AEM_3e_Chapter04

AEM_3e_Chapter04 - 4 The Laplace Transform EXERCISES 4.1...

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Unformatted text preview: 4 The Laplace Transform EXERCISES 4.1 Definition of the Laplace Transform 1 1. ∞ −e−st dt + {f (t)} = 0 e−st dt = 1 2 {f (t)} = 0 0 = e−st dt = {f (t)} = (2t + 1)e−st dt = 0 = {f (t)} = (sin t)e−st dt = − 0 = 6. 1 0+ 2 e−πs s +1 ∞ {f (t)} = 7. f (t) = − 1 e−πs/2 s2 + 1 − 0− 1 2 1 − s2 s = 0 1 2 (1 − 3e−s ) + 2 (1 − e−s ), s s 1 = 2 (e−πs + 1), s +1 1 e−πs/2 , s2 + 1 s>0 1 s 1 e−st cos t + 2 e−st sin t s2 + 1 s +1 =− ∞ π 0 s>0 ∞ π/2 s>0 0, 0 < t < 1 t, t > 1 ∞ {f (t)} = te−st dt = 1 8. f (t) = 0 1 − e−st s s 1 e−st sin t − 2 e−st cos t s2 + 1 s +1 π/2 =0− 0+ 1 1 1 − (0 − e−s ) = 2 (1 − e−s ), s s 1 − 0− 2 s +1 (cos t)e−st dt = s>0 2 2 1 − te−st − 2 e−st − e−st s s s 2 2 1 − e−s − 2 e−s − e−s s s s π 5. 1 1 1 − te−st − 2 e−st s s 1 − 0− 2 s 1 1 − e−s − 2 e−s s s ∞ s>0 4 = − (e−2s − 1), s 0 1 1 4. ∞ te−st dt + {f (t)} = 0 1 − e−st s 2 4 4e−st dt = − e−st s 1 3. 1 2 1 = e−s − , s s 1 1 1 = e−s − − 0 − e−s s s s 2. 1 −st e s 1 1 − te−st − 2 e−st s s ∞ = 1 1 −s 1 e + 2 e−s , s > 0 s s 0, 0<t<1 2t − 2, t > 1 ∞ {f (t)} = 2 1 1 1 (t − 1)e−st dt = 2 − (t − 1)e−st − 2 e−st s s 198 ∞ = 1 2 −s e , s>0 s2 s>0 4.1 1 − t, 0 < t < 1 0, t>1 9. The function is f (t) = 1 {f (t)} = ∞ (1 − t)e−st dt + 0 0, 0 < t < a 10. f (t) = c, a < t < b ; 0, t > b 11. ∞ {f (t)} = b {f (t)} = a ∞ et+7 e−st dt = e7 ∞ e−2t−5 e−st dt = e−5 ∞ {f (t)} = ∞ te4t e−st dt = ∞ 15. te(4−s)t dt = ∞ t2 e−2t e−st dt = b c = (e−sa − e−sb ), s > 0 s a ∞ =0− 0 e7 e7 = , 1−s s−1 ∞ e−5 −(s+2)t e s+2 = 0 e−5 , s+2 ∞ s > −2 0 t2 e−(s+2)t dt ∞ e−t (sin t)e−st dt = ∞ = 0 2 , (s + 2)3 (sin t)e−(s+1)t dt 0 ∞ −(s + 1) −(s+1)t 1 e e−(s+1)t cos t sin t − (s + 1)2 + 1 (s + 1)2 + 1 1 1 = 2 , s > −1 = (s + 1)2 + 1 s + 2s + 2 = ∞ {f (t)} = ∞ et (cos t)e−st dt = 0 0 (cos t)e(1−s)t dt 0 1−s 1 e(1−s)t cos t + e(1−s)t sin t (1 − s)2 + 1 (1 − s)2 + 1 1−s s−1 =− = 2 , s>1 (1 − s)2 + 1 s − 2s + 2 ∞ = 17. ∞ {f (t)} = 0 t(cos t)e−st dt 0 = = s>1 ∞ 1 1 e(4−s)t te(4−s)t − 4−s (4 − s)2 1 2 −(s+2)t 2 2 − te−(s+2)t − e−(s+2)t t e s+2 (s + 2)2 (s + 2)3 0 16. 0 0 − {f (t)} = 1 s>4 0 = e−(s+2)t dt = − 0 1 = , (4 − s)2 {f (t)} = e7 (1−s)t e 1−s 0 0 14. c ce−st dt = − e−st s e(1−s)t dt = 0 13. 1 1 − (1 − t)e−st + 2 e−st s s (1 − t)e−st dt = 0 0 ∞ {f (t)} = 1 0e−st dt = s>0 0 12. so 1 1 1 1 = 2 e−s + − 2 , s s s Definition of the Laplace Transform st s2 − 1 − 2 − s + 1 (s2 + 1)2 s2 − 1 2 (s2 + 1) , −st (cos t)e + 2s t + s2 + 1 (s2 + 1)2 s>0 199 ∞ −st (sin t)e 0 s > −2 4.1 Definition of the Laplace Transform ∞ {f (t)} = 18. t(sin t)e−st dt 0 t 2s − 2 − 2 + 1)2 s + 1 (s = = 2s , 2 (s2 + 1) −st (cos t)e − s2 − 1 st + 2 2+1 s (s2 + 1) ∞ −st (sin t)e 0 s>0 4! s5 19. {2t4 } = 2 21. {4t − 10} = 23. {t2 + 6t − 3} = 25. {t3 + 3t2 + 3t + 1} = 27. {1 + e4t } = 29. {1 + 2e2t + e4t } = 31. {4t2 − 5 sin 3t} = 4 33. {sinh kt} = 1 2 {ekt − e−kt } = 34. {cosh kt} = 1 2 {ekt + ekt } = 35. {et sinh t} = 36. {e−t cosh t} = 37. {sin 2t cos 2t} = 38. {cos2 t} = 5! s6 20. {t5 } = 22. {7t + 3} = 24. {−4t2 + 16t + 9} = −4 26. {8t3 − 12t2 + 6t − 1} = 8 28. {t2 − e−9t + 5} = 1 2 1 + + s s−2 s−4 30. {e2t − 2 + e−2t } = 1 2 1 − + s−2 s s+2 2 3 −5 2 s3 s +9 32. {cos 5t + sin 2t} = s 2 + s2 + 25 s2 + 4 4 10 − s2 s 2 6 3 + 2− s3 s s 3! 2 3 1 +3 3 + 2 + s4 s s s 1 1 + s s−4 et et − e−t 2 e−t 1 1 + cos 2t 2 2 2 16 9 + 2 + s3 s s 3! 2 6 1 − 12 3 + 2 − s4 s s s 5 2 1 + − s3 s+9 s 1 1 1 k − = 2 2 s−k s+k s − k2 s s2 − k 2 1 2t 1 e − 2 2 = et + e−t 2 1 sin 4t 2 3 7 + s2 s = = = = 1 1 −2t + e 2 2 1 1 − 2(s − 2) 2s = 1 1 + 2s 2(s + 2) 2 s2 + 16 1 1 s + 2s 2 s2 + 4 39. From the addition formula for the sine function, sin(4t + 5) = sin 4t cos 5 + cos 4t sin 5 so {sin(4t + 5)} = (cos 5) {sin 4t} + (sin 5) {cos 4t} = (cos 5) 40. From the addition formula for the cosine function, 4 s 4 cos 5 + (sin 5)s + (sin 5) 2 = . + 16 s + 16 s2 + 16 √ 3 π π 1 = cos t cos + sin t sin = cos t + sin t 6 6 2 2 √ π 3 1 cos t − = {cos t} + {sin t} 6 2 2 √ √ 1 1 1 3s + 1 3 s + = . = 2 s2 + 1 2 s2 + 1 2 s2 + 1 π cos t − 6 so s2 200 4.1 Definition of the Laplace Transform 41. (a) Using integration by parts for α > 0, ∞ Γ(α + 1) = ∞ tα e−t dt = −tα e−t 0 0 ∞ +α tα−1 e−t dt = αΓ(α). 0 (b) Let u = st so that du = s dt. Then ∞ {tα } = ∞ e−st tα dt = 0 42. (a) {t−1/2 } = 0 π s Γ(1/2) = s1/2 {t1/2 } = (b) α u s e−u 1 1 du = α+1 Γ(α + 1), s s √ π Γ(3/2) = 3/2 3/2 s 2s 2 2 {t3/2 } = (c) 1 −2/3 3t 43. Let F (t) = t1/3 . Then F (t) is of exponential order, but f (t) = F (t) = hence is not of exponential order. Let f (t) = 2tet cos et = α > −1. √ Γ(5/2) 3 π = 5/2 s5/2 4s is unbounded near t = 0 and 2 d sin et . dt This function is not of exponential order, but we can show that its Laplace transform exists. Using integration by parts we have 2 ∞ 2 {2tet cos et } = 2 d sin et dt e−st 0 ∞ = − sin 1 + s e−st sin et 2 dt = lim a→∞ e−st sin et dt = s 2 a a 0 +s e−st sin et dt 2 0 2 {sin et } − sin 1. 0 2 Since sin et is continuous and of exponential order, 2 {sin et } exists, and therefore 2 2 {2tet cos et } exists. 44. The relation will be valid when s is greater than the maximum of c1 and c2 . 2 45. Since et is an increasing function and t2 > ln M + ct for M > 0 we have et > eln M +ct = M ect for t sufficiently 2 large and for any c. Thus, et is not of exponential order. 46. Assuming that (c) of Theorem 4.1 is applicable with a complex exponent, we have {e(a+ib)t } = 1 1 (s − a) + ib s − a + ib . = = s − (a + ib) (s − a) − ib (s − a) + ib (s − a)2 + b2 By Euler’s formula, eiθ = cos θ + i sin θ, so {e(a+ib)t } = = = {eat eibt } = {eat cos bt} + i {eat (cos bt + i sin bt)} {eat sin bt} s−a b +i . (s − a)2 + b2 (s − a)2 + b2 Equating real and imaginary parts we get {eat cos bt} = s−a (s − a)2 + b2 and {eat sin bt} = b . (s − a)2 + b2 47. We want f (αx + βy) = αf (x) + βf (y) or m(αx + βy) + b = α(mx + b) + β(my + b) = m(αx + βy) + (α + β)b for all real numbers α and β. Taking α = β = 1 we see that b = 2b, so b = 0. Thus, f (x) = mx + b will be a linear transformation when b = 0. 201 4.1 Definition of the Laplace Transform {tn−1 } = (n − 1)!/sn . Then, using the definition of the Laplace transform and integration by 48. Assume that parts, we have ∞ ∞ 1 n ∞ −st n−1 e−st tn dt = − e−st tn + e t dt s s 0 0 0 n n (n − 1)! n! =0+ = n+1 . {tn−1 } = s s sn s {tn } = EXERCISES 4.2 The Inverse Transform and Transforms of Derivatives 1. 1 s3 = 1 2 2 s3 = 1 2 t 2 2. 1 s4 = 1 6 3! s4 = 1 3 t 6 3. 1 48 − 5 s2 s 4. 1 2 − 3 s s 5. (s + 1)3 s4 = 1 3 2 1 3! 1 +3· 2 + · 3 + · 4 s s 2 s 6 s 6. (s + 2)2 s3 = 1 2 1 +4· 2 +2· 3 s s s 7. 1 1 1 − + 2 s s s−2 = t − 1 + e2t 8. 4 1 6 − + s s5 s+8 = 9. 1 4s + 1 = 10. 1 5s − 2 = 11. 5 s2 + 49 = 12. 10s s2 + 16 = 10 cos 4t 13. 4s 4s2 + 1 = s s2 + 1/4 14. 1 4s2 + 1 = 1 1/2 · 2 s2 + 1/4 1 48 4! − · s2 24 s5 = 2 4· = 1 5! 4 3! 1 · 6 − · 4+ 2 s 6 s 120 s 1 s + 1/4 = 1 1 · 5 s − 2/5 1 −t/4 e 4 1 2t/5 e 5 = 5 7 · 7 s2 + 49 = 5 sin 7t 7 1 = cos t 2 = 2 1 5 t = 4t − t3 + 3 120 3 1 = 1 + 3t + t2 + t3 2 6 = 1 + 4t + 2t2 1 1 1 4! − + · s 4 s5 s+8 4· 1 4 = t − 2t4 1 1 sin t 2 2 202 1 = 4 + t4 − e−8t 4 4.2 15. 2s − 6 s2 + 9 = 16. s+1 s2 + 2 = 17. 1 s2 + 3s = 1 1 1 1 · − · 3 s 3 s+3 18. s+1 s2 − 4s = 1 1 5 1 − · + · 4 s 4 s−4 The Inverse Transform and Transforms of Derivatives s 3 −2· 2 = 2 cos 3t − 2 sin 3t +9 s +9 √ √ √ √ 2 2 s 1 +√ 2 = cos 2t + sin 2 t s2 + 2 s +2 2 2 2· s2 = 1 1 −3t − e 3 3 1 5 = − + e4t 4 4 s + 2s − 3 = 1 1 3 1 · + · 4 s−1 4 s+3 = 1 t 3 −3t e + e 4 4 20. 1 s2 + s − 20 = 1 1 1 1 · − · 9 s−4 9 s+5 = 1 4t 1 −5t e − e 9 9 21. 0.9s (s − 0.1)(s + 0.2) 22. s−3 √ √ (s − 3 )(s + 3 ) 23. s (s − 2)(s − 3)(s − 6) 24. s2 + 1 s(s − 1)(s + 1)(s − 2) 25. 1 s3 + 5s 26. s (s2 + 4)(s + 2) 19. 27. 28. 29. s2 (s2 s4 1 −9 = = 1 1 1 1 1 · − + · 2 s−2 s−3 2 s−6 = = = = 1 1 1 s · − 5 s 5 s2 + 5 = = 1 s 1 2 1 1 · + · − · 4 s2 + 4 4 s2 + 4 4 s + 2 2s − 4 s(s + 1)(s2 + 1) √ 1 1 − cos 5t 5 5 = 1 1 1 cos 2t + sin 2t − e−2t 4 4 4 4 3 s 3 − + + 2 + 2 s s+1 s +1 s +1 = = 6s + 3 (s2 + 1)(s2 + 4) 1 1 2t e − e3t + e6t 2 2 1 1 1 1 1 5 1 · − − · + · 2 s s−1 3 s+1 6 s−2 1 5 1 = − et − e−t + e2t 2 3 6 = −4 + 3e−t + cos t + 3 sin t √ √ 3 3 1 1 √ · 2 − √ · 2 6 3 s −3 6 3 s +3 1 (s2 + 1)(s2 + 4) = = 1 s(s2 + 5) = 2s − 4 + s)(s2 + 1) 1 1 + (0.6) · = 0.3e0.1t + 0.6e−0.2t s − 0.1 s + 0.2 √ √ √ √ √ 3 s − 3· 2 = cosh 3 t − 3 sinh 3 t 2−3 s s −3 (0.3) · = √ √ 1 1 = √ sinh 3 t − √ sin 3 t 6 3 6 3 1 1 1 1 · − · 3 s2 + 1 3 s2 + 4 1 1 = sin t − sin 2t 3 6 = 1 1 1 2 · − · 3 s2 + 1 6 s2 + 4 s 1 s 1 2 + −2· 2 − · s2 + 1 s2 + 1 s + 4 2 s2 + 4 1 = 2 cos t + sin t − 2 cos 2t − sin 2t 2 31. The Laplace transform of the initial-value problem is 1 s {y} − y(0) − {y} = . s 30. 2· 203 4.2 The Inverse Transform and Transforms of Derivatives Solving for {y} we obtain 1 1 {y} = − + . s s−1 Thus y = −1 + et . 