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Unformatted text preview: Chapter 4
Multiplying the ﬁrst equation by sin t and the second equation by cos t, and adding them together yield v2 (t) = tan2 t cos t = (sec2 t − 1) cos t = sec t − cos t. We ﬁnd v1 (t) from the ﬁrst equation in the system. v1 (t) = −v2 (t) tan t = −(sec t − cos t) tan t = sin t − Integrating, we get v1 (t) = v2 (t) = sin t − sin t cos2 t dt = − cos t − sec t, sin t . cos2 t (sec t − cos t) dt = ln  sec t + tan t − sin t, where we have taken zero integration constants. Therefore, yp (t) = −(cos t + sec t) cos t + (ln  sec t + tan t − sin t) sin t = sin t ln  sec t + tan t − 2, and a general solution is given by y (t) = c1 cos t + c2 sin t + sin t ln  sec t + tan t − 2. 13. The corresponding homogeneous equation in this problem is the same as that in Problem 1 (with y replaced by v ). Similarly to the solution of Problem 1, we conclude that v1 (t) = cos 2t and v2 (t) = sin 2t are two linearly independent solutions of the corresponding homogeneous equation, and a particular solution to the original equation can be found as vp (t) = u1 (t) cos 2t + u2 (t) sin 2t , where u1 (t) and u2 (t) satisfy u1 (t) cos 2t + u2 (t) sin 2t = 0, −2u1 (t) sin 2t + 2u2(t) cos 2t = sec4 2t. Multiplying the ﬁrst equation by sin 2t and the second equation by (1/2) cos 2t, and adding the results together, we get u2 (t) = 218 1 sec3 2t. 2 ...
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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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