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220_pdfsam_math 54 differential equation solutions odd

# 220_pdfsam_math 54 differential equation solutions odd - v...

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Chapter 4 7. The auxiliary equation in this problem is r 2 +16 = 0, which has the roots r = ± 4 i . Therefore, y 1 ( θ ) = cos 4 θ and y 2 ( θ ) = sin 4 θ are two linearly independent solutions, and a general solution to the corresponding homogeneous equation is given by y h ( θ ) = c 1 cos 4 θ + c 2 sin 4 θ. Using the variation of parameters method, we look for a particular solution to the original nonhomogeneous equation of the form y p ( θ ) = v 1 ( θ ) y 1 ( θ ) + v 2 ( θ ) y 2 ( θ ) = v 1 ( θ ) cos 4 θ + v 2 ( θ ) sin 4 θ. The system (9) on page 195 in the text becomes v 1 ( θ ) cos 4 θ + v 2 ( θ ) sin 4 θ = 0 , 4 v 1 ( θ ) sin 4 θ + 4 v 2 ( θ ) cos 4 θ = sec 4 θ. (4.7) Multiplying the first equation in (4.7) by sin 4 θ and the second equation by (1 / 4) cos 4 θ , and adding the resulting equations together, we get v 2 ( θ ) = 1 4 v 2 = 1 4 θ +
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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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