32. The Laplace transform of the initial-value problem is 2s Solving for {y} − 2y(0) + {y} = 0. {y} we obtain 6 3 = . 2s + 1 s + 1/2 {y} = Thus y = 3e−t/2 . 33. The Laplace transform of the initial-value problem is s Solving for {y} − y(0) + 6 {y} = 1 . s−4 {y} we obtain {y} = 1 2 1 1 19 1 + = · + · . (s − 4)(s + 6) s + 6 10 s − 4 10 s + 6 Thus y= 1 4t 19 −6t e + e . 10 10 34. The Laplace transform of the initial-value problem is s Solving for {y} − {y} = s2 2s . + 25 {y} we obtain {y} = 2s 1 1 1 s 5 5 = · − + · 2 . 2 + 25) 2 + 25 (s − 1)(s 13 s − 1 13 s 13 s + 25 Thus y= 1 t 1 5 e − cos 5t + sin 5t. 13 13 13 35. The Laplace transform of the initial-value problem is s2 Solving for {y} − sy(0) − y (0) + 5 [s {y} − y(0)] + 4 {y} = 0. {y} we obtain {y} = s+5 4 1 1 1 = − . s2 + 5s + 4 3 s+1 3 s+4 Thus y= 4 −t 1 −4t e − e . 3 3 36. The Laplace transform of the initial-value problem is s2 {y} − sy(0) − y (0) − 4 [s {y} − y(0)] = 204 3 6 − . s−3 s+1 4.2 Solving for The Inverse Transform and Transforms of Derivatives {y} we obtain {y} = = 6 3 s−5 − + 2 2 − 4s) 2 − 4s) (s − 3)(s (s + 1)(s s − 4s 5 1 2 3 1 11 1 · − − · + · . 2 s s − 3 5 s + 1 10 s − 4 Thus y= 5 3 11 − 2e3t − e−t + e4t . 2 5 10 37. The Laplace transform of the initial-value problem is s2 Solving for {y} − sy(0) + {y} = 2 . s2 + 2 {y} we obtain {y} = 2 10s 10s 2 2 + = 2 + − . (s2 + 1)(s2 + 2) s2 + 1 s + 1 s2 + 1 s2 + 2 Thus y = 10 cos t + 2 sin t − √ √ 2 sin 2 t. 38. The Laplace transform of the initial-value problem is {y} + 9 s2 Solving for {y} = 1 . s−1 {y} we obtain {y} = 1 1 1 1 1 1 s = · − · 2 − · 2 . 2 + 9) (s − 1)(s 10 s − 1 10 s + 9 10 s + 9 Thus y= 1 t 1 1 e − sin 3t − cos 3t. 10 30 10 39. The Laplace transform of the initial-value problem is 2 s3 {y} − s2 (0) − sy (0) − y (0) + 3 s2 Solving for {y} − sy(0) − y (0) − 3[s {y} − y(0)] − 2 {y} = 1 . s+1 {y} we obtain {y} = 2s + 3 1 1 5 1 8 1 1 1 = + − + . (s + 1)(s − 1)(2s + 1)(s + 2) 2 s + 1 18 s − 1 9 s + 1/2 9 s + 2 Thus y= 1 −t 5 8 1 e + et − e−t/2 + e−2t . 2 18 9 9 40. The Laplace transform of the initial-value problem is s3 {y} − s2 (0) − sy (0) − y (0) + 2 s2 Solving for {y} − sy(0) − y (0) − [s {y} − y(0)] − 2 {y} we obtain {y} = = s2 + 12 (s − 1)(s + 1)(s + 2)(s2 + 9) 13 1 13 1 16 1 3 s 1 3 − + + − . 60 s − 1 20 s + 1 39 s + 2 130 s2 + 9 65 s2 + 9 205 {y} = 3 . s2 + 9 4.2 The Inverse Transform and Transforms of Derivatives Thus y= 13 t 13 −t 16 −2t 1 3 e − e + e cos 3t − sin 3t. + 60 20 39 130 65 41. The Laplace transform of the initial-value problem is {y} + s Solving for {y} = s+3 . s2 + 6s + 13 {y} we obtain {y} = = s+3 1 1 1 s+1 = · − · 2 2 + 6s + 13) (s + 1)(s 4 s + 1 4 s + 6s + 13 1 1 1 · − 4 s+1 4 Thus y= s+3 2 − (s + 3)2 + 4 (s + 3)2 + 4 . 1 −t 1 −3t 1 e − e cos 2t + e−3t sin 2t. 4 4 4 42. The Laplace transform of the initial-value problem is s2 Solving for {y} − s · 1 − 3 − 2[s {y} − 1] + 5 {y} = (s2 − 2s + 5) {y} − s − 1 = 0. {y} we obtain {y} = s+1 s−1+2 s−1 2 = = + . s2 − 2s + 5 (s − 1)2 + 22 (s − 1)2 + 22 (s − 1)2 + 22 Thus y = et cos 2t + et sin 2t. 43. (a) Differentiating f (t) = teat we get f (t) = ateat + eat so f (0) = 0. Writing the equation as {teat } + a and solving for {ateat + eat } = s {eat } = s {teat }, where we have used {teat } {teat } we get {teat } = 1 s−a {eat } = 1 . (s − a)2 (b) Starting with f (t) = t sin kt we have f (t) = kt cos kt + sin kt f (t) = −k 2 t sin kt + 2k cos kt. Then {−k 2 t sin t + 2k cos kt} = s2 {t sin kt} where we have used f (0) = 0 and f (0) = 0. Writing the above equation as −k 2 and solving for {t sin kt} + 2k {cos kt} = s2 {t sin kt} {t sin kt} gives {t sin kt} = 44. Let f1 (t) = 1 and f2 (t) = 2k s2 + k 2 1, t ≥ 0, t = 1 0, t=1 {cos kt} = . Then s 2k 2ks = 2 . s2 + k 2 s2 + k 2 (s + k 2 )2 {f1 (t)} = 206 {f2 (t)} = 1/s, but f1 (t) = f2 (t). 4.3 Translation Theorems 45. For y − 4y = 6e3t − 3e−t the transfer function is W (s) = 1/(s2 − 4s). The zero-input response is s−5 s2 − 4s y0 (t) = 5 1 1 1 · − · 4 s 4 s−4 = = 5 1 4t − e , 4 4 and the zero-state response is 6 3 − (s − 3)(s2 − 4s) (s + 1)(s2 − 4s) y1 (t) = 27 1 2 5 1 3 1 · − + · − · 20 s − 4 s − 3 4 s 5 s + 1 = = 27 4t 5 3 e − 2e3t + − e−t . 20 4 5 {f (t)} = sF (s) − f (0). From 46. From Theorem 4.4, if f and f are continuous and of exponential order, Theorem 4.5, lims→∞ {f (t)} = 0 so lim [sF (s) − f (0)] = 0 and s→∞ For f (t) = cos kt, lim sF (s) = lim s s→∞ s→∞ s2 lim F (s) = f (0). s→∞ s = 1 = f (0). + k2 EXERCISES 4.3 Translation Theorems 1 (s − 10)2 2. te−6t = t3 e−2t = 3! (s + 2)4 4. t10 e−7t = 5. t et + e2t 2 6. e2t (t − 1)2 = 7. et sin 3t = 1. te10t = 3. 9. 10. 11. te2t + 2te3t + te4t = = t2 e2t − 2te2t + e2t = e3t 9 − 4t + 10 sin 1 (s + 2)3 = t 2 2 1 2 − + (s − 2)3 (s − 2)2 s−2 8. e−2t cos 4t = {cos 5t − et cos 5t + 3e−4t cos 5t} = = 9e3t − 4te3t + 10e3t sin 2 1 2 (s + 2)3 = 10! (s + 7)11 2 1 1 + + 2 2 (s − 2) (s − 3) (s − 4)2 3 (s − 1)2 + 9 {(1 − et + 3e−4t ) cos 5t} = 1 (s + 6)2 1 2 −2t t e 2 207 s+2 (s + 2)2 + 16 s s−1 3(s + 4) − + s2 + 25 (s − 1)2 + 25 (s + 4)2 + 25 t 5 9 4 + = − 2 s − 3 (s − 3)2 (s − 3)2 + 1/4 4.3 Translation Theorems 12. 1 (s − 1)4 = 1 6 3! (s − 1)4 = 1 3 t t e 6 1 (s − 3)2 + 12 s2 16. = 1 2 2 (s + 1)2 + 22 s + 4s + 5 = 1 s+2 −2 2 + 12 (s + 2) (s + 2)2 + 12 s2 15. 1 + 2s + 5 s2 14. 1 − 6s + 10 s2 13. 2s + 5 + 6s + 34 = = 2 = e3t sin t = 1 −t e sin 2t 2 = e−2t cos t − 2e−2t sin t 1 (s + 3) 5 − 2 + 52 (s + 3) 5 (s + 3)2 + 52 17. s (s + 1)2 = s+1−1 (s + 1)2 18. 5s (s − 2)2 = 5(s − 2) + 10 (s − 2)2 19. 2s − 1 s2 (s + 1)3 20. (s + 1)2 (s + 2)4 1 = 2e−3t cos 5t − e−3t sin 5t 5 1 1 − s + 1 (s + 1)2 = = et − te−t 5 10 + s − 2 (s − 2)2 = 5 5 3 1 4 2 − − − − s s2 s + 1 (s + 1)2 2 (s + 1)3 = 1 2 1 3! − + (s + 2)2 (s + 2)3 6 (s + 2)4 = = 5e2t + 10te2t 3 = 5 − t − 5e−t − 4te−t − t2 e−t 2 1 = te−2t − t2 e−2t + t3 e−2t 6 21. The Laplace transform of the differential equation is s Solving for {y} − y(0) + 4 {y} = 1 . s+4 {y} we obtain {y} = 1 2 + . 2 (s + 4) s+4 Thus y = te−4t + 2e−4t . 22. The Laplace transform of the differential equation is s Solving for {y} − {y} = 1 1 + . s (s − 1)2 {y} we obtain {y} = 1 1 1 1 1 =− + . + + 3 s(s − 1) (s − 1) s s − 1 (s − 1)3 Thus 1 y = −1 + et + t2 et . 2 23. The Laplace transform of the differential equation is s2 Solving for {y} − sy(0) − y (0) + 2 s {y} − y(0) + {y} we obtain {y} = s+3 2 1 + = . (s + 1)2 s + 1 (s + 1)2 Thus y = e−t + 2te−t . 208 {y} = 0. 4.3 Translation Theorems 24. The Laplace transform of the differential equation is {y} − sy(0) − y (0) − 4 [s s2 {y} − y(0)] + 4 {y} = 6 . (s − 2)4 1 1 5 2t 5! . Thus, y = t e . 20 (s − 2)6 20 25. The Laplace transform of the differential equation is Solving for {y} we obtain {y} − sy(0) − y (0) − 6 [s s2 Solving for {y} = {y} − y(0)] + 9 {y} = 1 . s2 {y} we obtain {y} = 1 + s2 1 10 1 2 1 1 1 2 + + = − . 2 (s − 3)2 2 s 27 s 9 s 27 s − 3 9 (s − 3)2 Thus y= 2 1 2 10 + t − e3t + te3t . 27 9 27 9 26. The Laplace transform of the differential equation is {y} − sy(0) − y (0) − 4 [s s2 Solving for {y} − y(0)] + 4 {y} = 6 . s4 {y} we obtain {y} = s5 − 4s4 + 6 13 1 3 1 9 1 3 2 1 3! 1 1 + − = + + + . s4 (s − 2)2 4 s 8 s2 4 s3 4 s4 4 s−2 8 (s − 2)2 Thus y= 3 9 1 1 13 3 + t + t2 + t3 + e2t − te2t . 4 8 4 4 4 8 27. The Laplace transform of the differential equation is s2 Solving for {y} − sy(0) − y (0) − 6 [s {y} − y(0)] + 13 {y} = 0. {y} we obtain {y} = − Thus 3 3 2 =− . s2 − 6s + 13 2 (s − 3)2 + 22 3 y = − e3t sin 2t. 2 28. The Laplace transform of the differential equation is 2 s2 Solving for {y} − sy(0) + 20 s {y} − y(0) + 51 {y} = 0. {y} we obtain {y} = 2s2 Thus 4s + 40 2s + 20 2(s + 5) 10 = = + . 2 + 1/2 2 + 1/2 + 20s + 51 (s + 5) (s + 5) (s + 5)2 + 1/2 √ √ √ y = 2e−5t cos(t/ 2 ) + 10 2 e−5t sin(t/ 2 ). 29. The Laplace transform of the differential equation is s2 {y} − sy(0) − y (0) − [s {y} − y(0)] = 209 s−1 . (s − 1)2 + 1 4.3 Translation Theorems Solving for {y} we obtain {y} = s(s2 1 1 1 1 s−1 1 1 = − + . 2+1 − 2s + 2) 2 s 2 (s − 1) 2 (s − 1)2 + 1 Thus 1 1 t 1 − e cos t + et sin t. 2 2 2 y= 30. The Laplace transform of the differential equation is {y} − sy(0) − y (0) − 2 [s s2 Solving for {y} − y(0)] + 5 {y} = 1 1 . + s s2 {y} we obtain {y} = = 4s2 + s + 1 −7s/25 + 109/25 7 1 1 1 + = + s2 (s2 − 2s + 5) 25 s 5 s2 s2 − 2s + 5 7 1 1 1 7 51 s−1 2 − + . + 25 s 5 s2 25 (s − 1)2 + 22 25 (s − 1)2 + 22 Thus y= 7 51 1 7 + t − et cos 2t + et sin 2t. 25 5 25 25 31. Taking the Laplace transform of both sides of the differential equation and letting c = y(0) we obtain {y } + {y} − sy(0) − y (0) + 2s 2 s {2y } + {y} = 0 {y} − 2y(0) + {y} = 0 {y} − cs − 2 + 2s {y} − 2c + {y} = 0 2 s2 {y} = cs + 2c + 2 2c + 2 cs + {y} = 2 (s + 1) (s + 1)2 s + 2s + 1 =c = s+1−1 2c + 2 + (s + 1)2 (s + 1)2 c c+2 . + s + 1 (s + 1)2 Therefore, y(t) = c 1 s+1 1 (s + 1)2 + (c + 2) = ce−t + (c + 2)te−t . To find c we let y(1) = 2. Then 2 = ce−1 + (c + 2)e−1 = 2(c + 1)e−1 and c = e − 1. Thus y(t) = (e − 1)e−t + (e + 1)te−t . 