I happened to go through reduction of order in my differential equations tutorial:

http://www.mathhelpforum.com/math-he...tml#post145671
Given that $\displaystyle y_1(x)$ is a solution, I state that the second solution is

$\displaystyle \displaystyle y_2=\left[\int\left(\frac{e^{-\int P(x)\,dx}}{y_1^2(x)}\right)\,dx\right]y_1(x)$

where in our case, $\displaystyle P(x)=\dfrac{2(2x-1)}{x(4x-1)}=\dfrac{2}{x}-\dfrac{4}{4x-1}$ and $\displaystyle y_1(x)=\dfrac{1}{x}$.

I leave it for you to verify that $\displaystyle y_2(x)=2x-1$.

I hope this helps (in addition to the other responses you have received) !!!