EDL2 : $(2x+1)y''+(4x-2)y'-8y=0$

Question

Solve this equation :

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

Attempt :

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First step : $\varphi_1$

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

I found $\varphi_1(x)= 4x^2+1$

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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?

Answer 1

Proposing $y = e^{a t}$ and substituting into the differential equation we have

$$ (2+a)(a+2a t)e^{a t} = 0\Rightarrow a+2 = 0 $$

but the differential equation is of second order so we have two conditions to satisfy and in consequence we need another independent solution.

Proposing now $y = b t^2+c t+ d$ and substituting into the differential equation we have _0 $$ 2(b-4d-c(1+2t)) = 0 \Rightarrow c = 0, b-4d = 0 \Rightarrow y = 4t^2+1 $$

and finally the general solution is

$$ y = C_1 e^{-2t}+C_2(4t^2+1) $$

Answer 2

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