32. Taking the Laplace transform of both sides of the differential equation and letting c = y (0) we obtain {y } + 2 s {8y } + {20y} = 0 {y} − y (0) + 8s {y} + 20 {y} = 0 {y} − c + 8s {y} + 20 {y} = 0 (s + 8s + 20) {y} = c s2 2 {y} = 210 c c = . s2 + 8s + 20 (s + 4)2 + 4 4.3 Translation Theorems Therefore, c (s + 4)2 + 4 y(t) = = c −4t e sin 2t = c1 e−4t sin 2t. 2 To find c we let y (π) = 0. Then 0 = y (π) = ce−4π and c = 0. Thus, y(t) = 0. (Since the differential equation is homogeneous and both boundary conditions are 0, we can see immediately that y(t) = 0 is a solution. We have shown that it is the only solution.) 33. Recall from Section 3.8 that mx = −kx − βx . Now m = W/g = 4/32 = 1 slug, and 4 = 2k so that k = 2 lb/ft. 8 Thus, the differential equation is x + 7x + 16x = 0. The initial conditions are x(0) = −3/2 and x (0) = 0. The Laplace transform of the differential equation is s2 Solving for 3 {x} + s + 7s 2 {x} + 21 + 16 2 {x} = 0. {x} we obtain √ √ −3s/2 − 21/2 15/2 3 s + 7/2 7 15 √ √ {x} = 2 =− − . s + 7s + 16 2 (s + 7/2)2 + ( 15/2)2 10 (s + 7/2)2 + ( 15/2)2 Thus √ √ √ 3 −7t/2 15 15 7 15 −7t/2 x=− e cos sin t− e t. 2 2 10 2 34. The differential equation is d2 q dq + 20 + 200q = 150, 2 dt dt The Laplace transform of this equation is s2 Solving for {q} + 20s q(0) = q (0) = 0. {q} + 200 150 . s {q} = {q} we obtain {q} = s(s2 150 3 1 3 s + 10 10 3 = − − . 2 + 102 + 20s + 200) 4 s 4 (s + 10) 4 (s + 10)2 + 102 Thus q(t) = 3 3 −10t 3 cos 10t − e−10t sin 10t − e 4 4 4 and i(t) = q (t) = 15e−10t sin 10t. 35. The differential equation is d2 q dq E0 + 2λ + ω 2 q = , 2 dt dt L The Laplace transform of this equation is s2 {q} + 2λs q(0) = q (0) = 0. {q} + ω 2 {q} = E0 1 L s or s2 + 2λs + ω 2 Solving for {q} = E0 1 . L s {q} and using partial fractions we obtain {q} = E0 L 1/ω 2 (1/ω 2 )s + 2λ/ω 2 − 2 s s + 2λs + ω 2 211 = E0 Lω 2 s + 2λ 1 − s s2 + 2λs + ω 2 . 4.3 Translation Theorems For λ > ω we write s2 + 2λs + ω 2 = (s + λ)2 − λ2 − ω 2 , so (recalling that ω 2 = 1/LC) {q} = E0 C 1 s+λ λ − − s (s + λ)2 − (λ2 − ω 2 ) (s + λ)2 − (λ2 − ω 2 ) . Thus for λ > ω, q(t) = E0 C 1 − e−λt cosh λ2 − ω 2 t − √ λ sinh − ω2 λ2 λ2 − ω 2 t . For λ < ω we write s2 + 2λs + ω 2 = (s + λ)2 + ω 2 − λ2 , so {q} = E0 C s+λ λ 1 − − s (s + λ)2 + (ω 2 − λ2 ) (s + λ)2 + (ω 2 − λ2 ) . Thus for λ < ω, q(t) = E0 C 1 − e−λt cos ω 2 − λ2 t − √ λ sin ω 2 − λ2 ω 2 − λ2 t . For λ = ω, s2 + 2λ + ω 2 = (s + λ)2 and {q} = 1/λ2 1/λ2 1/λ − − s s + λ (s + λ)2 E0 E0 1 = 2 L s(s + λ) L = E0 Lλ2 1 1 λ − − s s + λ (s + λ)2 . Thus for λ = ω, q(t) = E0 C 1 − e−λt − λte−λt . 36. The differential equation is dq 1 + q = E0 e−kt , q(0) = 0. dt C The Laplace transform of this equation is R Rs Solving for {q} + 1 C {q} = E0 1 . s+k {q} we obtain {q} = E0 C E0 /R = . (s + k)(RCs + 1) (s + k)(s + 1/RC) When 1/RC = k we have by partial fractions {q} = 1/(1/RC − k) 1/(1/RC − k) − s+k s + 1/RC E0 R Thus q(t) = = E0 1 R 1/RC − k E0 C e−kt − e−t/RC . 1 − kRC When 1/RC = k we have {q} = Thus q(t) = 37. (t − 1) 38. e2−t (t − 1) = (t − 2) = E0 1 . R (s + k)2 E0 −kt E0 −t/RC = . te te R R e−s s2 e−(t−2) (t − 2) = e−2s s+1 212 1 1 − s + k s + 1/RC . 4.3 Translation Theorems 39. t (t − 2) = {(t − 2) (t − 2) + 2 (t − 2)} = e−2s 2e−2s + 2 s s Alternatively, (16) of this section could be used: (t − 2)} = e−2s {t 40. (t − 1) = 3 (3t + 1) (t − 1) 1 2 + 2 s s {t + 2} = e−2s (t − 1) + 4 (t − 1) = . 3e−s 4e−s + s2 s Alternatively, (16) of this section could be used: (t − 1)} = e−s {(3t + 1) 41. (t − π) = cos 2t {cos 2(t − π) (t − π)} = {3t + 4} = e−s 3 4 + s2 s . se−πs s2 + 4 Alternatively, (16) of this section could be used: {cos 2t 42. t− sin t π 2 = (t − π)} = e−πs cos t − π 2 {cos 2(t + π)} = e−πs t− π 2 = {cos 2t} = e−πs s . s2 + 4 se−πs/2 s2 + 1 Alternatively, (16) of this section could be used: t− sin t π 2 43. e−2s s3 44. (1 + e−2s )2 s+2 45. e−πs s2 + 1 46. se−πs/2 s2 + 4 = cos 2 t − 47. e−s s(s + 1) = 48. e−2s s2 (s − 1) 49. (c) = e−πs/2 1 2 −2s · e 2 s3 = 2−4 56. 1− 57. t2 1 (t − 2)2 2 = sin(t − π) (t − π) = − sin t π 2 t− π 2 e−s e−s − s s+1 − = s . s2 + 1 (t − 2) + e−2(t−4) t− π 2 (t − 1) − e−(t−1) =− (t − 2) − (t − 2) 52. (b) 2 4 −3s − e s s 1 e−4s e−5s (t − 5) = − + s s s (t − 1)2 + 2t − 1 (t − 1) = (t − 1) (t − 2) + et−2 53. (a) (t − 1) = = 2 2 1 + 2+ s3 s s e−s 213 (t − 1)2 + 2(t − 1) − 1 (t − 2) 54. (d) (t − 3) = (t − 4) + (t − 4) (t − π) e−2s e−2s e−2s − 2 + s s s−1 51. (f ) {cos t} = e−πs/2 = e−2t + 2e−2(t−2) = − cos 2t = = e−πs/2 (t − 2) 1 2e−2s e−4s + + s+2 s+2 s+2 = 50. (e) 55. = π 2 sin t + (t − 1) 4.3 Translation Theorems Alternatively, by (16) of this section, (t − 1)} = e−s {t2 t− 3π 2 − cos t − 58. sin t 59. t−t 60. sin t − sin t 61. f (t) = (t − a) − (t − b) = 62. f (t) = (t − 1) + 3π 2 (t − 2) + = (t − 2) = t − (t − 2) (t − 2π) = 2 2 1 + 2+ s3 s s {t2 + 2t + 1} = e−s t− (t − 2) − 2 3π 2 =− se−3πs/2 s2 + 1 1 e−2s 2e−2s − 2 − 2 s s s (t − 2) = sin t − sin(t − 2π) . (t − 2π) = 1 e−2πs − 2 s2 + 1 s + 1 e−as e−bs − s s (t − 3) + · · · = e−s e−2s e−3s 1 e−s + + + ··· = s s s s 1 − e−s 63. The Laplace transform of the differential equation is {y} − y(0) + s Solving for {y} we obtain {y} = 5 −s e . s {y} = 5e−s 1 1 = 5e−s − . s(s + 1) s s+1 y=5 (t − 1) − 5e−(t−1) Thus (t − 1). 64. The Laplace transform of the differential equation is s Solving for {y} − y(0) + {y} = 1 2 −s − e . s s {y} we obtain {y} = 1 1 2e−s 1 1 1 − = − − 2e−s − . s(s + 1) s(s + 1) s s+1 s s+1 Thus y = 1 − e−t − 2 1 − e−(t−1) (t − 1). 65. The Laplace transform of the differential equation is s Solving for {y} − y(0) + 2 {y} = 1 s+1 − e−s 2 . 2 s s {y} we obtain {y} = 1 1 1 1 1 s+1 1 1 1 1 1 1 1 1 + − − e−s 2 =− + − e−s + . 2 2 + 2) s (s + 2) 4 s 2 s 4 s+2 4 s 2 s 4 s+2 s2 (s Thus 1 1 1 1 1 1 y = − + t + e−2t − + (t − 1) − e−2(t−1) 4 2 4 4 2 4 (t − 1). 66. The Laplace transform of the differential equation is s2 {y} − sy(0) − y (0) + 4 214 {y} = 1 e−s − . s s 4.3 Translation Theorems Solving for {y} we obtain {y} = 1−s 1 1 1 s 1 1 1 1 s 1 2 − e−s 2 = − − − e−s − . s(s2 + 4) s(s + 4) 4 s 4 s2 + 4 2 s2 + 4 4 s 4 s2 + 4 Thus y= 1 1 1 1 1 − cos 2t − sin 2t − − cos 2(t − 1) 4 4 2 4 4 (t − 1). 67. The Laplace transform of the differential equation is {y} − sy(0) − y (0) + 4 s2 Solving for {y} = e−2πs 1 . s2 + 1 {y} we obtain {y} = s2 s 1 2 1 1 + e−2πs − . 2+1 +4 3 s 6 s2 + 4 Thus y = cos 2t + 1 1 sin(t − 2π) − sin 2(t − 2π) 3 6 (t − 2π). 68. The Laplace transform of the differential equation is {y} − sy(0) − y (0) − 5 [s s2 Solving for {y} − y(0)] + 6 {y} = e−s . s {y} we obtain {y} = e−s = e−s 1 1 + s(s − 2)(s − 3) (s − 2)(s − 3) 1 1 1 1 1 1 1 1 − + − + . 6 s 2 s−2 3 s−3 s−2 s−3 Thus y= 1 1 2(t−1) 1 3(t−1) − e + e 6 2 3 (t − 1) − e2t + e3t . 69. The Laplace transform of the differential equation is s2 Solving for {y} − sy(0) − y (0) + {y} = e−2πs e−πs − . s s {y} we obtain {y} = e−πs 1 1 s s 1 − − e−2πs − + 2 . s s2 + 1 s s2 + 1 s +1 Thus y = [1 − cos(t − π)] (t − π) − [1 − cos(t − 2π)] (t − 2π) + sin t. 70. The Laplace transform of the differential equation is s2 Solving for {y} − sy(0) − y (0) + 4 s {y} − y(0) + 3 {y} = 1 e−2s e−4s e−6s − − + . s s s s {y} we obtain {y} = 1 1 1 1 1 1 1 1 1 1 1 1 − + − e−2s − + 3 s 2 s+1 6 s+3 3 s 2 s+1 6 s+3 − e−4s 1 1 1 1 1 1 1 1 1 1 1 1 − + + e−6s − + . 3 s 2 s+1 6 s+3 3 s 2 s+1 6 s+3 215 4.3 Translation Theorems Thus y= 1 1 −t 1 −3t 1 1 −(t−2) 1 −3(t−2) − + e − e + e − e 3 2 6 3 2 6 − 1 1 −(t−4) 1 −3(t−4) − e + e 3 2 6 (t − 4) + (t − 2) 1 1 −(t−6) 1 −3(t−6) − e + e 3 2 6 (t − 6). 71. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so that k = 16 lb/ft. Thus, the differential equation is x + 16x = f (t). The initial conditions are x(0) = 0, x (0) = 0. Also, since f (t) = 20t, 0 ≤ t < 5 t≥5 0, and 20t = 20(t − 5) + 100 we can write f (t) = 20t − 20t (t − 5) = 20t − 20(t − 5) (t − 5) − 100 (t − 5). The Laplace transform of the differential equation is {x} + 16 s2 Solving for {x} = 20 20 −5s 100 −5s − 2e − e . s2 s s {x} we obtain {x} = = 20 100 20 − e−5s − e−5s s2 (s2 + 16) s2 (s2 + 16) s(s2 + 16) 5 5 1 4 · 2− · 2 4 s 16 s + 16 1 − e−5s − 25 1 25 s · − · 2 e−5s . 4 s 4 s + 16 Thus x(t) = = 5 5 5 5 t− sin 4t − (t − 5) − sin 4(t − 5) 4 16 4 16 5 5 5 t− sin 4t − t 4 16 4 (t − 5) + 5 sin 4(t − 5) 16 (t − 5) − (t − 5) + 25 25 − cos 4(t − 5) 4 4 25 cos 4(t − 5) 4 (t − 5) (t − 5). 72. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so that k = 16 lb/ft. Thus, the differential equation is x + 16x = f (t). The initial conditions are x(0) = 0, x (0) = 0. Also, since f (t) = sin t, 0 ≤ t < 2π 0, t ≥ 2π and sin t = sin(t − 2π) we can write f (t) = sin t − sin(t − 2π) (t − 2π). The Laplace transform of the differential equation is s2 Solving for {x} + 16 {x} = 1 1 − e−2πs . s2 + 1 s2 + 1 {x} we obtain {x} = = 1 1 − e−2πs (s2 + 16) (s2 + 1) (s2 + 16) (s2 + 1) −1/15 1/15 −1/15 1/15 + − 2 + e−2πs . s2 + 16 s2 + 1 s + 16 s2 + 1 216 4.3 Translation Theorems Thus x(t) = − = 1 1 1 sin 4t + sin t + sin 4(t − 2π) 60 15 60 1 − 60 sin 4t + 1 15 (t − 2π) − 1 sin(t − 2π) 15 (t − 2π) sin t, 0 ≤ t < 2π t ≥ 2π. 0, 73. The differential equation is dq + 12.5q = 5 dt (t − 3). {q} + 5 2.5 2 −3s e . s The Laplace transform of this equation is s Solving for {q} = {q} we obtain 2 e−3s = s(s + 5) {q} = Thus q(t) = 2 5 2 1 2 1 · − · 5 s 5 s+5 2 (t − 3) − e−5(t−3) 5 e−3s . (t − 3). 