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# Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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+ 2 [s Y (s)− y(0)] + 2Y (s) = L [h(t)] . 252 Chapter 6. The Laplace Transform Applying the initial conditions, s2 Y (s) + 2s Y (s) + 2Y (s)− 1 = L [h(t)] . The forcing function can be written as h(t) = upi(t)− u2pi(t) . Its transform is L [h(t)] = e −pis − e−2pis s . Solving for Y (s), the transform of the solution is Y (s) = 1 s2 + 2s+ 2 + e−pis − e−2pis s(s2 + 2s+ 2) . First note that 1 s2 + 2s+ 2 = 1 (s+ 1)2 + 1 . Using partial fractions, 1 s(s2 + 2s+ 2) = 1 2 1 s − 1 2 (s+ 1) + 1 (s+ 1)2 + 1 . Taking the inverse transform, term-by-term, L [ 1 s2 + 2s+ 2 ] = L [ 1 (s+ 1)2 + 1 ] = e−t sin t . Now let G(s) = 1 s(s2 + 2s+ 2) . Then L−1 [G(s)] = 1 2 − 1 2 e−t cos t− 1 2 e−t sin t . Using Theorem 6.3.1, L−1 [ e−csG(s) ] = 1 2 uc(t)− 12e −(t−c) [cos(t− c) + sin(t− c)]uc(t) . Hence the solution of the IVP is y(t) = e−t sin t+ 1 2 upi(t)− 12e −(t−pi) [cos(t− pi) + sin(t− pi)]upi(t)− − 1 2 u2pi(t) + 1 2 e−(t−2pi) [cos(t− 2pi) + sin(t− 2pi)]u2pi(t) . That is, y(t) = e−t sin t+ 1 2 [upi(t)− u2pi(t)] + 12e −(t−pi) [cos t+ sin t] upi(t)+ + 1 2 e−(t−2pi) [cos t+ sin t]u2pi(t) . 6.4 253 (b) The solution starts out as free oscillation, due to the initial conditions. The am- plitude increases, as long as the forcing is present. Thereafter, the solution rapidly decays. 4.(a) Let h(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 [h(t)] . Applying the initial conditions, s2 Y (s) + 4Y (s) = L [h(t)] . The transform of the forcing function is L [h(t)] = 1 s2 + 1 + e−pis s2 + 1 . Solving for Y (s), the transform of the solution is Y (s) = 1 (s2 + 4)(s2 + 1) + e−pis (s2 + 4)(s2 + 1) . Using partial fractions, 1 (s2 + 4)(s2 + 1) = 1 3 [ 1 s2 + 1 − 1 s2 + 4 ] . It follows that L−1 [ 1 (s2 + 4)(s2 + 1) ] = 1 3 [ sin t− 1 2 sin 2t ] . Based on Theorem 6.3.1, L−1 [ e−pis (s2 + 4)(s2 + 1) ] = 1 3 [ sin(t− pi)− 1 2 sin(2t− 2pi) ] upi(t) . Hence the solution of the IVP is y(t) = 1 3 [ sin t− 1 2 sin 2t ] − 1 3 [ sin t+ 1 2 sin 2t ] upi(t) . 254 Chapter 6. The Laplace Transform (b) Since there is no damping term, the solution follows the forcing function, after which the response is a steady oscillation about y = 0. 5.(a) 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) + 3 [s Y (s)− y(0)] + 2Y (s) = L [f(t)] . Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2Y (s) = L [f(t)] . The transform of the forcing function is L [f(t)] = 1 s − e −10s s . Solving for the transform, Y (s) = 1 s(s2 + 3s+ 2) − e −10s s(s2 + 3s+ 2) . Using partial fractions, 1 s(s2 + 3s+ 2) = 1 2 [ 1 s + 1 s+ 2 − 2 s+ 1 ] . Hence L−1 [ 1 s(s2 + 3s+ 2) ] = 1 2 + e−2t 2 − e−t . Based on Theorem 6.3.1, L−1 [ e−10s s(s2 + 3s+ 2) ] = 1 2 [ 1 + e−2(t−10) − 2e−(t−10) ] u10(t) . Hence the solution of the IVP is y(t) = 1 2 [1− u10(t)] + e −2t 2 − e−t − 1 2 [ e−(2t−20) − 2e−(t−10) ] u10(t) . 6.4 255 (b) The solution increases to a temporary steady value of y = 1/2. After the forcing ceases, the response decays exponentially to y = 0 . 6.