Use variation parameters method to solve: $4x^2y'' + y = 8x^{1/2}$

Question

I've tried so hard and just get horrible, horrible equations. $$y_p = u_1(x)x^{1/2} + u_2(x)x^{1/2}\log x$$

Answer 1

You might try using reduction of order instead. Knowing $x^{1/2}$ is a solution, let $y = x^{1/2} u(x)$, and you should get $$ x v' + v = \dfrac{2}{x}$$ where $v = u'$.

EDIT: It's unfortunate that some texts restrict reduction of order to the homogeneous case. There's really no reason to do so. Anyway, if you insist on using variation of parameters, the Wronskian of $y_1 = x^{1/2}$ and $y_2 = x^{1/2} \log x$ is $1$, which should make things not so bad.