<s>[INST] Solve $x^2 y''+(-2x-x^3)y'+5y=0$ Ok so for me I am having trouble solving this equation to get 
$y=C_1y_1+C_2y_2$, but I'm having trouble dealing with the (-2x-x^3) part.  Usually I would isolate $y''$, then make $y=x^m$, then go from there, but I'm having trouble dealing with the $(-2x-x^3)$ part.  Thanks!
When I tried that approach, I got $[x^{3/2}x^{x^2/2}x^{(\sqrt{x^2+6x-11})/2}, x^{3/2}x^{x^2/2}x^{-(\sqrt{x^2+6x-11})/2}]$ form a basis, but it just seems too complicated when I need to solve
$y_1(1)=-4, y'_1(1)=5, y_2(1)=-5,y'_2(1)=5$
The point is to find the Wronskian, $w(x)=y_1(x)y'_2(x)-y_2(x)y'_1(x)$ for x>0
 [/INST] Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du}=4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}$
$\therefore x^2\left(4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}\right)+(-2x-x^3)2x\dfrac{dy}{du}+5y=0$
$4x^4\dfrac{d^2y}{du^2}-2x^2(x^2+1)\dfrac{dy}{du}+5y=0$
$4u^2\dfrac{d^2y}{du^2}-2u(u+1)\dfrac{dy}{du}+5y=0$
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