$\displaystyle x(1+2x)y'' + (1+6x)y' + 2y = \frac{1}{x^2}$

$\displaystyle y_1=\frac{1}{1+2x}$ is a solution to do related homogeneous equation, and I need to find a second linearly independent solution of the DE using reduction of order.

So, $\displaystyle y_2 = V(x)\frac{1}{1+2x}$

$\displaystyle y_2' = \frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}$

$\displaystyle y_2'' = \frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}$

Plugging $\displaystyle y_2$ into the homogeneous DE:

$\displaystyle x(1+2x)[\frac{V''}{1+2x} - \frac{4V'}{(1+2x)^2} + \frac{8V}{(1+2x)^3}] + (1+6x)[\frac{V'}{1+2x} - \frac{2V}{(1+2x)^2}] + 2[V(x)\frac{1}{1+2x}] = 0$

After some simplification, I get

$\displaystyle \frac{(4x^3+4x^2+x)V'' + (12x^3+3x)V' + (8x^2+8x)V}{(1+2x)^2} = 0$

I know that for reduction of order to work, the terms involving V need to go away, but I can't see a way to make that happen.