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204_pdfsam_math 54 differential equation solutions odd

204_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 y p ( θ ) = 2 ( A cos θ + B sin θ ) e θ + θ ( A cos θ + B sin θ ) e θ = 2 [( B A ) cos θ ( B + A ) sin θ ] e θ + θ ( A cos θ + B sin θ ) e θ . (Note that we did not evaluate the terms containing the factor θ because they give zero result when substituted into the original equation.) Therefore, y p + 2 y p + 2 y p = 2 [( B A ) cos θ ( B + A ) sin θ ] e θ + 2( A cos θ + B sin θ ) e θ = 2 ( B cos θ A sin θ ) e θ = e θ cos θ . Hence A = 0, B = 1 / 2, y p ( θ ) = (1 / 2) θe θ sin θ , and a general solution is given by y ( θ ) = 1 2 θe θ sin θ + ( c 1 cos θ + c 2 sin θ ) e θ . 23. The corresponding homogeneous equation, y y = 0, is separable. Solving yields dy dt = y dy y = dt ln | y | = t + c y = ± e c e t = Ce t , where C = 0 is an arbitrary constant. By inspection, y 0 is also a solution. Therefore, y h ( t ) = Ce t , where C is an arbitrary constant, is a general solution to the homogeneous equation. (Alternatively, one can apply the method of solving first order linear equations in
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