AEM_3e_Chapter_04

AEM_3e_Chapter_04 - 4 3. {f (t)} = The Laplace Transform...

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Unformatted text preview: 4 3. {f (t)} = The Laplace Transform EXERCISES 4.1 Definition of the Laplace Transform 1 0 ∞ 1 1 0 ∞ 1 te−st dt + e−st dt = 1 1 − te−st − 2 e−st s s 1 − e−st s = 6. {f (t)} = 1 1 − e−s − 2 e−s s s ∞ π/2 1 − 0− 2 s − 1 1 − (0 − e−s ) = 2 (1 − e−s ), s s ∞ π/2 s>0 (cos t)e−st dt = s 1 e−st cos t + 2 e−st sin t s2 + 1 s +1 s>0 1 1 e−πs/2 = − 2 e−πs/2 , s2 + 1 s +1 1 − t, 0 < t < 1 9. The function is f (t) = so 0, t>1 =0− 0+ 1 {f (t)} = 0 (1 − t)e−st dt + 1 ∞ 0e−st dt = 0 1 (1 − t)e−st dt = 1 1 − (1 − t)e−st + 2 e−st s s 1 0 1 1 1 = 2 e−s + − 2 , s ss 12. 15. {f (t)} = 0 ∞ s>0 ∞ 0 e−2t−5 e−st dt = e−5 e−t (sin t)e−st dt = 0 e−(s+2)t dt = − e−5 −(s+2)t e s+2 ∞ = 0 e−5 , s+2 s > −2 {f (t)} = 0 ∞ ∞ (sin t)e−(s+1)t dt ∞ 0 −(s + 1) −(s+1)t 1 e e−(s+1)t cos t sin t − 2+1 (s + 1) (s + 1)2 + 1 1 1 =2 , s > −1 = 2+1 (s + 1) s + 2s + 2 = 18. {f (t)} = 0 ∞ t(sin t)e−st dt (cos t)e −st = = t 2s −2 − 2 + 1)2 s + 1 (s 2s s>0 − s2 − 1 st + s2 + 1 (s2 + 1)2 ∞ (sin t)e −st 0 21. 24. 2, (s2 + 1) 4 10 {4t − 10} = 2 − s s {−4t2 + 16t + 9} = −4 2 16 9 + 2+ 3 s s s 57 4.1 Definition of the Laplace Transform 1 1 + s s−4 1 2 1 2t −2t {e − 2 + e } = −+ s−2 s s+2 1 1 1 1 k {sinh kt} = {ekt − e−kt } = − =2 2 2 s−k s+k s − k2 e−t et + e−t 2 = 1 1 −2t +e 22 = 1 1 + 2s 2(s + 2) 27. 30. 33. 36. {1 + e4t } = {e−t cosh t} = 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) {t −1/2 {sin 4t} + (sin 5) π s {t {cos 4t} = (cos 5) 1 /2 s2 4 s 4 cos 5 + (sin 5)s + (sin 5) 2 = . + 16 s + 16 s2 + 16 (c) {t 3/2 42. (a) Γ(1/2) } = 1 /2 = s (b) √ π Γ(3/2) } = 3 /2 = 3 /2 s 2s √ Γ(5/2) 3π } = 5 /2 = 5 /2 s 4s EXERCISES 4.2 The Inverse Transform and Transforms of Derivatives = t − 2t4 = 1 + 4t + 2t2 3. 6. 9. 12. 15. 18. 21. 24. 1 48 −5 s2 s (s + 2)2 s3 1 4s + 1 s2 10s + 16 = = = 1 4 1 48 4! · − s2 24 s5 1 1 2 +4· 2 +2· 3 s s s 1 s + 1/4 = 1 −t/4 e 4 = 10 cos 4t = = 2· s 3 −2· 2 +9 s +9 = 2 cos 3t − 2 sin 3t 15 = − + e4t 44 = 0.3e0.1t + 0.6e−0.2t 2s − 6 s2 + 9 s+1 s2 − 4s s2 11 5 1 −·+· 4 s 4 s−4 = = (0.3) · 0.9s (s − 0.1)(s + 0.2) 1 1 + (0.6) · s − 0.1 s + 0.2 s2 + 1 s(s − 1)(s + 1)(s − 2) 11 1 1 1 5 1 ·− −· +· 2 s s−1 3 s+1 6 s−2 1 1 5 = − et − e−t + e2t 2 3 6 2s − 4 s(s + 1)(s2 + 1) = 4 3 s 3 −+ +2 +2 s s+1 s +1 s +1 27. (s2 2s − 4 + s)(s2 + 1) = = −4 + 3e−t + cos t + 3 sin t 58 4.2 6s + 3 + 1)(s2 + 4) 2· The Inverse Transform and Transforms of Derivatives s 1 s 1 2 +2 −2· 2 − ·2 +1 s +1 s +4 2 s +4 1 = 2 cos t + sin t − 2 cos 2t − sin 2t 2 33. The Laplace transform of the initial-value problem is 30. (s2 = s2 s Solving for {y } we obtain {y } = Thus y= 1 2 1 1 19 1 + = · + · . (s − 4)(s + 6) s + 6 10 s − 4 10 s + 6 1 4t 19 −6t e+ e . 