74. The differential equation is dq + 10q = 30et − 30et (t − 1.5). dt The Laplace transform of this equation is 10 {q} − q0 + s Solving for {q} = 3 3e1.5 −1.5s . − e s − 1 s − 1.5 {q} we obtain {q} = q0 − 3 2 · 1 3 1 + · − 3e1.5 s+1 2 s−1 −2/5 2/5 + s+1 s − 1.5 e−1.5s . Thus q(t) = q0 − 3 2 3 6 e−t + et + e1.5 e−(t−1.5) − e1.5(t−1.5) 2 5 (t − 1.5). 75. (a) The differential equation is di 3π + 10i = sin t + cos t − dt 2 t− 3π 2 , i(0) = 0. The Laplace transform of this equation is s Solving for {i} + 10 {i} = 1 se−3πs/2 + 2 . s2 + 1 s +1 {i} we obtain {i} = = Thus i(t) = 1 s + e−3πs/2 (s2 + 1)(s + 10) (s2 + 1)(s + 10) 1 101 1 s 10 − + s + 10 s2 + 1 s2 + 1 + 1 101 −10 10s 1 + + s + 10 s2 + 1 s2 + 1 1 e−10t − cos t + 10 sin t 101 1 3π + −10e−10(t−3π/2) + 10 cos t − 101 2 217 + sin t − 3π 2 t− e−3πs/2 . 3π 2 . 4.3 Translation Theorems (b) i 0.2 1 2 5 4 3 t 6 -0.2 The maximum value of i(t) is approximately 0.1 at t = 1.7, the minimum is approximately −0.1 at 4.7. 76. (a) The differential equation is dq 1 + q = E0 [ dt 0.01 (t − 1) − (t − 3)], q(0) = 0 dq + 100q = E0 [ dt The Laplace transform of this equation is (t − 1) − (t − 3)], q(0) = 0. 50 or 50 {q} + 100 50s Solving for 1 −s 1 −3s . e − e s s {q} = E0 {q} we obtain {q} = E0 e−s e−3s E0 1 − = 50 s(s + 2) s(s + 2) 50 2 Thus q(t) = E0 100 1 − e−2(t−1) 1 1 − s s+2 e−s − 1 2 (t − 1) − 1 − e−2(t−3) 1 1 − s s+2 (t − 3) . q (b) 1 1 2 5 4 3 6 t The maximum value of q(t) is approximately 1 at t = 3. 77. The differential equation is d4 y = w0 [1 − (x − L/2)]. dx4 Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain EI s4 {y} − sy (0) − y (0) = w0 1 1 − e−Ls/2 . EI s Letting y (0) = c1 and y (0) = c2 we have {y} = c1 c2 w0 1 1 − e−Ls/2 + 4+ s3 s EI s5 so that y(x) = 1 1 1 w0 L c1 x2 + c2 x3 + x4 − x − 2 6 24 EI 2 4 x− To find c1 and c2 we compute y (x) = c1 + c2 x + 1 w0 L x2 − x − 2 EI 2 and 218 2 x− L 2 L 2 . e−3s . 4.3 Translation Theorems w0 L x− x− EI 2 y (x) = c2 + x− L 2 . Then y (L) = y (L) = 0 yields the system 1 w0 L2 − 2 EI c1 + c2 L + 2 L 2 = c1 + c2 L + w0 EI c2 + L 2 3 w0 L2 =0 8 EI = c2 + 1 w0 L = 0. 2 EI Solving for c1 and c2 we obtain c1 = 1 w0 L2 /EI and c2 = − 1 w0 L/EI. Thus 8 2 y(x) = w0 EI 1 1 1 1 2 2 L x − Lx3 + x4 − 16 12 24 24 x− 4 L 2 L 2 x− . 78. The differential equation is d4 y = w0 [ (x − L/3) − (x − 2L/3)]. dx4 Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain EI s4 {y} − sy (0) − y (0) = w0 1 −Ls/3 − e−2Ls/3 . e EI s Letting y (0) = c1 and y (0) = c2 we have c1 c2 w0 1 e−Ls/3 − e−2Ls/3 + 4+ s3 s EI s5 {y} = so that y(x) = 1 1 1 w0 c1 x2 + c2 x3 + 2 6 24 EI x− 4 L 3 x− L 3 − x− 2L 3 4 x− 2L 3 . To find c1 and c2 we compute y (x) = c1 + c2 x + 1 w0 2 EI x− 2 L 3 x− L 3 − x− 2L 3 2 x− 2L 3 and y (x) = c2 + w0 EI x− L 3 x− L 3 − x− 2L 3 x− 2L 3 . Then y (L) = y (L) = 0 yields the system 1 w0 2 EI 2L 3 2 − 2 L 3 c2 + c1 + c2 L + w0 EI = c1 + c2 L + 1 w0 L2 =0 6 EI 1 w0 L 2L L − = 0. = c2 + 3 3 3 EI Solving for c1 and c2 we obtain c1 = 1 w0 L2 /EI and c2 = − 1 w0 L/EI. Thus 6 3 y(x) = w0 EI 1 1 1 2 2 L x − Lx3 + 12 18 24 x− L 3 4 x− L 3 − x− 2L 3 79. The differential equation is EI d4 y 2w0 L L = −x+ x− dx4 L 2 2 219 x− L 2 . 4 x− 2L 3 . 4.3 Translation Theorems Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain s4 {y} − sy (0) − y (0) = 2w0 L 1 1 + 2 e−Ls/2 . − EIL 2s s2 s Letting y (0) = c1 and y (0) = c2 we have {y} = c1 c2 2w0 1 1 L + 4+ − 6 + 6 e−Ls/2 3 5 s s EIL 2s s s so that y(x) = = 1 1 2w0 L 4 1 5 1 c1 x2 + c2 x3 + x − x + 2 6 EIL 48 120 120 x− 5 L 2 5 1 5L 4 1 w0 L c1 x2 + c2 x3 + x − x5 + x − 2 6 60EIL 2 2 L 2 x− x− L 2 . To find c1 and c2 we compute y (x) = c1 + c2 x + w0 L 30Lx2 − 20x3 + 20 x − 60EIL 2 3 x− L 2 and y (x) = c2 + 2 w0 L 60Lx − 60x2 + 60 x − 60EIL 2 x− L 2 . Then y (L) = y (L) = 0 yields the system c1 + c2 L + w0 5 5w0 L2 30L3 − 20L3 + L3 = c1 + c2 L + =0 60EIL 2 24EI w0 w0 L c2 + [60L2 − 60L2 + 15L2 ] = c2 + = 0. 60EIL 4EI Solving for c1 and c2 we obtain c1 = w0 L2 /24EI and c2 = −w0 L/4EI. Thus y(x) = 5 w0 L2 2 5L 4 w0 L 3 w0 L x − x + x − x5 + x − 48EI 24EI 60EIL 2 2 x− 80. The differential equation is d4 y = w0 [1 − (x − L/2)]. dx4 Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain EI s4 {y} − sy (0) − y (0) = w0 1 1 − e−Ls/2 . EI s Letting y (0) = c1 and y (0) = c2 we have {y} = c1 c2 w0 1 1 − e−Ls/2 + 4+ s3 s EI s5 so that y(x) = 1 1 1 w0 L c1 x2 + c2 x3 + x4 − x − 2 6 24 EI 2 4 x− L 2 . To find c1 and c2 we compute y (x) = c1 + c2 x + 1 w0 L x2 − x − 2 EI 2 220 2 x− L 2 . L 2 . 4.3 Translation Theorems Then y(L) = y (L) = 0 yields the system 1 1 1 w0 c1 L2 + c2 L3 + L4 − 2 6 24 EI c1 + c2 L + 9 128 Solving for c1 and c2 we obtain c1 = y(x) = w0 EI L 2 4 = 1 w0 L2 − 2 EI 1 1 5w0 c1 L2 + c2 L3 + L4 = 0 2 6 128EI L 2 2 = c1 + c2 L + 3w0 2 L = 0. 8EI 57 w0 L2 /EI and c2 = − 128 w0 L/EI. Thus 19 1 1 9 2 2 L x − Lx3 + x4 − 256 256 24 24 x− L 2 4 x− L 2 . 81. (a) The temperature T of the cake inside the oven is modeled by where Tm dT = k(T − Tm ) dt is the ambient temperature of the oven. For 0 ≤ t ≤ 4, we have Tm = 70 + Hence for t ≥ 0, Tm = 300 − 70 t = 70 + 57.5t. 4−0 70 + 57.5t, 0 ≤ t < 4 t ≥ 4. 300, In terms of the unit step function, Tm = (70 + 57.5t)[1 (t − 4)] + 300 (t − 4) = 70 + 57.5t + (230 − 57.5t) (t − 4). The initial-value problem is then dT = k[T − 70 − 57.5t − (230 − 57.5t) dt (b) Let t(s) = (t − 4)], T (0) = 70. {T (t)}. Transforming the equation, using 230 − 57.5t = −57.5(t − 4) and Theorem 4.7, gives st(s) − 70 = k t(s) − or t(s) = 70 57.5 57.5 −4s − 2 + 2 e s s s 70 70k 57.5k 57.5k − − 2 + 2 e−4s . s − k s(s − k) s (s − k) s (s − k) After using partial functions, the inverse transform is then T (t) = 70 + 57.5 1 1 + t − ekt k k − 57.5 1 1 + t − 4 − ek(t−4) k k (t − 4). Of course, the obvious question is: What is k? If the cake is supposed to bake for, say, 20 minutes, then T (20) = 300. That is, 300 = 70 + 57.5 1 1 + 20 − e20k k k − 57.5 1 1 + 16 − e16k . k k But this equation has no physically meaningful solution. This should be no surprise since the model predicts the asymptotic behavior T (t) → 300 as t increases. Using T (20) = 299 instead, we find, with the help of a CAS, that k ≈ −0.3. 82. In order to apply Theorem 4.7 we need the function to have the form f (t − a) rewrite the functions given in the forms shown below. 221 (t − a). To accomplish this 4.3 Translation Theorems (a) 2t + 1 = 2(t − 1 + 1) + 1 = 2(t − 1) + 3 (b) et = et−5+5 = e5 et−5 (c) cos t = − cos(t − π) (d) t2 − 3t = (t − 2)2 + (t − 2) − 2 {tekti } = 1/(s − ki)2 . Then, using Euler’s formula, 83. (a) From Theorem 4.6 we have {tekti } = = {t cos kt + it sin kt} = {t cos kt} + i {t sin kt} 1 (s + ki)2 s2 − k 2 2ks = 2 = 2 +i 2 . 2 2 )2 (s − ki) (s + k (s + k 2 )2 (s + k 2 )2 Equating real and imaginary parts we have {t cos kt} = s2 − k 2 (s2 + k 2 )2 {t sin kt} = and (s2 2ks . + k 2 )2 (b) The Laplace transform of the differential equation is {x} + ω 2 s2 {x} we obtain Solving for {x} = s2 s . + ω2 {x} = s/(s2 + ω 2 )2 . Thus x = (1/2ω)t sin ωt. EXERCISES 4.4 Additional Operational Properties d ds 1 s + 10 = d ds s2 s +4 = d2 ds2 s2 1 −1 = 1. {te−10t } = − 3. {t cos 2t} = − 5. {t2 sinh t} = 6. {t2 cos t} = d2 ds2 s 2+1 s d ds 7. te2t sin 6t = − 8. te−3t cos 3t = − d ds = 2. s2 − 4 2 (s2 + 4) {t3 et } = (−1)3 d3 ds3 1 s−1 = 4. 1 (s + 10)2 {t sinh 3t} = − d ds 3 s2 − 9 = 6 (s − 1)4 6s 6s2 + 2 3 (s2 − 1) d ds 1 − s2 (s2 + 1)2 6 (s − 2)2 + 36 = s+3 (s + 3)2 + 9 = = 2s s2 − 3 3 (s2 + 1) 12(s − 2) 2 [(s − 2)2 + 36] (s + 3)2 − 9 2 [(s + 3)2 + 9] 9. The Laplace transform of the differential equation is s Solving for {y} + {y} = 2s . (s2 + 1)2 {y} we obtain {y} = 2s 1 1 1 s 1 1 1 s − + + =− + 2 . (s + 1)(s2 + 1)2 2 s + 1 2 s2 + 1 2 s2 + 1 (s2 + 1)2 (s + 1)2 222 2 (s2 − 9) 4.4 Thus 1 y(t) = − e−t − 2 1 = − e−t + 2 Additional Operational Properties 1 1 1 1 sin t + cos t + (sin t − t cos t) + t sin t 2 2 2 2 1 1 1 cos t − t cos t + t sin t. 2 2 2 10. The Laplace transform of the differential equation is {y} − s Solving for {y} = 2(s − 1) . ((s − 1)2 + 1)2 {y} we obtain {y} = 2 . ((s − 1)2 + 1)2 Thus y = et sin t − tet cos t. 11. The Laplace transform of the differential equation is {y} − sy(0) − y (0) + 9 s2 Letting y(0) = 2 and y (0) = 5 and solving for {y} = s . s2 + 9 {y} = {y} we obtain 2s + 5s + 19s − 45 5 s 2s + 2 + 2 = 2 . 