2nd order differential equation with variable coefficients Can someone give advice on how i can solve the following, 
$$
\xi\Phi''(\xi ) + ( 1+2\xi^2)\Phi'(\xi)+4\xi\Phi(\xi) = 0
$$
Thanks!
 A: Hint:
$\xi\Phi''(\xi)+(1+2\xi^2)\Phi'(\xi)+4\xi\Phi(\xi)=0$
$\xi^2\dfrac{d^2\Phi}{d\xi^2}+\xi(2\xi^2+1)\dfrac{d\Phi}{d\xi}+4\xi^2\Phi=0$
Let $x=\xi^2$ ,
Then $\dfrac{d\Phi}{d\xi}=\dfrac{d\Phi}{dx}\dfrac{dx}{d\xi}=2\xi\dfrac{d\Phi}{dx}$
$\dfrac{d^2\Phi}{d\xi^2}=\dfrac{d}{d\xi}\left(2\xi\dfrac{d\Phi}{dx}\right)=2\xi\dfrac{d}{d\xi}\left(\dfrac{d\Phi}{dx}\right)+2\dfrac{d\Phi}{dx}=2\xi\dfrac{d}{dx}\left(\dfrac{d\Phi}{dx}\right)\dfrac{dx}{d\xi}+2\dfrac{d\Phi}{dx}=2\xi\dfrac{d^2\Phi}{dx^2}2\xi+2\dfrac{d\Phi}{dx}=4\xi^2\dfrac{d^2\Phi}{dx^2}+2\dfrac{d\Phi}{dx}$
$\therefore\xi^2\left(4\xi^2\dfrac{d^2\Phi}{dx^2}+2\dfrac{d\Phi}{dx}\right)+2\xi^2(2\xi^2+1)\dfrac{d\Phi}{dx}+4\xi^2\Phi=0$
$4\xi^4\dfrac{d^2\Phi}{dx^2}+4\xi^2(\xi^2+1)\dfrac{d\Phi}{dx}+4\xi^2\Phi=0$
$x\dfrac{d^2\Phi}{dx^2}+(x+1)\dfrac{d\Phi}{dx}+\Phi=0$
It has a particular solution $\Phi=e^{-x}$ .
A: Three hints:


*

*Note that: $4\xi=\dfrac{d}{d\xi}\left[1+2\xi^2\right]$

*You could add (and substract) $\dfrac{d\Phi\left(\xi\right)}{d\xi}\dfrac{d\xi}{d\xi}$

*Apply the (reverse) product rule: $u\dfrac{dv}{d\xi}+\dfrac{du}{d\xi}v=\dfrac{d(uv)}{d\xi}$