10 10 6 3 − . s−3 s+1 {y } − y (0) + 6 {y } = 1 . s−4 36. The Laplace transform of the initial-value problem is s2 Solving for {y } we obtain {y } = = Thus y= 6 3 s−5 − +2 2 − 4s) 2 − 4s) (s − 3)(s (s + 1)(s s − 4s 51 2 3 1 11 1 ·− −· + · . 2 s s − 3 5 s + 1 10 s − 4 5 3 11 − 2e3t − e−t + e4t . 2 5 10 1 . s+1 {y } − sy (0) − y (0) − 4 [s {y } − y (0)] = 39. The Laplace transform of the initial-value problem is 2 s3 {y } − s2 (0) − sy (0) − y (0) + 3 s2 {y } we obtain {y } = Thus y= 2s + 3 11 5 1 8 1 11 = + − + . (s + 1)(s − 1)(2s + 1)(s + 2) 2 s + 1 18 s − 1 9 s + 1/2 9 s + 2 1 −t 5 8 1 e + et − e−t/2 + e−2t . 2 18 9 9 {y } − sy (0) − y (0) − 3[s {y } − y (0)] − 2 {y } = Solving for 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 } = s2 s+1 s−1+2 s−1 2 = = + . 2 + 22 2 + 22 2s + 5 (s − 1) (s − 1) (s − 1)2 + 22 y = et cos 2t + et sin 2t. Thus 59 4.2 4.3 Translation Theorems Transforms of Derivatives The Inverse Transform and EXERCISES 4.3 Translation Theorems 3. 6. 9. 12. 15. 18. t3 e−2t = 3! (s + 2)4 t2 e2t − 2te2t + e2t = 2 1 2 − + (s − 2)3 (s − 2)2 s−2 s2 s s−1 3(s + 4) − + 2 + 25 + 25 (s − 1) (s + 4)2 + 25 e2t (t − 1)2 = {(1 − et + 3e−4t ) cos 5t} = 1 (s − 1)4 s2 = 1 6 = {cos 5t − et cos 5t + 3e−4t cos 5t} = 3! (s − 1)4 = 13t te 6 s + 4s + 5 = 1 s+2 −2 2 + 12 (s + 2) (s + 2)2 + 12 5(s − 2) + 10 (s − 2)2 = = e−2t cos t − 2e−2t sin t = 5e2t + 10te2t 5s (s − 2)2 5 10 + s − 2 (s − 2)2 1 . s+4 21. The Laplace transform of the differential equation is s Solving for {y } we obtain {y } = Thus y = te−4t + 2e−4t . 24. The Laplace transform of the differential equation is s2 Solving for {y } − sy (0) − y (0) − 4 [s {y } = {y } − y (0)] + 4 {y } = 6 . (s − 2)4 1 2 . + (s + 4)2 s+4 {y } − y (0) + 4 {y } = 1 1 5 2t 5! . Thus, y = te . 20 (s − 2)6 20 27. The Laplace transform of the differential equation is {y } we obtain s2 Solving for {y } we obtain {y } = − Thus s2 3 3 2 . =− − 6s + 13 2 (s − 3)2 + 22 3 y = − e3t sin 2t. 2 1 1 + 2. ss {y } − sy (0) − y (0) − 6 [s {y } − y (0)] + 13 {y } = 0. 30. The Laplace transform of the differential equation is s2 {y } − sy (0) − y (0) − 2 [s {y } − y (0)] + 5 {y } = 60 4.3 Translation Theorems Solving for {y } we obtain {y } = = Thus y= s2 (s2 4s2 + s + 1 −7s/25 + 109/25 71 11 + = + − 2s + 5) 25 s 5 s2 s2 − 2s + 5 71 11 7 51 s−1 2 − + . + 2 2 + 22 25 s 5 s 25 (s − 1) 25 (s − 1)2 + 22 7 51 1 7 + t − et cos 2t + et sin 2t. 25 5 25 25 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 {x} we obtain 3 {x} + s + 7s 2 {x} + 21 + 16 2 {x} = 0. √ √ −3s/2 − 21/2 3 s + 7/2 15/2 7 15 √ √ {x} = 2 =− − . s + 7s + 16 2 (s + 7/2)2 + ( 15/2)2 10 (s + 7/2)2 + ( 15/2)2 √ √ √ 3 15 15 7 15 −7t/2 x = − e−7t/2 cos sin t− e t. 2 2 10 2 R Thus 36. The