(a) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 3 [s Y (s)− y(0)] + 2Y (s) = e −2s s . Applying the initial conditions, s2 Y (s) + 3s Y (s) + 2Y (s)− 1 = e −2s s . Solving for the transform, Y (s) = 1 s2 + 3s+ 2 + e−2s s(s2 + 3s+ 2) . Using partial fractions, 1 s2 + 3s+ 2 = 1 s+ 1 − 1 s+ 2 and 1 s(s2 + 3s+ 2) = 1 2 [ 1 s + 1 s+ 2 − 2 s+ 1 ] . Taking the inverse transform. term-by-term, the solution of the IVP is y(t) = e−t − e−2t + [ 1 2 − e−(t−2) + 1 2 e−2(t−2) ] u2(t) . 256 Chapter 6. The Laplace Transform (b) Due to the initial conditions, the response has a transient overshoot, followed by an exponential convergence to a steady value of ys = 1/2 . 7.(a) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + Y (s) = e −3pis s . Applying the initial conditions, s2 Y (s) + Y (s)− s = e −3pis s . Solving for the transform, Y (s) = s s2 + 1 + e−3pis s(s2 + 1) . Using partial fractions, 1 s(s2 + 1) = 1 s − s s2 + 1 . Hence Y (s) = s s2 + 1 + e−3pis [ 1 s − s s2 + 1 ] . Taking the inverse transform, the solution of the IVP is y(t) = cos t+ [1− cos(t− 3pi)]u3pi(t) = cos t+ [1 + cos t ]u3pi(t) . 6.4 257 (b) Due to initial conditions, the solution temporarily oscillates about y = 0 . After the forcing is applied, the response is a steady oscillation about ym = 1 . 9.(a) Let g(t) be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + Y (s) = L [g(t)] . Applying the initial conditions, s2 Y (s) + Y (s)− 1 = L [g(t)] . The forcing function can be written as g(t) = t 2 [1− u6(t)] + 3u6(t) = t2 − 1 2 (t− 6)u6(t) with Laplace transform L [g(t)] = 1 2s2 − e −6s 2s2 . Solving for the transform, Y (s) = 1 s2 + 1 + 1 2s2(s2 + 1) − e −6s 2s2(s2 + 1) . Using partial fractions, 1 2s2(s2 + 1) = 1 2 [ 1 s2 − 1 s2 + 1 ] . Taking the inverse transform, and using Theorem 6.3.1, the solution of the IVP is y(t) = sin t+ 1 2 [t− sin t]− 1 2 [(t− 6)− sin(t− 6)]u6(t) = 1 2 [t+ sin t]− 1 2 [(t− 6)− sin(t− 6)]u6(t). 258 Chapter 6. The Laplace Transform (b) The solution increases, in response to the ramp input, and thereafter oscillates about a mean value of ym = 3 . 11.(a) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 4Y (s) = e −pis s − e −3pis s . Applying the initial conditions, s2 Y (s) + 4Y (s) = e−pis s − e −3pis s . Solving for the transform, Y (s) = e−pis s(s2 + 4) − e −3pis s(s2 + 4) . Using partial fractions, 1 s(s2 + 4) = 1 4 [ 1 s − s s2 + 4 ] . Taking the inverse transform, and applying Theorem 6.3.1, y(t) = 1 4 [1− cos(2t− 2pi)]upi(t)− 14 [1− cos(2t− 6pi)]u3pi(t) = 1 4 [upi(t)− u3pi(t)]− 14 cos 2t · [upi(t)− u3pi(t)] . 6.4 259 (b) Since there is no damping term, the solution responds immediately to the forcing input. There is a temporary oscillation about y = 1/4 . 12.(a) Taking the Laplace transform of the ODE, we obtain s4 Y (s)− s3y(0)− s2y ′(0)− s y ′′(0)− y ′′′(0)− Y (s) = e −s s − e −2s s . Applying the initial conditions, s4Y (s)− Y (s) = e −s s − e −2s s . Solving for the transform of the solution, Y (s) = e−s s(s4 − 1) − e−2s s(s4 − 1) . Using partial fractions, 1 s(s4 − 1) = 1 4 [ −4 s + 1 s+ 1 + 1 s− 1 + 2s s2 + 1 ] . It follows that L−1 [ 1 s(s4 − 1) ] = 1 4 [−4 + e−t + et + 2 cos t] . Based on Theorem 6.3.1, the solution of the IVP is y(t) = − [u1(t)− u2(t)] + 14 [ e−(t−1) + e(t−1) + 2 cos(t− 1) ] u1(t)− − 1 4 [ e−(t−2) + e(t−2) + 2 cos(t− 2) ] u2(t) . 260 Chapter 6. The Laplace Transform (b) The solution increases without bound, exponentially. 13.(a) Taking the Laplace transform of the ODE, we obtain s4 Y (s)− s3y(0)− s2y ′(0)− s y ′′(0)− y ′′′(0)+ + 5 [ s2Y (s)− s y(0)− y ′(0)]+ 4Y (s) = 1