259_pdfsam_math 54 differential equation solutions odd

259_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Review Problems We apply the method of variation of parameters to find a particular solution to the original equation. We look for a particular solution of the form yp (θ) = v1 (θ) cos θ + v2 (θ) sin θ, where v1 (θ) and v2 (θ) satisfy the system (9), Section 4.6. That is, v1 cos θ + v2 sin θ = 0, −v1 sin θ + v2 cos θ = sec θ . Multiplying the first equation by sin θ, the second equation by cos θ, and adding them together yield v2 sin2 θ + v2 cos2 θ = sec θ cos θ ⇒ v2 = 1 ⇒ v2 (θ) = θ. From the first equation in the above system we also get v1 = −v2 tan θ = − tan θ ⇒ v1 (θ) = − tan θ dθ = ln | cos θ|, where we have taken the zero integration constant. So, yp (θ) = cos θ ln | cos θ| + θ sin θ , and y (θ) = c1 cos θ + c2 sin θ + cos θ ln | cos θ| + θ sin θ is a general solution to the original equation. Differentiating we find that y (θ) = −c1 sin θ + c2 cos θ − sin θ ln | cos θ| + θ cos θ. Substitution of y (θ) and y (θ) into the initial conditions yields y (0) = c1 = 1, y (0) = c2 = 2 ⇒ c1 = 1 , c2 = 2 , and so the answer is y (θ) = cos θ + 2 sin θ + cos θ ln | cos θ| + θ sin θ. 255 ...
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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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