differential equation is dq 1 + q = E0 e−kt , q (0) = 0. dt C The Laplace transform of this equation is Rs Solving for {q } we obtain {q } = E0 C E0 /R = . (s + k )(RCs + 1) (s + k )(s + 1/RC ) {q } + 1 C {q } = E 0 1 . s+k When 1/RC = k we have by partial fractions {q } = Thus q (t) = When 1/RC = k we have {q } = Thus q (t) = 39. t (t − 2) = {(t − 2) (t − 2) + 2 E0 1 . R (s + k )2 E0 R 1/(1/RC − k ) 1/(1/RC − k ) − s+k s + 1/RC = E0 1 R 1/RC − k 1 1 − s + k s + 1/RC . E0 C e−kt − e−t/RC . 1 − kRC E0 −kt E0 −t/RC = . te te R R e−2s 2e−2s + 2 s s 2 1 + s2 s . (t − 2)} = Alternatively, (16) of this section could be used: {t (t − 2)} = e−2s {t + 2} = e−2s 61 4.3 Translation Theorems π 2 π 2 π 2 se−πs/2 s2 + 1 π 2 s . s2 + 1 42. sin t t− = cos t − t− = Alternatively, (16) of this section in the text could be used: sin t e−πs s2 + 1 t− π 2 = e−πs/2 sin t + = e−πs/2 {cos t} = e−πs/2 45. 48. 51. (f ) 54. (d) 57. t2 = sin(t − π ) = − (t − π ) = − sin t (t − π ) =− (t − 2) − (t − 2) (t − 2) + et−2 (t − 2) e−2s s2 (s − 1) e−2s e−2s e−2s − 2+ s s s−1 (t − 1) = = (t − 1)2 + 2t − 1 2 2 1 + 2+ s3 s s e−s (t − 1) = (t − 1)2 + 2(t − 1) − 1 (t − 1) Alternatively, by (16) of this section in the text, {t2 (t − 1)} = e−s {t2 + 2t + 1} = e−s 2 2 1 + 2+ 3 s s s . 60. sin t − sin t (t − 2π ) = sin t − sin(t − 2π ) (t − 2π ) = 1 e−2πs −2 s2 + 1 s + 1 63. The Laplace transform of the differential equation is s Solving for {y } we obtain {y } = Thus y=5 (t − 1) − 5e−(t−1) (t − 1). {y } − y (0) + {y } = 5 −s e. s 5e−s 1 1 = 5e−s − . s(s + 1) s s+1 66. The Laplace transform of the differential equation is s2 Solving for {y } we obtain {y } = Thus y= 11 1 11 − cos 2t − sin 2t − − cos 2(t − 1) 44 2 44 (t − 1). 1−s 11 1 s 12 11 1 s 1 − e−s 2 = − − − e−s − . 2 + 4) 2+4 2+4 s(s s(s + 4) 4s 4s 2s 4 s 4 s2 + 4 {y } − sy (0) − y (0) + 4 {y } = 1 e−s − . s s 69. The Laplace transform of the differential equation is s2 {y } − sy (0) − y (0) + {y } = e−πs e−2πs − . s s 62 4.3 Translation Theorems Solving for {y } we obtain {y } = e−πs Thus y = [1 − cos(t − π )] (t − π ) − [1 − cos(t − 2π )] (t − 2π ) + sin t. 1 1 s s 1 − − e−2πs − +2 . s s2 + 1 s s2 + 1 s +1 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) = 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} we obtain {x} = = Thus x(t) = − = (s2 1 1 −2 e−2πs 2 + 1) + 16) (s (s + 16) (s2 + 1) {x} + 16 {x} = 1 1 − e−2πs . s2 + 1 s2 + 1 sin t, 0 ≤ t < 2π 0, t ≥ 2π −1/15 1/15 −1/15 1/15 + −2 + e−2πs . s2 + 16 s2 + 1 s + 16 s2 + 1 (t − 2π ) − 1 sin(t − 2π ) 15 (t − 2π ) 1 1 1 sin 4t + sin t + sin 4(t − 2π ) 60 15 60 1 − 60 sin 4t + 1 15 sin t, 0 ≤ t < 2π t ≥ 2π . 0, 75. (a) The differential equation is di 3π + 10i = sin t + cos t − dt 2 The Laplace transform of this equation is s Solving for {i} we obtain (s2 1 101 {i} + 10 {i} = s2 t− 3π 2 , i(0) = 0. 