2 + 9)2 (s s + 9 s + 9 (s + 9)2 3 2 Thus 5 1 sin 3t + t sin 3t. 3 6 y = 2 cos 3t + 12. The Laplace transform of the differential equation is s2 Solving for {y} − sy(0) − y (0) + {y} = 1 . s2 + 1 {y} we obtain {y} = s3 − s2 + s s 1 1 = 2 . − 2 + 2 2 + 1)2 (s s + 1 s + 1 (s + 1)2 Thus y = cos t − sin t + 1 1 sin t − t cos t 2 2 = cos t − 1 1 sin t − t cos t. 2 2 13. The Laplace transform of the differential equation is s2 {y} − sy(0) − y (0) + 16 {y} = {cos 4t − cos 4t (t − π)} or by (16) of Section 4.3 in the text, (s2 + 16) s − e−πs + 16 s =1+ 2 − e−πs s + 16 {y} = 1 + s2 Thus {y} = and y= {cos 4(t + π)} {cos 4t} = 1 + s2 s s − 2 e−πs . + 16 s + 16 1 s s − 2 e−πs + s2 + 16 (s2 + 16)2 (s + 16)2 1 1 1 sin 4t + t sin 4t − (t − π) sin 4(t − π) (t − π). 4 8 8 223 4.4 Additional Operational Properties 14. The Laplace transform of the differential equation is s2 {y} − sy(0) − y (0) + (s2 + 1) or Thus {y} = {y} = t− π + sin t 2 t− π 2 π 1 1 −πs/2 sin t + + e−πs/2 − e s s 2 1 1 −πs/2 =s+ − e + e−πs/2 {cos t} s s 1 1 s = s + − e−πs/2 + 2 e−πs/2 . s s s +1 {y} = s + s 1 1 s + − e−πs/2 + 2 e−πs/2 s2 + 1 s(s2 + 1) s(s2 + 1) (s + 1)2 = 1 s s + − − s2 + 1 s s2 + 1 = 1 − s 1 s − 2 s s +1 and y = 1 − 1 − cos t − = 1 − (1 − sin t) 15. 1− 1 s − s s2 + 1 e−πs/2 + π 2 (s2 e−πs/2 + s e−πs/2 (s2 + 1)2 s e−πs/2 + 1)2 π 1 π π + t− sin t − 2 2 2 2 π 1 π π t− − t− cos t t− . 2 2 2 2 t− 16. y 1 t− π 2 y 4 0.5 2 1 2 3 4 5 6 t 1 2 3 4 5 6 t -2 -0.5 -4 -1 17. From (7) of Section 4.2 in the text along with Theorem 4.8, dY d d {y } = − [s2 Y (s) − sy(0) − y (0)] = −s2 − 2sY + y(0), ds ds ds so that the transform of the given second-order differential equation is the linear first-order differential equation {ty } = − in Y (s): 4 3 4 or Y + Y =− 5. 3 s s s The solution of the latter equation is Y (s) = 4/s4 + c/s3 , so s2 Y + 3sY = − y(t) = {Y (s)} = 2 3 c 2 t + t . 3 2 18. From Theorem 4.8 in the text dY d d {y } = − [sY (s) − y(0)] = −s −Y ds ds ds so that the transform of the given second-order differential equation is the linear first-order differential equation in Y (s): 3 10 Y + − 2s Y = − . s s {ty } = − 224 4.4 Additional Operational Properties Using the integrating factor s3 e−s , the last equation yields 2 c 2 5 + 3 es . s3 s But if Y (s) is the Laplace transform of a piecewise-continuous function of exponential order, we must have, in view of Theorem 4.5, lims→∞ Y (s) = 0. In order to obtain this condition we require c = 0. Hence Y (s) = 5 s3 y(t) = 1 3! 6 = 5 4 s s s 19. 1 ∗ t3 = 21. e−t ∗ et cos t = t eτ dτ 23. = 0 {et } = t 24. cos τ dτ = 0 t 1 s e−τ cos τ dτ = = τ et−τ dτ = t 26. 1 s 1 s 0 0 22. {t sin t} = {et } = 0 1 s − 1 s+1 s+1 = s (s + 1)2 + 1 s (s2 + 2s + 2) d 1 ds s2 + 1 =− sin τ cos(t − τ ) dτ {sin t} = {cos t} = 0 t 29. t sin τ dτ =− 0 t 30. t τ e−τ dτ =− 0 31. 32. 33. 34. Using 1 s(s − 1) = 1 2 (s − 1) s = 1 3 (s − 1) s = 1 (s − a)2 d ds d ds 1 −2s 2 = 2 s (s2 + 1)2 (s2 + 1) 1 s2 (s − 1) t 28. 1 (s − 2)(s2 + 1) s 1 = 2 + 1) s +1 e−t cos t = {t} e2t ∗ sin t = s(s2 t 27. 2 s3 (s − 1)2 t2 ∗ tet = 1 s(s − 1) {cos t} = τ sin τ dτ 25. 5 2 t . 2 20. s−1 (s + 1) [(s − 1)2 + 1] 1 s = t s (s2 2 + 1) sin τ dτ =− d ds 1 1 s s2 + 1 τ e−τ dτ =− d ds 1 1 s (s + 1)2 0 t 0 1/(s − 1) s = 3s2 + 1 2 s2 (s2 + 1) = 3s + 1 + 1)3 s2 (s t eτ dτ = et − 1 = 0 1/s(s − 1) s t (eτ − 1)dτ = et − t − 1 = 0 1/s2 (s − 1) s t = 0 1 (eτ − τ − 1)dτ = et − t2 − t − 1 2 = teat , (8) in the text gives 1 s(s − a)2 35. (a) The result in (4) in the text is F (s) = t τ eaτ dτ = = 0 1 (ateat − eat + 1). a2 {F (s)G(s)} = f ∗ g, so identify (s2 2k 3 + k 2 )2 and 225 G(s) = 4s . s2 + k 2 4.4 Additional Operational Properties Then f (t) = sin kt − kt cos kt so 8k 3 s (s2 + k 2 )3 and g(t) = 4 cos kt t {F (s)G(s)} = f ∗ g = 4 = f (τ )g(t − τ )dt 0 t (sin kτ − kτ cos kτ ) cos k(t − τ )dτ. =4 0 Using a CAS to evaluate the integral we get 8k 3 s (s2 + k 2 )3 = t sin kt − kt2 cos kt. (b) Observe from part (a) that t(sin kt − kt cos kt) = and from Theorem 4.8 that 8k 3 s , + k 2 )3 (s2 tf (t) = −F (s). We saw in (5) in the text that {sin kt − kt cos kt} = 2k 3 /(s2 + k 2 )2 , so t(sin kt − kt cos kt) = − d 2k 3 8k 3 s = 2 . 2 + k 2 )2 ds (s (s + k 2 )3 36. The Laplace transform of the differential equation is s2 {y} + y 1 2s {y} = 2 . + (s + 1) (s2 + 1)2 50 Thus 1 2s + 2 (s2 + 1)2 (s + 1)3 and, using Problem 35 with k = 1, {y} = y= 5 -50 1 1 (sin t − t cos t) + (t sin t − t2 cos t). 2 4 37. The Laplace transform of the given equation is {f } + Solving for {f } we obtain {f } = s2 {t} {f } = {t}. 1 . Thus, f (t) = sin t. +1 38. The Laplace transform of the given equation is {f } = Solving for {f } we obtain {f } = {2t} − 4 {sin t} {f }. √ 2s2 + 2 5 8 2 1 + √ 2 = . s2 (s2 + 5) 5 s2 5 5 s +5 Thus f (t) = √ 2 8 t + √ sin 5 t. 5 5 5 39. The Laplace transform of the given equation is {f } = tet + 226 {t} {f }. 10 15 t 4.4 Solving for Additional Operational Properties {f } we obtain {f } = s2 1 1 3 1 2 1 1 1 = + . + − (s − 1)3 (s + 1) 8 s − 1 4 (s − 1)2 4 (s − 1)3 8 s+1 Thus 1 t 3 t 1 2 t 1 −t e + te + t e − e 8 4 4 8 f (t) = 40. The Laplace transform of the given equation is {f } + 2 Solving for {cos t} e−t + {f } = 4 {sin t}. {f } we obtain {f } = 4s2 + s + 5 4 2 7 = +4 . − (s + 1)3 s + 1 (s + 1)2 (s + 1)3 Thus f (t) = 4e−t − 7te−t + 4t2 e−t . 41. The Laplace transform of the given equation is {f } + Solving for {f } we obtain {f } = {1} {f } = {1}. 1 . Thus, f (t) = e−t . s+1 42. The Laplace transform of the given equation is {f } = Solving for e−t {cos t} + {f }. {f } we obtain {f } = s 1 + . s2 + 1 s2 + 1 Thus f (t) = cos t + sin t. 43. The Laplace transform of the given equation is {f } = = Solving for {1} + 1 8 1 + 2+ s s 3 {t} − {t3 } 8 3 t (t − τ )3 f (τ ) dτ 0 {f } = 1 16 1 + 2+ 4 s s s {f } we obtain {f } = s2 (s + 1) 1 1 3 1 1 2 1 s = + + + . 4 − 16 2+4 s 8 s+2 8 s−2 4 s 2 s2 + 4 Thus f (t) = 1 −2t 3 2t 1 1 + e + sin 2t + cos 2t. e 8 8 4 2 44. The Laplace transform of the given equation is {t} − 2 Solving for {f }. {f } we obtain {f } = {f } = et − e−t {f }. s2 − 1 1 1 1 3! = − . 4 2 2s 2 s 12 s4 227 4.4 Additional Operational Properties Thus f (t) = 1 1 t − t3 . 2 12 45. The Laplace transform of the given equation is {y} − y(0) = s Solving for {1} − {sin t} − {1} {y}. {f } we obtain {y} = s2 − s + 1 1 1 2s = 2 . − (s2 + 1)2 s + 1 2 (s2 + 1)2 Thus 1 t sin t. 2 y = sin t − 46. The Laplace transform of the given equation is {y} − y(0) + 6 s Solving for {f } we obtain {y} = {y} + 9 {1} {y} = 1 . Thus, y = te−3t . (s + 3)2 i 30 47. The differential equation is 0.1 t di 1 + 3i + dt 0.05 or di + 30i + 200 dt (t − 1) − i(τ )dτ = 100 (t − 2) 0 20 10 t i(τ )dτ = 1000 (t − 1) − 0.5 1 1.5 2 2.5 3 t (t − 2) , 0 -10 where i(0) = 0. The Laplace transform of the differential equation is {i} − y(0) + 30 s Solving for {1}. 200 {i} + s -20 -30 1000 −s {i} = (e − e−2s ). s {i} we obtain {i} = 1000e−s − 1000e−2s = s2 + 30s + 200 100 100 − (e−s − e−2s ). s + 10 s + 20 Thus i(t) = 100 e−10(t−1) − e−20(t−1) (t − 1) − 100 e−10(t−2) − e−20(t−2) i 48. The differential equation is 0.005 2 t di 1 +i+ dt 0.02 i(τ )dτ = 100 t − (t − 1) (t − 1) 0 or di + 200i + 10,000 dt 1.5 t i(τ )dτ = 20,000 t − (t − 1) (t − 1) , {i} + 200 {i} + 1 0 where i(0) = 0. The Laplace transform of the differential equation is s (t − 2). 10,000 s {i} = 20,000 1 1 − 2 e−s . 2 s s 228 0.5 0.5 1 1.5 2 t 4.4 Solving for Additional Operational Properties {i} we obtain {i} = 20,000 2 2 200 (1 − e−s ). (1 − e−s ) = − − 2 s(s + 100) s s + 100 (s + 100)2 Thus i(t) = 2 − 2e−100t − 200te−100t − 2 49. 50. {f (t)} = {f (t)} = a 1 1 − e−2as 1 1 − e−2as 2a e−st dt − 0 a (t − 1) + 2e−100(t−1) e−st dt = a e−st dt = 0 (t − 1) + 200(t − 1)e−100(t−1) (t − 1). (1 − e−as )2 1 − e−as = −2as ) s(1 − e s(1 + e−as ) 1 s(1 + e−as ) 51. Using integration by parts, 1 1 − e−bs {f (t)} = 52. 53. 54. {f (t)} = {f (t)} = {f (t)} = 1 1 1 − e−2s 0 π 1 1 − e−πs 0 e−st sin t dt = π 1 1 − bs bs e − 1 . 1 − e−s s2 (1 − e−2s ) eπs/2 + e−πs/2 πs 1 1 · πs/2 coth = 2 s2 + 1 e s +1 2 − e−πs/2 e−st sin t dt = 0 a −st a te dt = b s (2 − t)e−st dt = 1 0 1 1 − e−2πs 2 te−st dt + b 1 1 · s2 + 1 1 − e−πs 55. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is Ls From Problem 49 we have {i} + R {i} = {E(t)}. {E(t)} = (1 − e−s )/s(1 + e−s ). Thus (Ls + R) and {i} = {i} = 1 − e−s s(1 + e−s ) 1 1 − e−s 1 1 − e−s 1 = −s ) L s(s + R/L)(1 + e L s(s + R/L) 1 + e−s = 1 1 1 − (1 − e−s )(1 − e−s + e−2s − e−3s + e−4s − · · · ) R s s + R/L = 1 1 1 − (1 − 2e−s + 2e−2s − 2e−3s + 2e−4s − · · · ). R s s + R/L Therefore, 1 2 (t − 1) 1 − e−Rt/L − 1 − e−R(t−1)/L R R 2 2 + (t − 2) − 1 − e−R(t−2)/L 1 − e−R(t−3)/L R R ∞ 1 2 = 1 − e−R(t−n)/L (t − n). 1 − e−Rt/L + R R n=1 i(t) = 229 (t − 3) + · · · 4.4 Additional Operational Properties The graph of i(t) with L = 1 and R = 1 is shown below. i 1 0.5 1 2 3 4 t -0.5 -1 56. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is Ls {i} + R {i} = {E(t)}. From Problem 51 we have {E(t)} = 1 1 1 − s s es − 1 = 1 1 1 − . s2 s es − 1 Thus (Ls + R) {i} = 1 1 1 − 2 s s es − 1 and {i} = = 1 1 1 1 1 − L s2 (s + R/L) L s(s + R/L) es − 1 1 1 L 1 L 1 + − 2 R s R s R s + R/L − 1 1 1 − R s s + R/L e−s + e−2s + e−3s + · · · . Therefore i(t) = 1 R t− − = 1 R L L + e−Rt/L R R − 1 1 − e−R(t−2)/L R t− L L + e−Rt/L R R 1 (t − 1) 1 − e−R(t−1)/L R 1 (t − 2) − 1 − e−R(t−3)/L R (t − 3) − · · · ∞ − 1 1 − e−R(t−n)/L R n=1 (t − n). The graph of i(t) with L = 1 and R = 1 is shown below. i 1 0.5 1 2 3 4 t -0.5 -1 57. The differential equation is x + 2x + 10x = 20f (t), where f (t) is the meander function in Problem 49 with 230 4.4 Additional Operational Properties a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform we obtain 20 1 (1 − e−πs ) s 1 + e−πs 20 (1 − e−πs )(1 − e−πs + e−2πs − e−3πs + · · ·) = s 20 = (1 − 2e−πs + 2e−2πs − 2e−3πs + · · ·) s {x(t)} = (s2 + 2s + 10) ∞ 20 40 = (−1)n e−nπs . + s s n=1 Then ∞ {x(t)} = 20 40 (−1)n e−nπs + s(s2 + 2s + 10) s(s2 + 2s + 10) n=1 ∞ = 4 2 2s + 4 4s + 8 (−1)n − + − e−nπs s s2 + 2s + 10 n=1 s s2 + 2s + 10 = (s + 1) + 1 1 2 2(s + 1) + 2 − +4 − e−nπs (−1)n s (s + 1)2 + 9 s (s + 1)2 + 9 n=1 ∞ and ∞ 1 (−1)n 1 − e−(t−nπ) cos 3(t − nπ) x(t) = 2 1 − e−t cos 3t − e−t sin 3t + 4 3 n=1 1 − e−(t−nπ) sin 3(t − nπ) 3 (t − nπ). The graph of x(t) on the interval [0, 2π) is shown below. x 3 π 2π t −3 58. The differential equation is x + 2x + x = 5f (t), where f (t) is the square wave function with a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform, we obtain (s2 + 2s + 1) {x(t)} = 5 5 1 = (1 − e−πs + e−2πs − e−3πs + e−4πs − · · ·) −πs s 1+e s ∞ = Then {x(t)} = 5 s(s + 1)2 ∞ n=0 5 (−1)n e−nπs . s n=0 (−1)n e−nπs = 5 ∞ (−1)n n=0 231 1 1 1 − − s s + 1 (s + 1)2 e−nπs 4.4 Additional Operational Properties and ∞ x(t) = 5 (−1)n (1 − e−(t−nπ) − (t − nπ)e−(t−nπ) ) (t − nπ). n=0 The graph of x(t) on the interval [0, 4π) is shown below. x 5 2π 4π t −5 59. f (t) = − 1 t d [ln(s − 3) − ln(s + 1)] ds =− 1 t 1 1 − s−3 s+1 =− 1 3t e − e−t t 60. The transform of Bessel’s equation is − d 2 d [s Y (s) − sy(0) − y (0)] + sY (s) − y(0) − Y (s) = 0 ds ds or, after simplifying and using the initial condition, (s2 + 1)Y + sY = 0. This equation is both separable and √ linear. Solving gives Y (s) = c/ s2 + 1 . Now Y (s) = {J0 (t)}, where J0 has a derivative that is continuous and of exponential order, implies by Problem 46 of Exercises 4.2 that 1 = J0 (0) = lim sY (s) = c lim √ s→∞ s→∞ s2 s =c + k2 so c = 1 and Y (s) = √ 1 s2 + 1 or {J0 (t)} = √ 1 . s2 + 1 61. (a) Using Theorem 4.8, the Laplace transform of the differential equation is − d 2 d [s Y − sy(0) − y (0)] + sY − y(0) + [sY − y(0)] + nY ds ds d d = − [s2 Y ] + sY + [sY ] + nY ds ds dY dY = −s2 − 2sY + sY + s + Y + nY ds ds = (s − s2 ) dY ds + (1 + n − s)Y = 0. Separating variables, we find dY 1+n−s = ds = Y s2 − s n 1+n − s−1 s ln Y = n ln(s − 1) − (1 + n) ln s + c Y = c1 (s − 1)n . s1+n 232 ds 4.4 Additional Operational Properties Since the differential equation is homogeneous, any constant multiple of a solution will still be a solution, so for convenience we take c1 = 1. The following polynomials are solutions of Laguerre’s differential equation: n=0: L0 (t) = 1 s n=1: L1 (t) = s−1 s2 n=2: L2 (t) = (s − 1)2 s3 = 1 2 1 − 2+ 3 s s s n=3: L3 (t) = (s − 1)3 s4 = 1 3 1 3 + 3− 4 − s s2 s s n=4: L4 (t) = =1 1 1 − s s2 = (s − 1)4 = s5 2 1 = 1 − 4t + 3t2 − t3 + t4 . 3 24 =1−t 1 = 1 − 2t + t2 2 3 1 = 1 − 3t + t2 − t3 2 6 1 4 6 4 1 − + 3− 4+ 5 s s2 s s s (b) Letting f (t) = tn e−t we note that f (k) (0) = 0 for k = 0, 1, 2, . . . , n − 1 and f (n) (0) = n!. Now, by the first translation theorem, et dn n −t t e n! dtn where Y = 1 1 {et f (n) (t)} = {f (n) (t)} s→s−1 n! n! 1 n s = {tn e−t } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0) n! 1 n = {tn e−t } s n! s→s−1 n! 1 n (s − 1)n = = = Y, s n+1 n! (s + 1) sn+1 s→s−1 = s→s−1 {Ln (t)}. Thus Ln (t) = et dn n −t (t e ), n! dtn n = 0, 1, 2, . . . . 62. The output for the first three lines of the program are 9y[t] + 6y [t] + y [t] == t sin[t] 1 − 2s + 9Y + s2 Y + 6(−2 + sY ) == Y →− 2s (1 + s2 )2 −11 − 4s − 22s2 − 4s3 − 11s4 − 2s5 (1 + s2 )2 (9 + 6s + s2 ) The fourth line is the same as the third line with Y → removed. The final line of output shows a solution involving complex coefficients of eit and e−it . To get the solution in more standard form write the last line as two lines: euler={Eˆ(It)−>Cos[t] + I Sin[t], Eˆ(-It)−>Cos[t] - I Sin[t]} InverseLaplaceTransform[Y, s, t]/.euler//Expand We see that the solution is y(t) = 487 247 1 + t e−3t + (13 cos t − 15t cos t − 9 sin t + 20t sin t) . 250 50 250 63. The solution is y(t) = √ 1 t 1 −t/2 cos 15 t − e − e 6 6 233 √ 3/5 −t/2 sin 15 t. e 6 4.4 Additional Operational Properties 64. The solution is q(t) = 1 − cos t + (6 − 6 cos t) (t − 3π) − (4 + 4 cos t) (t − π). q 5 -5 Π 3Π t 5Π EXERCISES 4.5 The Dirac Delta Function 1. The Laplace transform of the differential equation yields {y} = 1 −2s e s−3 so that y = e3(t−2) (t − 2). 2. The Laplace transform of the differential equation yields {y} = 2 e−s + s+1 s+1 so that y = 2e−t + e−(t−1) (t − 1). 3. The Laplace transform of the differential equation yields 1 {y} = 2 1 + e−2πs s +1 so that y = sin t + sin t (t − 2π). 4. The Laplace transform of the differential equation yields 1 4 {y} = e−2πs 2 + 16 4 s so that 1 1 y = sin 4(t − 2π) (t − 2π) = sin 4t 4 4 5. The Laplace transform of the differential equation yields 1 {y} = 2 e−πs/2 + e−3πs/2 s +1 so that π π 3π y = sin t − t− + sin t − 2 2 2 π π = − cos t t− + cos t t− . 2 2 234 (t − 2π). t− 3π 2 4.5 The Dirac Delta Function 6. The Laplace transform of the differential equation yields {y} = s 1 + (e−2πs + e−4πs ) s2 + 1 s2 + 1 so that (t − 2π) + y = cos t + sin t[ (t − 4π)]. 7. The Laplace transform of the differential equation yields {y} = 1 1 1 1 1 (1 + e−s ) = − (1 + e−s ) s2 + 2s 2 s 2 s+2 so that y= 1 1 −2t 1 1 −2(t−1) + − e − e 2 2 2 2 (t − 1). 8. The Laplace transform of the differential equation yields {y} = s+1 1 3 1 1 1 1 1 −2s 3 1 1 1 + e−2s = − − − e + 2 − 2) s(s − 2) 4 s−2 4 s 2 s 2 s−2 2 s s2 (s so that y= 3 2t 3 1 1 1 e − − t + e2(t−2) − 4 4 2 2 2 (t − 2). 9. The Laplace transform of the differential equation yields {y} = 1 e−2πs (s + 2)2 + 1 so that y = e−2(t−2π) sin t (t − 2π). 10. The Laplace transform of the differential equation yields {y} = 1 e−s (s + 1)2 so that y = (t − 1)e−(t−1) (t − 1). 11. The Laplace transform of the differential equation yields {y} = = s2 4+s e−πs + e−3πs + 2 + 4s + 13 s + 4s + 13 2 3 3 s+2 1 e−πs + e−3πs + + 2 + 32 2 + 32 3 (s + 2) (s + 2) 3 (s + 2)2 + 32 so that y= 2 −2t 1 sin 3t + e−2t cos 3t + e−2(t−π) sin 3(t − π) e 3 3 1 + e−2(t−3π) sin 3(t − 3π) (t − 3π). 3 (t − π) 12. The Laplace transform of the differential equation yields {y} = 1 e−2s + e−4s + 2 (s − 6) (s − 1) (s − 1)(s − 6) =− 1 1 1 1 1 1 1 1 1 1 + − + − + 25 s − 1 5 (s − 1)2 25 s − 6 5 s−1 5 s−6 235 e−2s + e−4s 4.5 The Dirac Delta Function so that y=− 1 t 1 t 1 1 1 e − te + e6t + − et−2 + e6(t−2) 25 5 25 5 5 1 1 (t − 2) + − et−4 + e6(t−4) 5 5 (t − 4). 13. The Laplace transform of the differential equation yields {y} = 1 2 1 3! 1 P0 3! −Ls/2 y (0) + y (0) + e 3 4 2 s 6 s 6 EI s4 so that y= 1 1 1 P0 y (0)x2 + y (0)x3 + 2 6 6 EI 3 L 2 L 2 . L 2 . 0≤x< x− L 2 x− Using y (L) = 0 and y (L) = 0 we obtain y= = 3 1 P0 L 2 1 P 0 3 1 P 0 L x − x + x− 4 EI 6 EI 6 EI 2 P0 L x2 − 1 x3 , 0 ≤ x < L EI 4 6 2 P0 L2 4EI 1 L x− 2 12 x− L 2 L ≤ x ≤ L. 2 , 14. From Problem 13 we know that y= 1 1 1 P0 y (0)x2 + y (0)x3 + 2 6 6 EI x− 3 L 2 x− Using y(L) = 0 and y (L) = 0 we obtain y= = 1 P0 L 2 1 x − 16 EI 12 P0 L x2 − EI 16 P0 EI P0 3 1 P0 x + EI 6 EI x− L 2 3 x− 1 3 x , 12 1 L 2 x − x3 16 12 + 1 P0 6 EI x− L 2 3 , L 2 L ≤ x ≤ L. 2 15. You should disagree. Although formal manipulations of the Laplace transform lead to y(t) = 1 e−t sin 3t in both 3 cases, this function does not satisfy the initial condition y (0) = 0 of the second initial-value problem. 236 4.6 Systems of Linear Differential Equations EXERCISES 4.6 Systems of Linear Differential Equations 1. Taking the Laplace transform of the system gives s s {x} = − {x} + {y} − 1 = 2 {y} {x} so that {x} = 1 1 1 1 1 = − (s − 1)(s + 2) 3 s−1 3 s+2 {y} = 1 2 2 1 1 1 + = + . s s(s − 1)(s + 2) 3 s−1 3 s+2 and Then x= 1 t 1 −2t e − e 3 3 and y= 2 t 1 −2t e + e . 3 3 2. Taking the Laplace transform of the system gives s s 1 s−1 1 {x} − 2 s {x} − 1 = 2 {y} + {y} − 1 = 8 so that {y} = and y= s3 + 7s2 − s + 1 1 1 8 1 173 1 53 1 = − + − 2 − 16) s(s − 1)(s 16 s 15 s − 1 96 s − 4 160 s + 4 1 8 173 4t 53 −4t − et + e − e . 16 15 96 160 Then x= 1 1 173 4t 53 −4t 1 1 y + t = t − et + e + e . 8 8 8 15 192 320 3. Taking the Laplace transform of the system gives s {x} + 1 = s {x} − 2 {y} {y} − 2 = 5 {x} − {y} so that {x} = −s − 5 s 5 3 =− 2 − 2+9 s s + 9 3 s2 + 9 and x = − cos 3t − 5 sin 3t. 3 Then y= 1 7 1 x − x = 2 cos 3t − sin 3t. 2 2 3 237 4.6 Systems of Linear Differential Equations 4. Taking the Laplace transform of the system gives {x} + s {y} = 1 s {x} + (s − 1) {y} = 1 s−1 (s + 3) (s − 1) so that {y} = 5s − 1 4 1 1 1 1 1 + + =− 3s(s − 1)2 3 s 3 s − 1 3 (s − 1)2 {x} = 1 − 2s 1 1 1 1 1 1 − − = . 