34_pdfsam_math 54 differential equation solutions odd

34_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 19. For a general solution, separate variables and integrate: dy = y sin θ dy y =s in θdθ Z dy y = Z sin θdθ ln | y | = cos θ + C 1 ⇒| y | = e cos θ + C 1 = Ce cos θ y = Ce cos θ (because at the initial point, θ = π , y ( π ) < 0). We substitute now the initial condition, y ( π )= 3, and obtain 3= y ( π )= Ce cos π = Ce C =3 e 1 . Hence, the answer is given by y = 3 e 1 e cos θ = 3 e 1 cos θ . 21. Separate variables to obtain 1 2 ( y +1) 1 / 2 dy =cos xdx. Integrating, we have ( y +1) 1 / 2 =s in x + C. Using the fact that y ( π
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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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