1 se−3πs/2 +2 . +1 s +1 {i} = = Thus i(t) = 1 s +2 e−3πs/2 + 1)(s + 10) (s + 1)(s + 10) 1 s 10 − + s + 10 s2 + 1 s2 + 1 + 1 101 −10 10s 1 + + s + 10 s2 + 1 s2 + 1 e−3πs/2 . 1 e−10t − cos t + 10 sin t 101 1 3π + −10e−10(t−3π/2) + 10 cos t − 101 2 + sin t − 3π 2 t− 3π 2 . 63 4.3 Translation Theorems (b) i 0.2 1 -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. 2 3 4 5 6 t 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 {y } = so that y (x) = 1 1 1 w0 c1 x2 + c2 x3 + 2 6 24 EI x− L 3 4 c1 c2 w0 1 e−Ls/3 − e−2Ls/3 + 4+ s3 s EI s5 x− L 3 − x− 2L 3 4 x− 2L 3 . To find c1 and c2 we compute y (x) = c1 + c2 x + and y (x) = c2 + w0 EI x− L 3 x− L 3 − x− 2L 3 x− 2L 3 . 1 w0 2 EI x− L 3 2 x− L 3 − x− 2L 3 2 x− 2L 3 Then y (L) = y (L) = 0 yields the system c1 + c2 L + 1 w0 2 EI 2L 3 2 − L 3 w0 EI 2 = c1 + c2 L + 1 w0 L2 =0 6 EI c2 + 1 w0 L 2L L − = c2 + = 0. 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 22 1 1 L x − Lx3 + 12 18 24 x− L 3 4 x− L 3 − x− 2L 3 4 x− 2L 3 . 81. (a) The temperature T of the cake inside the oven is modeled by 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 where Tm 70 + 57.5t, 0 ≤ t < 4 300, t ≥ 4. 64 4.4 In terms of the unit step function, Tm = (70 + 57.5t)[1 − (t − 4)] + 300 Additional Operational Properties (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) − 70 57.5 57.5 −4s − 2+ 2e s s s or t(s) = 70 70k 57.5k 57.5k − − + e−4s . s − k s(s − k ) s2 (s − k ) s2 (s − k ) 1 1 + t − 4 − ek(t−4) k k After using partial functions, the inverse transform is then T (t) = 70 + 57.5 1 1 + t − ekt k k − 57.5 (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. EXERCISES 4.4 Additional Operational Properties s2 − 4 (s2 + 4) d ds 2 3. {t cos 2t} = − {t2 cos t} = d ds s 2+4 s s2 s +1 = 6. d2 ds2 = 1 − s2 (s2 + 1)2 = 2s s2 − 3 (s2 + 1) 3 9. The Laplace transform of the differential equation is s Solving for {y } we obtain {y } = Thus 2s 11 s 11 1s 1 =− +2 . − + + (s + 1)(s2 + 1)2 2 s + 1 2 s2 + 1 2 s2 + 1 (s2 + 1)2 (s + 1)2 1 y (t) = − e−t − 2 1 = − e−t + 2 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 {y } + {y } = (s2 2s . + 1)2 65 4.4 Additional Operational Properties 12. The Laplace transform of the differential equation is s2 Solving for {y } we obtain {y } = Thus y = cos t − sin t + 15. y 1 0.5 1 -0.5 -1 2 3 4 5 6 t {y } − sy (0) − y (0) + {y } = s2 1 . +1 s3 − s2 + s s 1 1 =2 . −2 +2 2 + 1)2 (s s + 1 s + 1 (s + 1)2 1 1 sin t − t cos t 2 2 = cos t − 1 1 sin t − t cos t. 2 2 18. From Theorem 4.8 in the text {ty } = − d ds {y } = − dY d [sY (s) − y (0)] = −s −Y ds ds so that the transform of the given second-order differential equation is the linear first-order differential equation in Y (s): 10 3 Y+ − 2s Y = − . s s Using the integrating factor s3 e−s , the last equation yields 2 Y (s) = 5 c2 + 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 (t) = e−t ∗ et cos t = t 5 s3 = 52 t. 2 21. 