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s − e −pis s . Applying the initial conditions, s4Y (s) + 5s2Y (s) + 4Y (s) = 1 s − e −pis s . Solving for the transform of the solution, Y (s) = 1 s(s4 + 5s2 + 4) − e −pis s(s4 + 5s2 + 4) . Using partial fractions, 1 s(s4 + 5s2 + 4) = 1 12 [ 3 s + s s2 + 4 − 4s s2 + 1 ] . It follows that L−1 [ 1 s(s4 + 5s2 + 4) ] = 1 12 [3 + cos 2t− 4 cos t] . Based on Theorem 6.3.1, the solution of the IVP is y(t) = 1 4 [1− upi(t)] + 112 [cos 2t− 4 cos t]− − 1 12 [cos 2(t− pi)− 4 cos(t− pi)]upi(t) . That is, y(t) = 1 4 [1− upi(t)] + 112 [cos 2t− 4 cos t]− − 1 12 [cos 2t+ 4 cos t]upi(t) . 6.4 261 (b) After an initial transient, the solution oscillates about ym = 0 . 14. The specified function is defined by f(t) = 0, 0 ≤ t < t0 h k (t− t0), t0 ≤ t < t0 + k h, t ≥ t0 + k which can conveniently be expressed as f(t) = h k (t− t0)ut0(t)− h k (t− t0 − k)ut0+k(t) . 15. The function is defined by g(t) = 0, 0 ≤ t < t0 h k (t− t0), t0 ≤ t < t0 + k −hk (t− t0 − 2k), t0 + k ≤ t < t0 + 2k 0, t ≥ t0 + 2k which can also be written as g(t) = h k (t− t0)ut0(t)− 2h k (t− t0 − k)ut0+k(t) + h k (t− t0 − 2k)ut0+2k(t). 16.(d) From part (c) in the Student Solutions Manual, the solution is u(t) = 4k u3/2(t)h ( t− 3 2 ) − 4k u5/2(t)h ( t− 5 2 ) , where h(t) = 1 4 − √ 7 84 e−t/8 sin ( 3 √ 7 t 8 ) − 1 4 e−t/8 cos ( 3 √ 7 t 8 ) . Due to the damping term, the solution will decay to zero. The maximum will occur shortly after the forcing ceases. By plotting the various solutions, it appears that the solution will reach a value of y = 2 , as long as k > 2.51 . 262 Chapter 6. The Laplace Transform (e) Based on the graph, and numerical calculation, |u(t)| < 0.1 for t > 25.6773 . 17. We consider the initial value problem y ′′ + 4y = 1 k [(t− 5)u5(t)− (t− 5− k)u5+k(t)] , with y(0) = y ′(0) = 0 . (a) The specified function is defined by f(t) = 0, 0 ≤ t < 5 1 k (t− 5), 5 ≤ t < 5 + k 1, t ≥ 5 + k (b) Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 4Y (s) = e −5s ks2 − e −(5+k)s ks2 . Applying the initial conditions, s2 Y (s) + 4Y (s) = e−5s ks2 − e −(5+k)s ks2 . 6.4 263 Solving for the transform, Y (s) = e−5s ks2(s2 + 4) − e −(5+k)s ks2(s2 + 4) . Using partial fractions, 1 s2(s2 + 4) = 1 4 [ 1 s2 − 1 s2 + 4 ] . It follows that L−1 [ 1 s2(s2 + 4) ] = 1 4 t− 1 8 sin 2t . Using Theorem 6.3.1, the solution of the IVP is y(t) = 1 k [h(t− 5)u5(t)− h(t− 5− k)u5+k(t)] , in which h(t) = 14 t− 18 sin 2t . (c) Note that for t > 5 + k , the solution is given by y(t) = 1 4 − 1 8k sin(2t− 10) + 1 8k sin(2t− 10− 2k) = 1 4 − sin k 4k cos(2t− 10− k) . So for t > 5 + k , the solution oscillates about ym = 1/4 , with an amplitude of A = |sin(k)| 4k . (a) k = 1 (b) k = 3 (c) k = 5 264 Chapter 6. The Laplace Transform 18.(a) (b) The forcing function can be expressed as fk(t) = 1 2k [u4−k(t)− u4+k(t)] . Taking the Laplace transform of both sides of the ODE, we obtain s2 Y (s)− s y(0)− y ′(0) + 1 3 [s Y (s)− y(0)] + 4Y (s) = e −(4−k)s 2ks − e −(4+k)s 2ks . Applying the initial conditions, s2 Y (s) + 1 3 s Y (s) + 4Y (s) = e−(4−k)s 2ks − e −(4+k)s 2ks . Solving for the transform, Y (s) = 3 e−(4−k)s 2ks(3s2 + s+ 12) − 3 e −(4+k)s 2ks(3s2 + s+ 12) . Using partial fractions, 1 s(3s2 + s+ 12) = 1 12 [ 1 s − 1 + 3s 3s2 + s+ 12 ] = 1 12 [ 1 s − 1 6 1 + 6(s+ 16 ) (s+ 16 ) 2 + 14336 ] . Let H(s) = 1 8k [ 1 s − 1 6 (s+ 16 ) 2 + 14336 − s+ 1 6 (s+ 16 ) 2 + 14336 ] . It follows that h(t) = L−1 [H(s)] = 1 8k − e −t/6 8k [ 1√ 143 sin (√ 143 t 6 ) + cos (√ 143 t 6 )] . Based on Theorem 6.3.1, the solution of the IVP is y(t) = h(t− 4 + k)u4−k(t)− h(t− 4− k)u4+k(t) . 