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Unformatted text preview: Exercises 4.4
To choose s, we ﬁnd the roots of the characteristic equation, which is r 2 + 9 = 0. Since r = ±3i are its simple roots, we take s = 1. Thus yp (t) = t A3 t3 + A2 t2 + A1 t + A0 cos 3t + t B3 t3 + B2 t2 + B1 t + B0 sin 3t. 29. The characteristic equation r 2 − 6r + 9 = (r − 3)2 = 0 has a double root r = 3. Therefore, a particular solution is of the form yp (t) = t2 A6 t6 + A5 t5 + A4 t4 + A3 t3 + A2 t2 + A1 t + A0 e3t . 31. From the form of the righthand side, we conclude that a particular solution should be of the form yp (t) = ts A3 t3 + A2 t2 + A1 t + A0 cos t + B3 t3 + B2 t2 + B1 t + B0 sin t e−t . Since r = −1 ± i are simple roots of the characteristic equation, r 2 + 2r + 2 = 0, we should take s = 1. Therefore, yp (t) = t A3 t3 + A2 t2 + A1 t + A0 cos t + B3 t3 + B2 t2 + B1 t + B0 sin t e−t . 33. The righthand side of the equation suggests that yp (t) = ts (A cos t + B sin t). By inspection, we see that r = i is not a root of the corresponding auxiliary equation, r 3 − r 2 + 1 = 0. Thus, with s = 0, yp (t) = A cos t + B sin t, yp (t) = −A sin t + B cos t, yp (t) = −A cos t − B sin t, yp (t) = A sin t − B cos t, and substitution into the original equation yields (A sin t − B cos t) − (−A cos t − B sin t) + (A cos t + B sin t) = sin t ⇒ ⇒ (2A − B ) cos t + (A + 2B ) sin t = sin t 2A − B = 0, A + 2B = 1 ⇒ A = 1/5, B = 2/5 ⇒ yp (t) = 2 1 cos t + sin t. 5 5 195 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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