24. s−1 (s + 1) [(s − 1)2 + 1] 1 s {cos t} = {t} d ds s(s2 s 1 =2 + 1) s +1 cos τ dτ 0 t = = 27. 0 τ et−τ dτ t {et } = t 0 1 s2 (s − 1) =− = 0 30. 33. t 0 τ e−τ dτ 1 − 1) =− = τ e−τ dτ d ds t 1 1 s (s + 1)2 = s2 (s 3s + 1 + 1)3 s3 (s 1/s2 (s − 1) s 1 (eτ − τ − 1)dτ = et − t2 − t − 1 2 66 4.4 Additional Operational Properties 36. The Laplace transform of the differential equation is s2 Thus {y } = and, using 8k 3 s (s2 + k 2 )3 = t sin kt − kt2 cos kt 1 2s +2 (s2 + 1)2 (s + 1)3 {y } + 1 2s {y } = 2 . +2 (s + 1) (s + 1)2 50 y 5 - 50 10 15 t from Problem 35 with k = 1 we have y= 1 1 (sin t − t cos t) + (t sin t − t2 cos t). 2 4 39. The Laplace transform of the given equation is {f } = Solving for {f } we obtain {f } = Thus f (t) = 1 t 3 t 1 2 t 1 −t e + te + t e − e 8 4 4 8 s2 11 3 1 2 1 11 = + . + − (s − 1)3 (s + 1) 8 s − 1 4 (s − 1)2 4 (s − 1)3 8 s+1 tet + {t } {f }. 42. The Laplace transform of the given equation is {f } = Solving for {f } we obtain {f } = Thus f (t) = cos t + sin t. 45. The Laplace transform of the given equation is s Solving for {f } we obtain {y } = Thus y = sin t − 1 t sin t. 2 s2 − s + 1 1 1 2s =2 . − (s2 + 1)2 s + 1 2 (s2 + 1)2 {y } − y (0) = {1} − {sin t} − {1} {y }. s2 s 1 +2 . +1 s +1 {cos t} + e−t {f }. 67 4.4 Additional Operational Properties 48. The differential equation is di 1 0.005 + i + dt 0.02 or di + 200i + 10,000 dt 0 t t i 2 i(τ )dτ = 100 t − (t − 1) 0 (t − 1) 1.5 i(τ )dτ = 20,000 t − (t − 1) (t − 1) , 1 0.5 0.5 1 1.5 2t where i(0) = 0. The Laplace transform of the differential equation is s {i} + 200 {i} + 10,000 s {i} = 20,000 1 1 − 2 e−s . 2 s s Solving for {i} we obtain {i} = 20,000 2 2 200 (1 − e−s ). (1 − e−s ) = − − s(s + 100)2 s s + 100 (s + 100)2 Thus i(t) = 2 − 2e−100t − 200te−100t − 2 51. Using integration by parts, {f (t)} = 1 1 − e−bs b 0 (t − 1) + 2e−100(t−1) (t − 1) + 200(t − 1)e−100(t−1) (t − 1). a −st a te dt = b s 1 1 − bs ebs − 1 . 54. {f (t)} = 1 1 − e−2πs π 0 e−st sin t dt = 1 1 · s2 + 1 1 − e−πs 57. The differential equation is x + 2x + 10x = 20f (t), where f (t) is the meander function in Problem 49 with a = π . Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform we obtain (s2 + 2s + 10) {x(t)} = 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 = Then {x(t)} = 20 40 + (−1)n e−nπs s(s2 + 2s + 10) s(s2 + 2s + 10) n=1 4 2s + 4 4s + 8 2 − + − e−nπs (−1)n s s2 + 2s + 10 n=1 s s2 + 2s + 10 1 2 2(s + 1) + 2 (s + 1) + 1 (−1)n − +4 − e−nπs s (s + 1)2 + 9 s (s + 1)2 + 9 n=1 ∞ ∞ ∞ 20 40 (−1)n e−nπs . + s s n=1 ∞ = = 68 4.5 and The Dirac Delta Function 