3s(s − 1)2 3 s 3 s − 1 3 (s − 1)2 and Then x= 1 1 t 1 t − e − te 3 3 3 1 1 4 y = − + et + tet . 3 3 3 and 5. Taking the Laplace transform of the system gives (2s − 2) (s − 3) {x} + s {x} + (s − 3) 1 s 2 {y} = s {y} = so that {x} = −s − 3 1 1 5 1 2 =− + − s(s − 2)(s − 3) 2 s 2 s−2 s−3 {y} = 3s − 1 1 1 5 1 8 1 =− − + . s(s − 2)(s − 3) 6 s 2 s−2 3 s−3 and Then 1 5 x = − + e2t − 2e3t 2 2 1 5 8 y = − − e2t + e3t . 6 2 3 and 6. Taking the Laplace transform of the system gives (s + 1) {x} − (s − 1) {y} = −1 s {x} + (s + 2) {y} = 1 so that s + 1/2 s + 1/2 √ = s2 + s + 1 (s + 1/2)2 + ( 3/2)2 √ √ −3/2 3/2 √ {x} = 2 . =− 3 2 + ( 3/2)2 s +s+1 (s + 1/2) {y} = and Then √ −t/2 y=e 3 cos t 2 and √ x = − 3 e−t/2 sin 7. Taking the Laplace transform of the system gives (s2 + 1) − {x} − {x} + (s2 + 1) {y} = −2 {y} = 1 so that {x} = −2s2 − 1 1 1 3 1 =− 2 − s4 + 2s2 2 s 2 s2 + 2 and 238 √ 3 t. 2 4.6 Systems of Linear Differential Equations √ 1 3 x = − t − √ sin 2 t. 2 2 2 Then √ 1 3 y = x + x = − t + √ sin 2 t. 2 2 2 8. Taking the Laplace transform of the system gives {x} + {y} = 1 {x} − (s + 1) {y} = 1 (s + 1) 4 so that {x} = s+2 s+1 2 1 = + s2 + 2s + 5 (s + 1)2 + 22 2 (s + 1)2 + 22 and {y} = Then s2 −s + 3 s+1 2 =− +2 . 2 + 22 + 2s + 5 (s + 1) (s + 1)2 + 22 1 x = e−t cos 2t + e−t sin 2t 2 and y = −e−t cos 2t + 2e−t sin 2t. 9. Adding the equations and then subtracting them gives d2 x 1 = t2 + 2t 2 dt 2 d2 y 1 = t2 − 2t. dt2 2 Taking the Laplace transform of the system gives 1 1 3! 1 4! {x} = 8 + + 5 s 24 s 3 s4 and {y} = so that x=8+ 1 4 1 3 t + t 24 3 1 4! 1 3! − 5 24 s 3 s4 and y= 1 4 1 3 t − t . 24 3 10. Taking the Laplace transform of the system gives (s − 4) (s + 2) {x} + s3 {x} − 2s3 {y} = s2 6 +1 {y} = 0 so that {x} = 4 4 1 4 s 8 1 = − − (s − 2)(s2 + 1) 5 s − 2 5 s2 + 1 5 s2 + 1 {y} = 2s + 4 2 1 1 1 2 6 s 8 1 = − 2 −2 3 + − + . s3 (s − 2)(s2 + 1) s s s 5 s − 2 5 s2 + 1 5 s2 + 1 and Then x= and 4 2t 4 8 e − cos t − sin t 5 5 5 1 6 8 y = 1 − 2t − 2t2 + e2t − cos t + sin t. 5 5 5 239 4.6 Systems of Linear Differential Equations 11. Taking the Laplace transform of the system gives {x} + 3(s + 1) s2 s2 so that {x} = − {x} + 3 {y} = 2 {y} = 1 (s + 1)2 2s + 1 1 1 1 2 1 = + 2+ . − 3 + 1) s s 2 s s+1 s3 (s Then 1 x = 1 + t + t2 − e−t 2 and y= 1 1 1 1 −t 1 te − x = te−t + e−t − . 3 3 3 3 3 12. Taking the Laplace transform of the system gives {x} + 2 {y} = 2e−s s {x} + (s + 1) {y} = 1 e−s + 2 s (s − 4) −3 so that {x} = −1/2 1 + e−s (s − 1)(s − 2) (s − 1)(s − 2) = 1 1 1 1 1 1 − + e−s − + 2 s−1 2 s−2 s−1 s−2 {y} = e−s −s/2 + 2 s/4 − 1 + + e−s s (s − 1)(s − 2) (s − 1)(s − 2) and = 1 1 1 3 1 1 3 1 − + e−s − + . 4 s−1 2 s−2 s 2 s−1 s−2 Then x= 1 t 1 2t e − e + −et−1 + e2(t−1) 2 2 y= 3 t 1 2t 3 e − e + 1 − et−1 + e2(t−1) 4 2 2 and (t − 1) 13. The system is x1 = −3x1 + 2(x2 − x1 ) x2 = −2(x2 − x1 ) x1 (0) = 0 x1 (0) = 1 x2 (0) = 1 x2 (0) = 0. Taking the Laplace transform of the system gives {x1 } − 2 {x2 } = 1 {x1 } + (s2 + 2) {x2 } = s (s2 + 5) −2 240 (t − 1). 4.6 so that and Systems of Linear Differential Equations √ 2 s 1 1 2 s 4 6 s2 + 2s + 2 {x1 } = 4 = + − + √ 2 2+6 2+1 2+1 2+6 s + 7s 5 s 5 s 5 s 5 6 s +6 √ 6 s3 + 5s + 2 4 s 2 1 1 s 2 {x2 } = 2 = + + − √ 2 . 2 + 6) 2+1 2+1 2+6 (s + 1)(s 5 s 5 s 5 s 5 6 s +6 Then x1 = √ √ 2 4 1 2 cos t + sin t − cos 6 t + √ sin 6 t 5 5 5 5 6 x2 = √ √ 4 2 2 1 cos t + sin t + cos 6 t − √ sin 6 t. 5 5 5 5 6 and 14. In this system x1 and x2 represent displacements of masses m1 and m2 from their equilibrium positions. Since the net forces acting on m1 and m2 are −k1 x1 + k2 (x2 − x1 ) and − k2 (x2 − x1 ) − k3 x2 , respectively, Newton’s second law of motion gives m1 x1 = −k1 x1 + k2 (x2 − x1 ) m2 x2 = −k2 (x2 − x1 ) − k3 x2 . Using k1 = k2 = k3 = 1, m1 = m2 = 1, x1 (0) = 0, x1 (0) = −1, x2 (0) = 0, and x2 (0) = 1, and taking the Laplace transform of the system, we obtain {x1 } − {x2 } = −1 {x1 } − (2 + s2 ) {x2 } = −1 (2 + s2 ) so that 1 +3 and √ 1 x1 = − √ sin 3 t 3 and {x1 } = − Then s2 {x2 } = s2 1 . +3 √ 1 x2 = √ sin 3 t. 3 15. (a) By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop we have E(t) = Ri1 + L1 i2 and E(t) = Ri1 + L2 i3 or L1 i2 + Ri2 + Ri3 = E(t) and L2 i3 + Ri2 + Ri3 = E(t). (b) Taking the Laplace transform of the system 0.01i2 + 5i2 + 5i3 = 100 0.0125i3 + 5i2 + 5i3 = 100 gives {i2 } + 500 {i3 } = 10,000 s {i2 } + (s + 400) {i3 } = 8,000 s (s + 500) 400 so that {i3 } = Then i3 = 80 80 −900t − e 9 9 s2 and 8,000 80 1 80 1 = − . + 900s 9 s 9 s + 900 i2 = 20 − 0.0025i3 − i3 = 241 100 100 −900t . − e 9 9 4.6 Systems of Linear Differential Equations (c) i1 = i2 + i3 = 20 − 20e−900t 16. (a) Taking the Laplace transform of the system i2 + i3 + 10i2 = 120 − 120 (t − 2) −10i2 + 5i3 + 5i3 = 0 gives {i2 } + s (s + 10) −10s {i2 } + 5(s + 1) {i3 } = 120 1 − e−2s s {i3 } = 0 so that {i2 } = 48 120(s + 1) 60 12 1 − e−2s = − + (3s2 + 11s + 10)s s + 5/3 s + 2 s {i3 } = 240 240 240 1 − e−2s = − 3s2 + 11s + 10 s + 5/3 s + 2 1 − e−2s and 1 − e−2s . Then i2 = 12 + 48e−5t/3 − 60e−2t − 12 + 48e−5(t−2)/3 − 60e−2(t−2) (t − 2) and i3 = 240e−5t/3 − 240e−2t − 240e−5(t−2)/3 − 240e−2(t−2) (t − 2). (b) i1 = i2 + i3 = 12 + 288e−5t/3 − 300e−2t − 12 + 288e−5(t−2)/3 − 300e−2(t−2) (t − 2) 17. Taking the Laplace transform of the system i2 + 11i2 + 6i3 = 50 sin t i3 + 6i2 + 6i3 = 50 sin t gives (s + 11) 6 {i2 } + 6 {i2 } + (s + 6) 50 +1 50 {i3 } = 2 s +1 {i3 } = s2 so that {i2 } = 20 1 375 1 145 s 85 1 50s =− + + + . 2 + 1) 2+1 2+1 (s + 2)(s + 15)(s 13 s + 2 1469 s + 15 113 s 113 s Then i2 = − and i3 = 85 20 −2t 375 −15t 145 e e cos t + sin t + + 13 1469 113 113 25 11 30 −2t 250 −15t 280 1 810 + − sin t − i2 − i2 = e e cos t + sin t. 3 6 6 13 1469 113 113 18. Taking the Laplace transform of the system 0.5i1 + 50i2 = 60 0.005i2 + i2 − i1 = 0 242 4.6 Systems of Linear Differential Equations gives s −200 120 s {i2 } = 0 {i1 } + 100 {i2 } = {i1 } + (s + 200) so that {i2 } = 24,000 6 6 1 6 s + 100 100 − . = − s(s2 + 200s + 20,000) 5 s 5 (s + 100)2 + 1002 5 (s + 100)2 + 1002 Then i2 = and 6 6 −100t 6 cos 100t − e−100t sin 100t − e 5 5 5 i1 = 0.005i2 + i2 = 6 6 −100t cos 100t. − e 5 5 19. Taking the Laplace transform of the system 2i1 + 50i2 = 60 0.005i2 + i2 − i1 = 0 gives {i1 } + 50 2s −200 {i1 } + (s + 200) 60 s {i2 } = 0 {i2 } = so that {i2 } = 6,000 s(s2 + 200s + 5,000) √ √ 6 1 6 6 2 s + 100 50 2 √ √ = − . − 5 s 5 (s + 100)2 − (50 2 )2 5 (s + 100)2 − (50 2 )2 Then and √ √ √ 6 6 −100t 6 2 −100t i2 = − e e cosh 50 2 t − sinh 50 2 t 5 5 5 √ √ √ 6 6 −100t 9 2 −100t i1 = 0.005i2 + i2 = − e e cosh 50 2 t − sinh 50 2 t. 5 5 10 20. (a) Using Kirchhoff’s first law we write i1 = i2 + i3 . Since i2 = dq/dt we have i1 − i3 = dq/dt. Using Kirchhoff’s second law and summing the voltage drops across the shorter loop gives E(t) = iR1 + 1 q, C so that i1 = 1 1 q. E(t) − R1 R1 C Then dq 1 1 = i1 − i3 = q − i3 E(t) − dt R1 R1 C and R1 dq 1 + q + R1 i3 = E(t). dt C 243 (1) 4.6 Systems of Linear Differential Equations Summing the voltage drops across the longer loop gives E(t) = i1 R1 + L di3 + R2 i3 . dt Combining this with (1) we obtain i1 R1 + L or L di3 1 + R2 i3 = i1 R1 + q dt C di3 1 + R2 i3 − q = 0. dt C (b) Using L = R1 = R2 = C = 1, E(t) = 50e−t (t − 1) = 50e−1 e−(t−1) the Laplace transform of the system we obtain {i3 } = (t − 1), q(0) = i3 (0) = 0, and taking 50e−1 −s e s+1 (s + 1) {q} + (s + 1) {i3 } − {q} = 0, {q} = 50e−1 e−s (s + 1)2 + 1 so that and q(t) = 50e−1 e−(t−1) sin(t − 1) (t − 1) = 50e−t sin(t − 1) (t − 1). 21. (a) Taking the Laplace transform of the system 4θ1 + θ2 + 8θ1 = 0 θ1 + θ2 + 2θ2 = 0 gives {θ1 } + s2 4 s2 + 2 s2 {θ1 } + s2 + 2 {θ2 } = 3s {θ2 } = 0 so that 3s2 + 4 s2 + 4 or {θ2 } = Then θ2 = θ1 = (b) θ1 s 3 s 1 − . 2 + 4/3 2 s 2 s2 + 4 2 1 3 cos √ t − cos 2t 2 2 3 so that {θ2 } = −3s3 θ1 = −θ2 − 2θ2 and 3 2 1 cos √ t + cos 2t. 4 4 3 θ2 2 2 1 1 −1 3π 6π t −1 −2 −2 244 3π 6π t 4.6 Systems of Linear Differential Equations Mass m2 has extreme displacements of greater magnitude. Mass m1 first passes through its equilibrium position at about t = 0.87, and mass m2 first passes through its equilibrium position at about t = 0.66. √ √ The motion of the pendulums is not periodic since cos(2t/ 3 ) has period 3 π, cos 2t has period π, and √ the ratio of these periods is 3 , which is not a rational number. (c) The Lissajous curve is plotted for 0 ≤ t ≤ 30. θ2 2 1 -1-0.5 0.5 1 θ1 -1 -2 (d) t=0 t=1 t=2 t=3 t=4 t=7 t=8 θ1 θ2 1 2 3 4 5 6 7 8 9 10 t=5 t=6 t -0.2111 -0.6585 0.4830 -0.1325 -0.4111 0.8327 0.0458 -0.9639 0.3534 0.4370 0.8263 0.6438 -1.9145 0.1715 1.6951 -0.8662 -0.3186 0.9452 -1.2741 -0.3502 t=9 (e) Using a CAS to solve θ1 (t) = θ2 (t) we see that θ1 = θ2 (so that the double pendulum is straight out) when t is about 0.75 seconds. t=10 t=0.75 (f ) To make a movie of the pendulum it is necessary to locate the mass in the plane as a function of time. Suppose that the upper arm is attached to the origin and that the equilibrium position lies along the 245 4.6 Systems of Linear Differential Equations negative y-axis. Then mass m1 is at (x, (t), y1 (t)) and mass m2 is at (x2 (t), y2 (t)), where x1 (t) = 16 sin θ1 (t) and y1 (t) = −16 cos θ1 (t) x2 (t) = x1 (t) + 16 sin θ2 (t) and y2 (t) = y1 (t) − 16 cos θ2 (t). and A reasonable movie can be constructed by letting t range from 0 to 10 in increments of 0.1 seconds. CHAPTER 4 REVIEW EXERCISES 1 1. {f (t)} = 0 (2 − t)e−st dt = 1 4 2. ∞ te−st dt + {f (t)} = e−st dt = 2 1 2 − 2 e−s s2 s 1 −2s − e−4s e s 3. False; consider f (t) = t−1/2 . 