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Unformatted text preview: v 1 ( ) = 1 4 tan 4 v 1 ( ) = 1 4 Z tan 4 d = 1 16 ln  cos 4  + c 4 . Taking c 3 = c 4 = 0, we obtain y p ( ) = cos 4 16 ln  cos 4  + 1 4 sin 4 y ( ) = c 1 cos 4 + c 2 sin 4 + 4 sin 4 + cos 4 16 ln  cos 4  . 9. In this problem, the corresponding homogeneous equation is the same as that in Problem 1. Hence y 1 ( t ) = cos 2 t and y 2 ( t ) = sin 2 t are two linearly independent solutions, and a general solution to the homogeneous equation is given by y h ( t ) = c 1 cos 2 t + c 2 sin 2 t, 216...
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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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