Unformatted text preview: v 1 ( θ ) = − 1 4 tan 4 θ ⇒ v 1 ( θ ) = − 1 4 Z tan 4 θ dθ = 1 16 ln  cos 4 θ  + c 4 . Taking c 3 = c 4 = 0, we obtain y p ( θ ) = cos 4 θ 16 ln  cos 4 θ  + 1 4 θ sin 4 θ ⇒ y ( θ ) = c 1 cos 4 θ + c 2 sin 4 θ + θ 4 sin 4 θ + cos 4 θ 16 ln  cos 4 θ  . 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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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Cos, Elementary algebra, linearly independent solutions, corresponding homogeneous equation

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