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# Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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1)4 . Using partial fractions, s2 − 4s+ 7 (s− 1)4 = 4 (s− 1)4 − 2 (s− 1)3 + 1 (s− 1)2 . Note that L [ tn ] = (n!)/sn+1 and L [eatf(t)] = L [f(t)]s→s−a . Hence the solution of the IVP is y(t) = L−1 [ s2 − 4s+ 7 (s− 1)4 ] = 2 3 t3et − t2et + t et. 18. Taking the Laplace transform of the ODE, we obtain s4 Y (s)− s3y(0)− s2y ′(0)− s y ′′(0)− y ′′′(0)− Y (s) = 0 . Applying the initial conditions, s4Y (s)− Y (s)− s3 − s = 0 . Solving for the transform of the solution, Y (s) = s s2 − 1 . By inspection, it follows that y(t) = L−1 [ s s2−1 ] = cosh t . 19. Taking the Laplace transform of the ODE, we obtain s4 Y (s)− s3y(0)− s2y ′(0)− s y ′′(0)− y ′′′(0)− 4Y (s) = 0 . Applying the initial conditions, s4Y (s)− 4Y (s)− s3 + 2s = 0 . Solving for the transform of the solution, Y (s) = s s2 + 2 . It follows that y(t) = L−1 [ s s2+2 ] = cos √ 2 t . 20. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + ω2 Y (s) = s s2 + 4 . Applying the initial conditions, s2 Y (s) + ω2 Y (s)− s = s s2 + 4 . 6.2 237 Solving for Y (s), the transform of the solution is Y (s) = s (s2 + ω2)(s2 + 4) + s s2 + ω2 . Using partial fractions on the first term, s (s2 + ω2)(s2 + 4) = 1 4− ω2 [ s s2 + ω2 − s s2 + 4 ] . First note that L−1 [ s s2 + ω2 ] = cos ωt and L−1 [ s s2 + 4 ] = cos 2t . Hence the solution of the IVP is y(t) = 1 4− ω2 cos ωt− 1 4− ω2 cos 2t+ cos ωt = = 5− ω2 4− ω2 cos ωt− 1 4− ω2 cos 2t . 21. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0)− 2 [s Y (s)− y(0)] + 2Y (s) = s s2 + 1 . Applying the initial conditions, s2 Y (s)− 2s Y (s) + 2Y (s)− s+ 2 = s s2 + 1 . Solving for Y (s), the transform of the solution is Y (s) = s (s2 − 2s+ 2)(s2 + 1) + s− 2 s2 − 2s+ 2 . Using partial fractions on the first term, s (s2 − 2s+ 2)(s2 + 1) = 1 5 [ s− 2 s2 + 1 − s− 4 s2 − 2s+ 2 ] . Thus we can write Y (s) = 1 5 s s2 + 1 − 2 5 1 s2 + 1 + 2 5 2s− 3 s2 − 2s+ 2 . For the last term, we note that s2 − 2s+ 2 = (s− 1)2 + 1 . So that 2s− 3 s2 − 2s+ 2 = 2(s− 1)− 1 (s− 1)2 + 1 . We know that L−1 [ 2 ξ ξ2 + 1 − 1 ξ2 + 1 ] = 2 cos t− sin t . Based on the translation property of the Laplace transform, L−1 [ 2s− 3 s2 − 2s+ 2 ] = et(2 cos t− sin t). 238 Chapter 6. The Laplace Transform Combining the above, the solution of the IVP is y(t) = 1 5 cos t− 2 5 sin t+ 2 5 et(2 cos t− sin t) . 23. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 2 [s Y (s)− y(0)] + Y (s) = 4 s+ 1 . Applying the initial conditions, s2 Y (s) + 2s Y (s) + Y (s)− 2s− 3 = 4 s+ 1 . Solving for Y (s), the transform of the solution is Y (s) = 4 (s+ 1)3 + 2s+ 3 (s+ 1)2 . First write 2s+ 3 (s+ 1)2 = 2(s+ 1) + 1 (s+ 1)2 = 2 s+ 1 + 1 (s+ 1)2 . We note that L−1 [ 4 ξ3 + 2 ξ + 1 ξ2 ] = 2 t2 + 2 + t . So based on the translation property of the Laplace transform, the solution of the IVP is y(t) = 2 t2e−t + t e−t + 2 e−t. 25. Let f(t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + Y (s) = L [f(t)] . Applying the initial conditions, s2 Y (s) + Y (s) = L [f(t)] . Based on the definition of the Laplace transform, L [f(t)] = ∫ ∞ 0 f(t) e−stdt = ∫ 1 0 t e−stdt = 1 s2 − e −s s − e −s s2 . Solving for the transform, Y (s) = 1 s2(s2 + 1) − e−s s+ 1 s2(s2 + 1) . Using partial fractions, 1 s2(s2 + 1) = 1 s2 − 1 s2 + 1 and s s2(s2 + 1) = 1 s − s s2 + 1 . 