2 1 2 G(s) = Using the result in Problem 25(b), L−1 [G(2s)] = t 2 , 1 2 in which g (t) = u2 (t). Hence L−1 [G(2s)] = L−1 [F (s)] = u2 (t/2) = u4 (t). It follows that 1 t/2 e u4 (t). 2 30. By deﬁnition of the Laplace transform, ∞ L [ f (t)] = 0 e−st u1 (t)dt . 1 − e−s . s That is, 1 L [ f (t)] = 0 e−st dt = Sec 6.3, Pg 330 31. First write the function as f (t) = u0 (t) − u1 (t) + u2 (t) − u3 (t) . It follows that 1 L [ f (t)] = 0 e−st dt + 2 3 e−st dt . 6.3 That is, L [ f (t)] = 249 1 − e−s e−2s − e−3s 1 − e−s + e−2s − e−3s + = . s s s 32. The transform may be computed directly. On the other hand, using the translation property of the transform, 1 L [f (t)] = + s That is, L [f (t)] = 2n+1 k=1 1 e−ks = (−1) s s k 2n+1 (−e−s )k = k=0 1 1 − (−e−s )2n+2 . s 1 + e− s 1 − (e−2s )n+1 . s(1 + e−s ) 35. The given function is periodic, with T = 2 . Using the result of Problem 34, L [f (t)] = 1 1 − e−2s L [f (t)] = 2 0 e−st f (t)dt = 1 1 − e−2s 1 0 e−st dt . That is, 1 − e−s 1 = . s(1 − e−2s ) s(1 + e−s ) 37. The function is periodic, with T = 1 . Using the result of Problem 34, L [f (t)] = 1 1 − e−s 1 0 t e−st dt . It follows that L [f (t)] = 1 − e−s (1 + s) . s2 (1 − e−s ) 38. The function is periodic, with T = π . Using the result of Problem 34, L [f (t)] = 1 1 − e−πs π 0 sin t · e−st dt . 1 + e−πs . 1 + s2 We ﬁrst calculate 0 π sin t · e−st dt = Hence L [f (t)] = 1 + e−πs . (1 − e−πs )(1 + s2 ) 6.3 That is, L [ f (t)] = 249 1 − e−s e−2s − e−3s 1 − e−s + e−2s − e−3s + = . s s s 32. The transform may be computed directly. On the other hand, using the translation property of the transform, 1 L [f (t)] = + s That is, L [f (t)] = 2n+1 k=1 1 e−ks = (−1) s s k 2n+1 (−e−s )k = k=0 1 1 − (−e−s )2n+2 . s 1 + e− s 1 − (e−2s )n+1 . s(1 + e−s ) 35. The given function is periodic, with T = 2 . Using the result of Problem 34, L [f (t)] = 1 1 − e−2s L [f (t)] = 2 0 e−st f (t)dt = 1 1 − e−2s 1 0 e−st dt . That is, 1 − e−s 1 = . s(1 − e−2s ) s(1 + e−s ) Sec 6.3, Pg 331 37. The function is periodic, with T = 1 . Using the result of Problem 34, L [f (t)] = 1 1 − e−s 1 0 t e−st dt . It follows that L [f (t)] = 1 − e−s (1 + s) . s2 (1 − e−s ) 38. The function is periodic, with T = π . Using the result of Problem 34, L [f (t)] = 1 1 − e−πs π 0 sin t · e−st dt . 1 + e−πs . 1 + s2 We ﬁrst calculate 0 π sin t · e−st dt = Hence L [f (t)] = 1 + e−πs . (1 − e−πs )(1 + s2 ) 6.4 (b) 257 Due to initial conditions, the solution temporarily oscillates about y = 0 . After the forcing is applied, the response is a steady oscillation about ym = 1 . Sec 6.4, Pg 336 9.(a) Let g(t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s) − s y (0) − y (0) + Y (s) = L [g (t)] . Applying the initial conditions, s2 Y (s) + Y (s) − 1 = L [g (t)] . The forcing function can be written as g (t) = with Laplace transform L [g (t)] = t t 1 [1 − u6 (t)] + 3 u6 (t) = − (t − 6)u6 (t) 2 22 1 e − 6s − . 2s2 2s2 Solving for the transform, Y (s) = Using partial fractions, 1 2s2 (s2 + 1) = 11 1 −2 . 2 2s s +1 s2 1 1 e−6s + 22 − 22 . +1 2s (s + 1) 2s (s + 1) Taking the inverse transform, and using Theorem 6.3.1, the solution of the IVP is y (t) = sin t + 1 1 [t − sin t] − [(t − 6) − sin(t − 6)] u6 (t) 2 2 1 1 = [t + sin t] − [(t − 6) − sin(t − 6)] u6 (t). 