Solve the second order ODE $(2x^2 + 1)y'' − 4xy' + 4y = 0$ I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.
Given $y_1=x$ is a solution
$$(2x^2 + 1)y'' − 4xy' + 4y = 0$$
Find the second solution $y_2$ for equation.
 A: Via reduction of order, assume $y_2=y_1v=xv$ is a solution. Then you have derivatives
$$\begin{cases}y_2=xv\\{y_2}'=xv'+v\\{y_2}''=xv''+2v'\end{cases}$$
Substitute into the ODE:
$$(2x^2+1)\left(xv''+2v'\right)-4x\left(xv'+v\right)+4xv$$
or, grouping together the same-order derivatives,
$$x(2x^2+1)v''+\left(2\left(2x^2+1\right)-4x^2\right)v'+(4x-4x)v=0$$
and simplifying a bit gives
$$x(2x^2+1)v''+2v'=0$$
As you can see, the order of the ODE has been reduced. Substituting $z=v'$, you have a separable ODE linear in $z$.
$$x\left(2x^2+1\right)z'+2z=0$$
(I can add more details on finding the solution upon request, but this should hopefully address whatever previous mistake you made.)
A: Hint
By the reduction of order method, if you know a solution $y_1(x)$ of the homogeneous equation 
$$y^{''}(x)+p(x)y^{'}(x)+q(x)=0$$
then the other solution is given by
$$y_2(x)=y_1(x)\int \frac{e^{-\int{p(x)dx}}}{y_1^2(x)}dx$$

Solution
So in your example $p(x)=-\frac{4x}{2x^2+1}$ and $y_1(x)=x$. Hence
$$\begin{align}
y_2(x) &= x \int \frac{e^{\int{\frac{4x}{2x^2+1}dx}}}{x^2}dx \\
&= x \int \frac{e^{\ln(2x^2+1)}}{x^2}dx \\
&= x \int \frac{2x^2+1}{x^2}dx \\
&= x \int (2+\frac{1}{x^2})dx \\
&= x (2x-\frac{1}{x}) \\
&= 2x^2-1 
\end{align}$$

