Differential Equations Homework 6.1, 6.2, 6.3

Homework: 6.1, 6.2, 6.3 6.1, #19 Compute L(t2 sin(at)). Note the following, from Euler’s Formula: t2 eiat = t2 cos(at) + it2 sin(at) so that, by the linearity of the Laplace transform: L(t2 eiat ) = L(t2 cos(at)) + iL(t2 sin(at)) so that we need to compute L(t2 eiat ), then find the imaginary part of our answer. Now: Z 2 iat L(t e ) = ∞ t2 e−t(s−ia) dt 0 so that integration by parts yields: −t2 −t(s−ia) e s − ia ∞ − 0 2t e−t(s−ia) (s − ia)2 ∞ − 0 2 e−t(s−ia) (s − ia)3 ∞ 0 After evaluating the limits, we see that: L(t2 eiat ) = 2 (s − ia)3 which is, after multiplying top and bottom by (s + ia)3 , 2s3 − 6a2 s 6as2 − 2a3 L(t e ) = 2 +i 2 (s + a2 )3 (s + a2 )3 2 iat Therefore, L(t2 sin(at)) = 6as2 − 2a3 (s2 + a2 )3 NOTE: We can verify this by computing the Laplace transform by using Table Entry 19: (−t)n f (t) ↔ F (n) (s) 6.1, #21 A straightforward computation shows this integral converges: Z ∞ π π 1 T dt = lim tan−1 (t) 0 = − 0 = 2 T →∞ 1+t 2 2 0 6.1, #23 Does the integral converge or diverge? Z 1 ∞ et dt t2 First, we note that: et et et = lim = lim =∞ t→∞ t2 t→∞ 2t t→∞ 2 Therefore, we suspect that the integral diverges. However, we will not be able to find a simple antiderivative. We must instead compare this integral to an integral that is Z ∞ 1 dt = 1, so this function won’t work). known to diverge. (Side Remark: Note that t2 1 lim 1 t From our previous computation, we know that te2 goes to infinity. If we can show that, t for some constant M and some time T , te2 > M for all t ≥ T , then we will be almost done! t et . THIS IS VERY NICE! It says that f (t) is increasing If f (t) = te2 , then f 0 (t) = t−2 t3 for all time after t = 2. Therefore, f (t) ≥ f (2), for all t ≥ 2. Therefore: et e2 > , for all t > 2 t2 4 so that: Z ∞ 2 Z e e2 ∞ et dt > dt = 1 dt t2 4 4 2 2 2 The last integral above diverges. Our conclusion: The integral Z Z 1 ∞ ∞ et dt diverges. t2 6.2, #8 Compute the inverse Laplace transform. 8s2 − 4s + 12 A Bs + C = + 2 2 s(s + 4) s s +4 where: A+B =0 C = −4 4A = 12 From which A = 3, B = 5, C = −4, so: 1 s 2 8s2 − 4s + 12 =3 +5 2 −2 2 2 s(s + 4) s s +4 s +4 so that the inverse transform is: 3 + 5 cos(2t) − 2 sin(2t) 6.2, #15 Solve y 00 − 2y 0 − 2y = 0, y(0) = 2, y 0 (0) = 0. First, we compute the transform of both sides to get: s2 Y − 2s − 2(sY − 2) − 2Y = 0 2s − 4 Y = 2 s − 2s − 2 √ 2 We note that the roots of s − 2s − 2 are −1 ± 3, which is messy to work with. We will instead complete the square and use the table entries with hyperbolic sines and cosines (Note: Only do this as a ”last resort”, always try to get nice factors first). √ 2s − 4 (s − 1) 2 3 =2 −√ 2 2 s − 2s − 2 (s − 1) − 3 3 (s − 1)2 − 3 so that y(t) = e t   √ 2 2 cosh( 3t) − √ sinh( 3t) 3 √ 2 6.3, #6 Sketch f (t) = u1 (t)(t − 1) − u2 (t)2(t − 2) + u3 (t)(t − 3) To begin, write f (t) piecewise:  0,    t − 1, f (t) = (t − 1) − 2t + 2,    (t − 1) − 2t + 2 + t − 3, 0≤t<1 1≤t<2 2≤t<3 t≥3 which simplifies to:  0,    t − 1, f (t) = 3 − t,    0, 6.3, #14 Invert: 0≤t<1 1≤t<2 2≤t<3 t≥3 e−2s s2 +s−2 First, think of this as e−2s H(s), so that the inverse transform is: u2 (t)h(t − 2) Now we need to find h(t): See if the denominator is easily factorable, and in this case it is: s2 +s−2 = (s+2)(s−1). Therefore, 1 A B H(s) = 2 = + s +s−2 s+2 s−1 When we solve for A, B, we get: A = − 13 , B = 13 . Therefore, h(t) = −1 −2t 1 t e + e 3 3 6.3, #20 This problem could be solved by a variety of means- the answer is 2(2t)n 3