2 2 6.5 (b) 273 Sec 6.5, Pg 343 9.(a) Taking the initial conditions into consideration, the transform of the ODE is s2 Y (s) + Y (s) = Solving for the transform, Y ( s) = Using partial fractions, 3 e−(3π/2)s e−2πs e−(π/2)s + − . 2 + 1) 2+1 s( s s s(s2 + 1) 1 1 s = −2 . + 1) s s +1 e−(π/2)s e−2πs + 3 e−(3π/2)s − . s s s(s2 Hence s e−(π/2)s 3 e−(3π/2)s e−2πs s e−2πs e−(π/2)s − + − +2 . 2+1 2+1 s s s s s +1 Based on Theorem 6.3.1, the solution of the IVP is Y (s) = y (t) = uπ/2 (t) − cos(t − π 3π )uπ/2 (t) + 3 sin(t − )u3π/2 (t)− 2 2 − u2π (t) + cos(t − 2π )u2π (t) . That is, y (t) = [1 − sin(t)] uπ/2 (t) + 3 cos(t) u3π/2 (t) − [1 − cos(t)] u2π (t) . 286 Chapter 6. The Laplace Transform 3. It follows directly that t f ∗ f (t) = 0 sin(t − τ ) sin(τ ) dτ = 1 (sin t − t cos t). 2 = 1 2 t [cos(t − 2τ ) − cos(t)] dτ = 0 The range of the resulting function is R . Sec 6.6, Pg 350 5. We have L [e−t ] = 1/(s + 1) and L [sin t] = 1/(s2 + 1). Based on Theorem 6.6.1, t L 0 e−(t−τ ) sin(τ ) dτ = 1 1 1 ·2 = . s+1 s +1 (s + 1)(s2 + 1) 6. Let g (t) = t and h(t) = et . Then f (t) = g ∗ h (t) . Applying Theorem 6.6.1, t L 0 g (t − τ )h(τ ) dτ = 1 1 1 · =2 . s2 s − 1 s (s − 1) 7. We have f (t) = g ∗ h (t) , in which g (t) = sin t and h(t) = cos t . The transform of the convolution integral is t L 0 g (t − τ )h(τ ) dτ = 1 s s · =2 . s2 + 1 s2 + 1 (s + 1)2 9. It is easy to see that L− 1 1 = e−t s+1 and L−1 s2 s = cos 2t . +4 Applying Theorem 6.6.1, L−1 s = (s + 1)(s2 + 4) t 0 e−(t−τ ) cos 2τ dτ . 10. We ﬁrst note that L− 1 1 = t e− t (s + 1)2 1 (s + 1)2 (s2 + 4) and L− 1 1 1 = sin 2t . s2 + 4 2 Based on the convolution theorem, L− 1 1 2 1 = 2 = t 0 t 0 (t − τ )e−(t−τ ) sin 2τ dτ τ e−τ sin(2t − 2τ ) dτ . 11. Let g (t) = L−1 [G(s)]. Since L−1 1/(s2 + 1) = sin t , the inverse transform of the product is L−1 G(s) = s2 + 1 t t g (t − τ ) sin τ dτ = 0 0 sin(t − τ ) g (τ ) dτ . 288 Chapter 6. The Laplace Transform Based on the elementary properties of the Laplace transform, L−1 1 4s2 + 4s + 17 t 0 = 1 −t/2 e sin 2t . 8 Applying the convolution theorem, the solution of the IVP is y (t) = 1 8 e−(t−τ )/2 sin 2(t − τ ) g (τ ) dτ . Sec 6.6, Pg 351 17. Taking the initial conditions into consideration, the transform of the ODE is s2 Y (s) − 2s + 3 + 4 [s Y (s) − 2] + 4 Y (s) = G(s). Solving for the transform of the solution, Y ( s) = We can write 2s + 5 G(s) + . (s + 2)2 (s + 2)2 2s + 5 2 1 = + . (s + 2)2 s + 2 (s + 2)2 2 = 2e−2t s+2 1 = t e − 2t . (s + 2)2 It follows that L− 1 and L−1 Based on the convolution theorem, the solution of the IVP is y (t) = 2e−2t + t e−2t + 0 t (t − τ )e−2(t−τ ) g (τ ) dτ . 19. The transform of the ODE (given the speciﬁed initial conditions) is s4 Y (s) − Y (s) = G(s). Solving for the transform of the solution, Y (s) = First write G(s) . s4 − 1 1 1 1 1 = −2 . s4 − 1 2 s2 − 1 s +1 L−1 It follows that 1 s4 − 1 t = 1 [sinh t − sin t] . 2 Based on the convolution theorem, the solution of the IVP is y (t) = 1 2 [sinh(t − τ ) − sin(t − τ )] g (τ ) dτ . 0 20. Taking the initial conditions into consideration, the transform of the ODE is s4 Y (s) − s3 + 5s2 Y (s) − 5s + 4 Y (s) = G(s). ...
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