6.4 265 (c) (a) k = 2 (b) k = 1 (c) k = 1/2 As the parameter k decreases, the solution remains null for a longer period of time. Since the magnitude of the impulsive force increases, the initial overshoot of the response also increases. The duration of the impulse decreases. All solutions eventually decay to y = 0 . 19.(a) (c) From part (b) in the Student Solutions Manual, u(t) = 1− cos t+ 2 n∑ k=1 (−1)k [1− cos(t− kpi)]ukpi(t) . 266 Chapter 6. The Laplace Transform 21.(a) (b) Taking the Laplace transform of both sides of the ODE, we obtain s2 U(s)− s u(0)− u ′(0) + U(s) = 1 s + n∑ k=1 (−1)ke−kpis s . Applying the initial conditions, s2 U(s) + U(s) = 1 s + n∑ k=1 (−1)ke−kpis s . Solving for the transform, U(s) = 1 s(s2 + 1) + n∑ k=1 (−1)ke−kpis s(s2 + 1) . Using partial fractions, 1 s(s2 + 1) = 1 s − s s2 + 1 . Let h(t) = L−1 [ 1 s(s2 + 1) ] = 1− cos t . Applying Theorem 6.3.1, term-by-term, the solution of the IVP is u(t) = h(t) + n∑ k=1 (−1)kh(t− kpi)ukpi(t) . Note that h(t− kpi) = u0(t− kpi)− cos(t− kpi) = ukpi(t)− (−1)k cos t . Hence u(t) = 1− cos t+ n∑ k=1 (−1)k ukpi(t)− (cos t) n∑ k=1 ukpi(t) . 6.4 267 (c) The ODE has no damping term. Each interval of forcing adds to the energy of the system, so the amplitude will increase. For n = 15, g(t) = 0 when t > 15pi. There- fore the oscillation will eventually become steady, with an amplitude depending on the values of u(15pi) and u ′(15pi) . (d) As n increases, the interval of forcing also increases. Hence the amplitude of the transient will increase with n . Eventually, the forcing function will be constant. In fact, for large values of t , g(t) = { 1 , n even 0 , n odd Further, for t > npi , u(t) = 1− cos t− n cos t− 1− (−1) n 2 . Hence the steady state solution will oscillate about 0 or 1 , depending on n , with an amplitude of A = n+ 1 . In the limit, as n → ∞ , the forcing function will be a periodic function, with period 2pi . From Problem 33, in Section 6.3 , L [g(t)] = 1 s(1 + e−s) . As n increases, the duration and magnitude of the transient will increase without bound. 22.(a) Taking the initial conditions into consideration, the transform of the ODE is s2 U(s) + 0.1 sU(s) + U(s) = 1 s + n∑ k=1 (−1)ke−kpis s . Solving for the transform, U(s) = 1 s(s2 + 0.1s+ 1) + n∑ k=1 (−1)ke−kpis s(s2 + 0.1s+ 1) . 268 Chapter 6. The Laplace Transform Using partial fractions, 1 s(s2 + 0.1s+ 1) = 1 s − s+ 0.1 s2 + 0.1s+ 1 . Since the denominator in the second term is irreducible, write s+ 0.1 s2 + 0.1s+ 1 = (s+ 0.05) + 0.05 (s+ 0.05)2 + (399/400) . Let h(t) = L−1 [ 1 s − (s+ 0.05) (s+ 0.05)2 + (399/400) − 0.05 (s+ 0.05)2 + (399/400) ] = 1− e−t/20 [ cos( √ 399 20 t) + 1√ 399 sin( √ 399 20 t) ] . Applying Theorem 6.3.1, term-by-term, the solution of the IVP is u(t) = h(t) + n∑ k=1 (−1)kh(t− kpi)ukpi(t) . For odd values of n, the solution approaches y = 0 . For even values of n, the solution approaches y = 1 . (b) The solution is a sum of damped sinusoids, each of frequency ω = √ 399 /20 ≈ 1 . Each term has an initial amplitude of approximately 1 . For any given n , the 6.4 269 solution contains n+ 1 such terms. Although the amplitude will increase with n, the amplitude will also be bounded by n+ 1 . (c) Suppose that the forcing function is replaced by g(t) = sin t . Based on the methods in Chapter 3, the general solution of the differential equation is u(t)