Solve using variation of parameters: y\'\'-2y\'-8y=csc(x) Solution So for variat

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So for variation of parameters, let y = u cos x + v sin x. Differentiating, y' = u' cos x - u sin x + v' sin x + v cos x. Letting u' cos x + v' sin x = 0 (*), we have y' = -u sin x + v cos x. Differentiate again: y'' = -u' sin x - u cos x + v' cos x - v sin x. Plug this into the DE: [-u' sin x - u cos x + v' cos x - v sin x] + [u cos x + v sin x] = 1/x ==> -u' sin x + v' cos x = 1/x (**). -------------------- Now, we have two DE's in u' and v': u' cos x + v' sin x = 0 -u' sin x + v' cos x = 1/x Solving for u' and v' yields u' = -sin(x)/x and v' = cos(x)/x. Therefore, we obtain (ignoring arbitrary constants for a particular solution): u = ?(t = 0 to x) -sin t dt/t, and v = ?(t = 0 to x) cos t dt/t. Hence, the general solution is y = A cos x + B sin x + [cos x * ?(t = 0 to x) -sin t dt/t + sin x * ?(t = 0 to x) cos t dt/t].

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