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 0 1 and v 0 2 (see (9) on page 195 in the text): v 0 1 ( θ )cos4 θ + v 0 2 ( θ )sin4 θ =0 4 v 0 1 ( θ θ +4 v 0 2 ( θ θ =tan4 θ. Multiplying the Frst equation by sin 4 θ and the second equation by (1 / 4) cos 4 θ , and adding the resulting equations together, we get v 0 2 ( θ )= 1 4 sin 4 θ v 2 = 1 16 cos 4 θ + c 3 . ±rom the Frst equation in the above system we also obtain v 0 1 ( θ v 0 2 ( θ )tan4 θ = 1 4 sin 2 4 θ cos 4 θ = 1 4 (sec 4 θ cos 4 θ ) v 1 ( θ 1 4 Z (sec 4 θ cos 4 θ ) = 1 16 ln | sec 4 θ +tan4 θ | + 1 16 sin 4 θ + c 4 . Taking c 3 = c 4 =0,weobtain y p ( θ ± 1 16 ln | sec 4 θ θ | + 1 16 sin 4 θ ² cos 4 θ + ± 1 16 cos θ ² sin 4 θ = 1 16 (cos 4 θ )ln | sec 4 θ 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 θ | sec 4 θ θ | . 25. Since the auxiliary equation, 4
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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.

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