Math 3400 Homework 5 Section 3.6 2. Solve by using variation of parameters and the method of undetermined coefficients. (N H) y 00 − y 0 − 2y = 2e−t . The associated homogeneous equation is (H) y 00 − y 0 − 2y = 0. The characteristic equation is r2 − r − 2 = (r − 2)(r + 1) = 0, r = 2, −1. Take y2 (t) = e−t . y1 (t) = e2t , The Wronskian for these solutions is W (y1 , y2 ) = y1 y20 − y10 y2 = −et − 2et = −3et . The variation of parameters formula give a particular solution Z t Z t y2 (s)g(s) y1 (s)g(s) yp (t) = −y1 (t) ds + y2 (t) ds. W (y1 , y2 )(s) W (y1 , y2 )(s) In this case we get 2t yp (t) = −e 2 = e2t 3 Z t e−3s t Z t 2s −s 2e−s e−s 2e e −t ds + e ds s −3e −3es Z 2 −t t 2 2 ds − e 1 ds = − e−t − te−t . 3 9 3 Z Since e−t is a solution of (H), another particular solution is 2 yp (t) = − te−t . 3 The method of undetermined coefficients leads to the third order homogeneous equation (D + 1)(D − 2)(D + 1)y = 0. The root r = −1 is repeated, so y(t) = c1 e2t + c2 e−t + c3 te−t . Putting this form into (NH) leads to [−2c3 e−t + c3 te−t ] − [c3 e−t − c3 te−t ] − 2c3 te−t = 2e−t , 1 or 2 c3 = − . 3 3. Solve by using variation of parameters and the method of undetermined coefficients. (N H) y 00 + 2y 0 + y = 3e−t . The associated homogeneous equation is (H) y 00 + 2y 0 + y = 0. The characteristic equation is r2 + 2r + 1 = (r + 1)2 = 0, r = 2, −1. Take y1 (t) = e−t , y2 (t) = te−t . The Wronskian for these solutions is W (y1 , y2 ) = y1 y20 − y10 y2 = e−t [e−t − te−t ] + te−t e−t = e−2t . The variation of parameters formula give a particular solution Z t Z t y2 (s)g(s) y1 (s)g(s) yp (t) = −y1 (t) ds + y2 (t) ds. W (y1 , y2 )(s) W (y1 , y2 )(s) In this case we get t Z t −s −s 3se−s e−s 3e e −t yp (t) = −e ds + te ds −2s e e−2s Z t Z t 3 3 −t −t = −3e s ds + 3te 1 ds = − te−t + 3t2 e−t = t2 e−t . 2 2 The method of undetermined coefficients leads to the third order homogeneous equation −t Z (D + 1)3 y = 0. The root r = −1 has multiplicity 3, so y(t) = c1 e−t + c2 te−t + c3 t2 e−t . Putting this form into (NH) leads to 3 c3 = . 2 4. 2 5. Find the general solution (N H) y 00 + y = tan(t). The associated homogeneous equation is (H) y 00 + y = 0. The homogeneous equation has a basis of solutions y1 (t) = cos(t), y2 (t) = sin(t), with Wronskian W (y1 , y2 ) = y1 y20 − y10 y2 = cos2 (t) + sin2 (t) = 1. The variation of parameters formula give a particular solution Z t Z t y2 (s)g(s) y1 (s)g(s) yp (t) = −y1 (t) ds + y2 (t) ds. W (y1 , y2 )(s) W (y1 , y2 )(s) In this case we get Z t yp (t) = − cos(t) Z t sin(s) tan(s) ds + sin(t) cos(s) tan(s) ds t Z t sin2 (s) = − cos(t) sin(s) ds. ds + sin(t) cos(s) Consulting a table of integrals leads to the general solution Z y(t) = c1 cos(t) + c2 sin(t) − cos(t) log(tan(t) + sec(t)). 14.Solve t2 y 00 − t(t + 2)y 0 + (t + 2)y = 2t3 . This is not in standard form. Rewrite it as (t + 2) 0 (t + 2) y 00 − y + y = 2t. t t2 We’re given the solutions of the homogeneous equation y1 = t, y2 = tet , with Wronskian W (y1 , y2 ) = y1 y20 − y10 y2 = t[et + tet ] − tet = t2 et . The variation of parameters formula gives Z t Z y2 (s)g(s) ds + y2 (t) yp (t) = −y1 (t) W (y1 , y2 )(s) t y1 (s)g(s) ds. W (y1 , y2 )(s) In this case we get Z t 2 2s2 es 2s t yp (t) = −t ds + te ds = −2t2 − 2t. 2 s 2 s se se Since t is a solution of the associated homogeneous equation, another particular solution Z t is yp (t) = −2t2 . 15. 3