4. False, since f (t) = (et )10 = e10t . 5. True, since lims→∞ F (s) = 1 = 0. (See Theorem 4.5 in the text.) 6. False; consider f (t) = 1 and g(t) = 1. 1 s+7 7. e−7t = 8. te−7t = 9. {sin 2t} = 10. 1 (s + 7)2 2 s2 + 4 11. {t sin 2t} = − 12. {sin 2t 20 s6 14. 1 3s − 1 15. 1 (s − 5)3 s2 d 2 4s = 2 ds s2 + 4 (s + 4)2 (t − π)} = 13. 16. 2 (s + 3)2 + 4 e−3t sin 2t = 1 5! 6 s6 = 1 −5 {sin 2(t − π) = 1 2 s2 2 e−πs +4 1 5 t 6 1 1 3 s − 1/3 = = = (t − π)} = 2 (s − 5)3 = 1 t/3 e 3 = 1 2 5t t e 2 1 1 1 1 √ + √ √ − √ 2 5 s+ 5 2 5 s− 5 246 √ 1 √ 1 = − √ e− 5 t + √ e 5 t 2 5 2 5 CHAPTER 4 REVIEW EXERCISES 17. s − 10s + 29 s2 18. 1 −5s e s2 19. s + π −s e s2 + π 2 5 s−5 2 + 2 + 22 (s − 5) 2 (s − 5)2 + 22 = = (t − 5) (t − 5) s π e−s + 2 e−s s2 + π 2 s + π2 = = cos π(t − 1) 1 L2 s2 + n2 π 2 20. = 5 = e5t cos 2t + e5t sin 2t 2 (t − 1) + sin π(t − 1) 1 L L2 nπ 21. te8t f (t) = − = 1 nπ sin t Lnπ L e−5t exists for s > −5. 22. nπ/L s2 + (n2 π 2 )/L2 (t − 1) 23. {eat f (t − k) d F (s − 8). ds (t − k)} = e−ks t eaτ f (τ ) dτ 24. = 0 1 s {eat f (t)} = t eat = f (τ ) dτ 0 {eat f (t)} = e−k(s−a) F (s − a) F (s − a) , whereas s t f (τ ) dτ 0 25. f (t) {ea(t+k) f (t)} = e−ks eak = s→s−a F (s) s = s→s−a F (s − a) . s−a (t − t0 ) 26. f (t) − f (t) 27. f (t − t0 ) (t − t0 ) (t − t0 ) 28. f (t) − f (t) (t − t0 ) + f (t) (t − t1 ) 29. f (t) = t − [(t − 1) + 1] (t − 1) + (t − 1) − (t − 4) = t − (t − 1) (t − 1) − (t − 4) 1 1 1 − 2 e−s − e−4s 2 s s s 1 1 1 −4(s−1) et f (t) = − e−(s−1) − e 2 2 (s − 1) (s − 1) s−1 {f (t)} = (t − π) − sin t (t − 3π) = − sin(t − π) (t − π) + sin(t − 3π) 1 1 {f (t)} = − 2 e−πs + 2 e−3πs s +1 s +1 1 1 et f (t) = − e−π(s−1) + e−3π(s−1) (s − 1)2 + 1 (s − 1)2 + 1 30. f (t) = sin t 31. f (t) = 2 − 2 (t − 2) + [(t − 2) + 2] (t − 2) = 2 + (t − 2) 2 1 {f (t)} = + 2 e−2s s s 2 1 t e−2(s−1) e f (t) = + s − 1 (s − 1)2 32. f (t) = t − t (t − 2) (t − 1) + (2 − t) (t − 1) − (2 − t) (t − 2) = t − 2(t − 1) 1 2 1 {f (t)} = 2 − 2 e−s + 2 e−2s s s s 1 2 1 t e f (t) = − e−(s−1) + e−2(s−1) (s − 1)2 (s − 1)2 (s − 1)2 247 (t − 3π) (t − 1) + (t − 2) (t − 2) CHAPTER 4 REVIEW EXERCISES 33. Taking the Laplace transform of the differential equation we obtain {y} = so that 5 2 1 + (s − 1)2 2 (s − 1)3 1 y = 5tet + t2 et . 2 34. Taking the Laplace transform of the differential equation we obtain {y} = = (s − 1 − 8s + 20) 1)2 (s2 6 6 5 1 1 1 s−4 2 − + + 2 2 + 22 169 s − 1 13 (s − 1) 169 (s − 4) 338 (s − 4)2 + 22 so that y= 6 t 1 6 4t 5 4t e + tet − e cos 2t + e sin 2t. 169 13 169 338 35. Taking the Laplace transform of the given differential equation we obtain s3 + 6s2 + 1 1 2 − e−2s − e−2s + 1)(s + 5) s2 (s + 1)(s + 5) s(s + 1)(s + 5) 6 1 1 1 3 1 13 1 =− · + · 2 + · − · 25 s 5 s 2 s + 1 50 s + 5 6 1 1 1 1 1 1 1 − − · + · 2+ · − · e−2s 25 s 5 s 4 s + 1 100 s + 5 {y} = s2 (s − so that y=− 1 1 1 2 1 1 · − · + · 5 s 2 s + 1 10 s + 5 e−2s 6 13 4 1 3 1 + t + e−t − e−5t − (t − 2) − (t − 2) (t − 2) 25 5 2 50 25 5 1 −(t−2) 9 −5(t−2) + e (t − 2) − (t − 2). e 4 100 36. Taking the Laplace transform of the differential equation we obtain {y} = s3 + 2 2 + 2s + s2 −s − 3 e − 5) s (s − 5) s3 (s =− 2 1 1 2 127 1 1 2 37 2 1 37 1 12 1 1 − + − + − − − − e−s 125 s 25 s2 5 s3 125 s − 5 125 s 25 s2 5 s3 125 s − 5 so that y=− 2 127 5t 37 37 5(t−1) 2 1 12 1 − t − t2 + e − − − (t − 1) − (t − 1)2 + e 125 25 5 125 125 25 5 125 37. Taking the Laplace transform of the integral equation we obtain {y} = so that 1 1 2 1 + + s s2 2 s3 1 y(t) = 1 + t + t2 . 2 38. Taking the Laplace transform of the integral equation we obtain ( {f })2 = 6 · 6 s4 or 248 {f } = ±6 · 1 s2 (t − 1). CHAPTER 4 REVIEW EXERCISES so that f (t) = ±6t. 39. Taking the Laplace transform of the system gives s 4 {x} + {y} = {x} + s 1 +1 s2 {y} = 2 so that {x} = s2 − 2s + 1 1 1 1 1 9 1 =− + + . s(s − 2)(s + 2) 4 s 8 s−2 8 s+2 Then 1 1 9 x = − + e2t + e−2t 4 8 8 and y = −x + t = 9 −2t 1 2t − e + t. e 4 4 40. Taking the Laplace transform of the system gives s2 {x} + s2 2s {x} + s2 1 s−2 1 {y} = − s−2 {y} = so that {x} = 2 1 1 1 1 1 − + = s(s − 2)2 2 s 2 s − 2 (s − 2)2 {y} = −s − 2 1 3 1 1 1 3 1 − − =− + . s2 (s − 2)2 4 s 2 s2 4 s − 2 (s − 2)2 and Then x= 1 1 2t − e + te2t 2 2 3 3 1 and y = − − t + e2t − te2t . 4 2 4 41. The integral equation is t i(τ ) dτ = 2t2 + 2t. 10i + 2 0 Taking the Laplace transform we obtain {i} = 4 2 + 2 s3 s s s+2 9 2 9 2 45 9 = 2 =− + 2 + =− + 2 + . 10s + 2 s (5s + 2) s s 5s + 1 s s s + 1/5 Thus i(t) = −9 + 2t + 9e−t/5 . 42. The differential equation is 1 d2 q dq + 10 + 100q = 10 − 10 2 dt2 dt Taking the Laplace transform we obtain {q} = = s(s2 (t − 5). 20 1 − e−5s + 20s + 200) 1 1 1 1 s + 10 10 − − 10 s 10 (s + 10)2 + 102 10 (s + 10)2 + 102 249 1 − e−5s CHAPTER 4 REVIEW EXERCISES so that q(t) = 1 1 1 − e−10t cos 10t − e−10t sin 10t 10 10 10 1 1 1 −10(t−5) − cos 10(t − 5) − e−10(t−5) sin 10(t − 5) − e 10 10 10 (t − 5). 43. Taking the Laplace transform of the given differential equation we obtain {y} = 2w0 EIL 1 1 L 4! 5! 5! −sL/2 − + e · · · 48 s5 120 s6 120 s6 + c1 2! c2 3! + · · 2 s3 6 s4 so that y= 2w0 L 4 1 5 1 x − x + EIL 48 120 120 x− L 2 5 x− L 2 + c1 2 c2 3 x + x 2 6 where y (0) = c1 and y (0) = c2 . Using y (L) = 0 and y (L) = 0 we find c2 = −w0 L/4EI. c1 = w0 L2 /24EI, Hence y= w0 L L2 3 L3 2 1 1 − x5 + x4 − x + x + 12EIL 5 2 2 4 5 x− L 2 5 x− L 2 . 44. In this case the boundary conditions are y(0) = y (0) = 0 and y(π) = y (π) = 0. If we let c1 = y (0) and c2 = y (0) then s4 {y} − s3 y(0) − s2 y (0) − sy(0) − y (0) + 4 and {y} = {y} = {δ(t − π/2)} c1 2s c2 4 w0 4 · + · + · e−sπ/2 . 2 s4 + 4 4 s4 + 4 4EI s4 + 4 From the table of transforms we get c1 c2 y= sin x sinh x + (sin x cosh x − cos x sinh x) 2 4 w0 π π π π + sin x − cosh x − − cos x − sinh x − 4EI 2 2 2 2 x− π 2 x− π . 2 Using y(π) = 0 and y (π) = 0 we find c1 = Hence y= w0 sinh π 2 , EI sinh π c2 = − w0 cosh π 2 . EI sinh π w0 sinh π w0 cosh π 2 2 sin x sinh x − (sin x cosh x − cos x sinh x) 2EI sinh π 4EI sinh π + w0 π π π π sin x − cosh x − − cos x − sinh x − 4EI 2 2 2 2 45. (a) With ω 2 = g/l and K = k/m the system of differential equations is θ1 + ω 2 θ1 = −K(θ1 − θ2 ) θ2 + ω 2 θ2 = K(θ1 − θ2 ). Denoting the Laplace transform of θ(t) by Θ(s) we have that the Laplace transform of the system is (s2 + ω 2 )Θ1 (s) = −KΘ1 (s) + KΘ2 (s) + sθ0 (s2 + ω 2 )Θ2 (s) = KΘ1 (s) − KΘ2 (s) + sψ0 . 250 CHAPTER 4 REVIEW EXERCISES If we add the two equations, we get Θ1 (s) + Θ2 (s) = (θ0 + ψ0 ) s s2 + ω 2 which implies θ1 (t) + θ2 (t) = (θ0 + ψ0 ) cos ωt. This enables us to solve for first, say, θ1 (t) and then find θ2 (t) from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt. Now solving (s2 + ω 2 + K)Θ1 (s) − KΘ2 (s) = sθ0 −kΘ1 (s) + (s2 + ω 2 + K)Θ2 (s) = sψ0 gives [(s2 + ω 2 + K)2 − K 2 ]Θ1 (s) = s(s2 + ω 2 + K)θ0 + Ksψ0 . Factoring the difference of two squares and using partial fractions we get Θ1 (s) = θ0 + ψ0 s s s(s2 + ω 2 + K)θ0 + Ksψ0 θ0 − ψ0 = , + 2 + ω 2 )(s2 + ω 2 + 2K) 2 + ω2 2 + ω 2 + 2K (s 2 s 2 s so θ0 + ψ0 θ0 − ψ0 cos ωt + cos 2 2 Then from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt we get θ1 (t) = θ2 (t) = θ0 + ψ0 θ0 − ψ0 cos ωt − cos 2 2 ω 2 + 2K t. ω 2 + 2K t. (b) With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = θ0 , θ2 (0) = 0 we have θ1 (t) = θ0 cos ωt, θ2 (t) = θ0 cos ωt. Physically this means that both pendulums swing in the same direction as if they were free since the spring exerts no influence on the motion (θ1 (t) and θ2 (t) are free of K). With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = −θ0 , θ2 (0) = 0 we have θ1 (t) = θ0 cos ω 2 + 2K t, θ2 (t) = −θ0 cos ω 2 + 2K t. Physically this means that both pendulums swing in the opposite directions, stretching and compressing the spring. The amplitude of both displacements is |θ0 |. Moreover, θ1 (t) = θ0 and θ2 (t) = −θ0 at precisely the same times. At these times the spring is stretched to its maximum. 251 ...
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This note was uploaded on 05/05/2008 for the course MATH 240 taught by Professor Storm during the Spring '08 term at UPenn.

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