1 x(t) = 2 1 − e−t cos 3t − e−t sin 3t + 4 (−1)n 1 − e−(t−nπ) cos 3(t − nπ ) 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 EXERCISES 4.5 The Dirac Delta Function 3. The Laplace transform of the differential equation yields {y } = so that y = sin t + sin t 6. The Laplace transform of the differential equation yields {y } = so that y = cos t + sin t[ (t − 2π ) + (t − 4π )]. s 1 + (e−2πs + e−4πs ) s2 + 1 s2 + 1 (t − 2π ). 1 1 + e−2πs s2 + 1 9. The Laplace transform of the differential equation yields {y } = so that y = e−2(t−2π) sin t (t − 2π ). 1 e−2πs (s + 2)2 + 1 69 4.5 The Dirac Delta Function 12. The Laplace transform of the differential equation yields {y } = 1 e−2s + e−4s + (s − 1)2 (s − 6) (s − 1)(s − 6) 1 1 1 1 11 11 1 1 − +− + + 2 25 s − 1 5 (s − 1) 25 s − 6 5 s−1 5 s−6 e−2s + e−4s =− so that y=− 1t 1t 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). EXERCISES 4.6 Systems of Linear Differential Equations 3. Taking the Laplace transform of the system gives s s so that {x} = and x = − cos 3t − Then y= −s − 5 s 53 =− 2 − 2+9 s s + 9 3 s2 + 9 5 sin 3t. 3 {x} + 1 = {y } − 2 = 5 {x} − 2 {x} − {y } {y } 1 7 1 x − x = 2 cos 3t − sin 3t. 2 2 3 {x} − (s − 1) {x} + (s + 2) {y } = −1 {y } = 1 6. Taking the Laplace transform of the system gives (s + 1) s so that s + 1/2 s + 1/2 √ = +s+1 (s + 1/2)2 + ( 3/2)2 √ √ −3/2 3/2 √ {x} = 2 =− 3 . s +s+1 (s + 1/2)2 + ( 3/2)2 {y } = s2 y=e −t/2 and Then 3 cos t 2 √ and √ x = − 3 e−t/2 sin √ 3 t. 2 9. Adding the equations and then subtracting them gives d2 x 1 = t2 + 2t dt2 2 d2 y 1 = t2 − 2t. 2 dt 2 70 4.6 Taking the Laplace transform of the system gives Systems of Linear Differential Equations and 1 1 4! 1 3! {x} = 8 + + 5 s 24 s 3 s4 {y } = 1 4! 1 3! − 24 s5 3 s4 and y= 1 4 13 t − t. 24 3 so that x=8+ 1 4 13 t+ t 24 3 12. Taking the Laplace transform of the system gives (s − 4) −3 so that {x} = = and {y } = = Then x= and y= 1 t 1 2t e − e + −et−1 + e2(t−1) 2 2 (t − 1) (t − 1). −1/2 1 + e−s (s − 1)(s − 2) (s − 1)(s − 2) 11 11 1 1 − + e−s − + 2 s−1 2 s−2 s−1 s−2 e−s −s/2 + 2 s/4 − 1 + + e−s s (s − 1)(s − 2) (s − 1)(s − 2) 31 11 1 131 − + e−s − + . 4 s−1 2 s−2 s 2 s−1 s−2 {x} + 2 {y } = {y } = 2e−s s 1 e−s + 2 s {x} + (s + 1) 3 t 1 2t 3 e − e + 1 − et−1 + e2(t−1) 4 2 2 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 (s + 500) 400 so that {i3 } = Then i3 = 80 80 −900t −e 9 9 s2 {i2 } + 500 {i3 } = {i3 } = 10,000 s 8,000 s {i2 } + (s + 400) 8,000 80 1 80 1 = − . + 900s 9s 9 s + 900 i2 = 20 − 0.0025i3 − i3 = 100 100 −900t . − e 9 9 and 71 4.6 Systems of Linear Differential Equations (c) i1 = i2 + i3 = 20 − 20e−900t 18. Taking the Laplace transform of the system 0.5i1 + 50i2 = 60 0.005i2 + i2 − i1 = 0 gives s −200 so that {i2 } = Then i2 = and 6 6 −100t 6 cos 100t − e−100t sin 100t −e 55 5 6 6 −100t −e cos 100t. 55 {i1 } + 100 120 s {i2 } = 0 {i2 } = {i1 } + (s + 200) 24,000 6 61 6 s + 100 100 − . = − s(s2 + 200s + 20,000) 5 s 5 (s + 100)2 + 1002 5 (s + 100)2 + 1002 i1 = 0.005i2 + i2 = CHAPTER 4 REVIEW EXERCISES 3. False; consider f (t) = t−1/2 . 