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255_pdfsam_math 54 differential equation solutions odd

255_pdfsam_math 54 differential equation solutions odd -...

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Review Problems we get the following system for v 1 and v 2 (see (9) on page 195 in the text): 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 = tan 4 θ. Multiplying the first equation 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 sin 4 θ v 2 = 1 16 cos 4 θ + c 3 . From the first equation in the above system we also obtain v 1 ( θ ) = v 2 ( θ ) tan 4 θ = 1 4 sin 2 4 θ cos 4 θ = 1 4 (sec 4 θ cos 4 θ ) v 1 ( θ ) = 1 4 (sec 4 θ cos 4 θ ) = 1 16 ln | sec 4 θ + tan 4 θ | + 1 16 sin 4 θ + c 4 . Taking c 3 = c 4 = 0, we obtain y p ( θ ) = 1 16 ln | sec 4 θ + tan 4 θ | + 1 16 sin 4 θ cos 4 θ + 1 16 cos θ sin 4 θ = 1 16 (cos 4 θ ) ln | sec 4 θ + tan 4 θ | , and a general solution to the original equation is y ( θ ) = c 1 cos 4 θ + c 2 sin 4 θ 1 16 (cos 4
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