6.2 239 We find, by inspection, that L−1 [ 1 s2(s2 + 1) ] = t− sin t . Referring to Line 13 , in Table 6.2.1 , L [uc(t)f(t− c)] = e−csL [f(t)] . Let L [ g(t) ] = s+ 1 s2(s2 + 1) = 1 s + 1 s2 − s s2 + 1 − 1 s2 + 1 . Then g(t) = 1 + t− cos t− sin t . It follows, therefore, that L−1 [ e−s · s+ 1 s2(s2 + 1) ] = u1(t) [ 1 + (t− 1)− cos(t− 1)− sin(t− 1) ] . Combining the above, the solution of the IVP is y(t) = t− sin t− u1(t) [ t− cos(t− 1)− sin(t− 1) ] . 26. Let f(t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 4Y (s) = L [f(t)] . Applying the initial conditions, s2 Y (s) + 4Y (s) = L [f(t)] . Based on the definition of the Laplace transform, L [f(t)] = ∫ ∞ 0 f(t) e−stdt = ∫ 1 0 t e−stdt+ ∫ ∞ 1 e−stdt = 1 s2 − e −s s2 . Solving for the transform, Y (s) = 1 s2(s2 + 4) − e−s 1 s2(s2 + 4) . Using partial fractions, 1 s2(s2 + 4) = 1 4 [ 1 s2 − 1 s2 + 4 ] . We find that L−1 [ 1 s2(s2 + 4) ] = 1 4 t− 1 8 sin 2t . Referring to Line 13 , in Table 6.2.1 , L [uc(t)f(t− c)] = e−csL [f(t)] . It follows that L−1 [ e−s · 1 s2(s2 + 4) ] = u1(t) [ 1 4 (t− 1)− 1 8 sin 2(t− 1) ] . 240 Chapter 6. The Laplace Transform Combining the above, the solution of the IVP is y(t) = 1 4 t− 1 8 sin 2t− u1(t) [ 1 4 (t− 1)− 1 8 sin 2(t− 1) ] . 28.(a) Assuming that the conditions of Theorem 6.2.1 are satisfied, F ′(s) = d ds ∫ ∞ 0 e−stf(t)dt = ∫ ∞ 0 ∂ ∂s [ e−stf(t) ] dt = = ∫ ∞ 0 [−t e−stf(t)] dt = ∫ ∞ 0 e−st [−tf(t)] dt . (b) Using mathematical induction, suppose that for some k ≥ 1 , F (k)(s) = ∫ ∞ 0 e−st [ (−t)kf(t)] dt . Differentiating both sides, F (k+1)(s) = d ds ∫ ∞ 0 e−st [ (−t)kf(t)] dt = ∫ ∞ 0 ∂ ∂s [ e−st(−t)kf(t)] dt = = ∫ ∞ 0 [−t e−st(−t)kf(t)] dt = ∫ ∞ 0 e−st [ (−t)k+1f(t)] dt . 29. We know that L [ eat ] = 1 s− a . Based on Problem 28, L [−t eat] = d ds [ 1 s− a ] . Therefore, L [ t eat ] = 1 (s− a)2 . 31. Based on Problem 28, L [(−t)n] = d n dsn L [ 1 ] = d n dsn [ 1 s ] . Therefore, L [ tn ] = (−1)n (−1) nn! sn+1 = n! sn+1 . 33. Using the translation property of the Laplace transform, L [ eat sin bt ] = b (s− a)2 + b2 . Therefore, L [ t eat sin bt ] = − d ds [ b (s− a)2 + b2 ] = 2b(s− a) ((s− a)2 + b2)2 . 6.2 241 34. Using the translation property of the Laplace transform, L [ eat cos bt ] = s− a (s− a)2 + b2 . Therefore, L [ t eat cos bt ] = − d ds [ s− a (s− a)2 + b2 ] = (s− a)2 − b2 ((s− a)2 + b2)2 . 35.(a) Taking the Laplace transform of the given Bessel equation, L [ t y ′′] + L [ y ′] + L [ t y ] = 0 . Using the differentiation property of the transform, − d ds L [ y ′′] + L [ y ′]− d ds L [ y ] = 0 . That is, − d ds [ s2Y (s)− s y(0)− y ′(0)]+ s Y (s)− y(0)− d ds Y (s) = 0 . It follows that (1 + s2)Y ′(s) + s Y (s) = 0 . (b) We obtain a first-order linear ODE in Y (s): Y ′(s) + s s2 + 1 Y (s) = 0 , with integrating factor µ(s) = e ∫ s s2+1 ds = √ s2 + 1 . The first-order ODE can be written as d ds [√ s2 + 1 · Y (s) ] = 0 , with solution Y (s) = c√ s2 + 1 . (c) In order to obtain negative powers of s , first write 1√ s2 + 1 = 1 s [ 1 + 1 s2 ]−1/2 . Expanding (1 + 1s2 ) −1/2 in a binomial series, 1√ 1 + (1/s2) = 1− 1 2 s−2 + 1 · 3 2 · 4 s −4 − 1 · 3 · 5 2 · 4 · 6 s −6 + · · · , 242 Chapter 6. The Laplace Transform valid for s−2 < 1 . Hence, we can formally express Y (s) as Y (s) = c [ 1 s − 1 2 1 s3 + 1 · 3 2 · 4 1 s5