EDL2 : $(2x+1)y''+(4x-2)y'-8y=0$ 
Solve this equation :
$(2x+1)y''+(4x-2)y'-8y=0 \qquad $ on $ \left]\dfrac{1}{2},+\infty\right[$ and on $\mathbb{R}$

Attempt :

First step : $\varphi_1$
The first step was to find a solution in that form : $\varphi_1(x)=ax^2+bx+c$
I found $\varphi_1(x)= 4x^2+1$

Second step : $\varphi_2$
We consider now this equation : $y''+\dfrac{4x-2}{2x+1}y'-\dfrac{8}{2x+1}y=0$
Let $W(x)=\begin{array}{|ll|}\varphi_1(x)&\varphi_2(x)\\\varphi_1'(x)&\varphi_2'(x)\end{array}$
As we know that $W'(x)=-\dfrac{4x-2}{2x+1}W(x)$ we have got $\displaystyle W(x) =\exp\left[\int_{x_0}^x -\dfrac{4s-2}{2s+1}\cdot ds\right]=\lambda(2x+1)^2e^{-2x}$
And as $\left(\dfrac{\varphi_2(x)}{\varphi_1(x)}\right)' =\dfrac{W(x)}{(\varphi_1(x))^2}$
So now I must compute $\displaystyle \int \dfrac{W(x)}{(\varphi_1(x))^2}\cdot  dx = \int \dfrac{\lambda(2x+1)^2}{(4x^2+1)^2}\cdot e^{-2x}\cdot  dx=\gamma(x)$
Thus $\dfrac{\varphi_2(x)}{\varphi_1(x)}=\gamma(x)\iff \varphi_2(x)=\varphi_1(x)\cdot \gamma(x)$
But I can't find $\displaystyle \gamma(x)=\int \dfrac{\lambda(2x+1)^2}{(4x^2+1)^2}\cdot e^{-2x}\cdot  dx$
May I have a hint to solve this integral?

For the integral consider
$$\left(\frac {e^{-2x}}{4x^2+1}\right)'=-2e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}$$
$$\left(\frac {e^{-2x}}{4x^2+1}\right)=-2\int e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}dx$$
$$\phi_2=-\frac 12 e^{-2x} $$
or
$$\phi_2=e^{-2x} $$
Another look at the differential equation
$$(2x+1)y''+(4x-2)y'-8y=0 \qquad$$
$$(2x+1)(y''+2y')-4(y'+2y)=0 \qquad$$
Substitute $z=y'+2y$
$$(2x+1)z'-4z=0 \qquad$$
Which is separable and easy to integrate 