6. False; consider f (t) = 1 and g (t) = 1. 9. 12. 15. 18. 21. 24. 0 t {sin 2t} = {sin 2t 2 s2 + 4 {sin 2(t − π ) 2 (s − 5)3 (t − 5) (t − π )} = = 1 2 5t te 2 2 e−πs s2 + 4 (t − π )} = = 1 2 1 (s − 5)3 1 −5s e s2 = (t − 5) e−5t exists for s > −5. t eaτ f (τ ) dτ eat 0 = = 1 s {eat f (t)} = t F (s − a) , whereas s = s→s−a f (τ ) dτ (t − t0 ) f (τ ) dτ 0 F (s) s = s→s−a F (s − a) . s−a 27. f (t − t0 ) 72 CHAPTER 4 REVIEW EXERCISES (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 e−π(s−1) + e−3π(s−1) et f (t) = − 2+1 (s − 1) (s − 1)2 + 1 5 1 2 + 2 (s − 1) 2 (s − 1)3 (t − 3π ) 30. f (t) = sin t 33. Taking the Laplace transform of the differential equation we obtain {y } = so that 1 y = 5tet + t2 et . 2 36. Taking the Laplace transform of the differential equation we obtain {y } = s3 + 2 2 + 2s + s2 −s −3 e s3 (s − 5) s (s − 5) 21 12 127 1 12 37 21 37 1 12 1 1 − + − + − −− − e−s 2 3 2 3 125 s 25 s 5s 125 s − 5 125 s 25 s 5s 125 s − 5 =− so that y=− 2 2 1 12 1 127 5t 37 37 5(t−1) − t − t2 + e−− − (t − 1) − (t − 1)2 + e 125 25 5 125 125 25 5 125 (t − 1). 39. Taking the Laplace transform of the system gives s 4 so that {x} = Then s2 − 2s + 1 11 1 1 91 =− + + . s(s − 2)(s + 2) 4 s 8 s−2 8 s+2 and y = −x + t = 9 −2t 1 2t − e + t. e 4 4 {x} + {x} + s {y } = 1 +1 s2 {y } = 2 11 9 x = − + e2t + e−2t 48 8 42. The differential equation is 1 d2 q dq + 10 + 100q = 10 − 10 2 dt2 dt Taking the Laplace transform we obtain {q } = = so that q (t) = 20 1 − e−5s s(s2 + 20s + 200) 11 1 s + 10 10 1 − − 2 + 102 10 s 10 (s + 10) 10 (s + 10)2 + 102 1 − e−5s (t − 5). 1 1 1 − e−10t cos 10t − e−10t sin 10t 10 10 10 1 1 1 − − e−10(t−5) cos 10(t − 5) − e−10(t−5) sin 10(t − 5) 10 10 10 (t − 5). 73 CHAPTER 4 REVIEW EXERCISES 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) = −K Θ1 (s) + K Θ2 (s) + sθ0 (s2 + ω 2 )Θ2 (s) = K Θ1 (s) − K Θ2 (s) + sψ0 . If we add the two equations, we get Θ1 (s) + Θ2 (s) = (θ0 + ψ0 ) 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) − K Θ2 (s) = sθ0 −k Θ1 (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) = so s(s2 + ω 2 + K )θ0 + Ksψ0 θ0 − ψ0 θ0 + ψ0 s s + = , 2 + ω 2 )(s2 + ω 2 + 2K ) 2 + ω2 2 + ω 2 + 2K (s 2 s 2 s θ1 (t) = ω 2 + 2K t. s2 s + ω2 θ0 + ψ0 θ0 − ψ0 cos ωt + cos 2 2 Then from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt we get θ2 (t) = θ0 + ψ0 θ0 − ψ0 cos ωt − cos 2 2 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. 74 ...
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