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
 A: It seems the following.
By Liouville-Ostrogradski formula,
$$W(x)=W(1)e^{-\int_1^x \frac {-2t-t^3}{t^2}dt}=
5e^{\int_1^x 2t^{-1}+t dt}=
5e^{2\ln t+t^2/2|^x_1}=
5e^{2\ln x+x^2/2-1/2}=
5x^2e^{(x^2-1)/2}.$$
A: the wronskian $w= y_1y_2' - y_2y_1'$ of the equation $ay'' + by' + cy = 0$ satisfies the able's equation $$ a\frac{dw}{dy} + b w = 0.$$  we have $ay_j'' + by_j' + cy_j = 0, \,  j = 1, 2.$  considering $$a\frac{dw}{dy} = ay_1y_2'' -ay_2y_1'' = -y_1(by_2'+cy_2) + y_2(by_2'+cy_2) =-bw$$  establishes the abel's equation.
in your case you have $$\frac{dw}{w} = -\frac{b}{a} = \frac{(2x+x^3)dx}{x^2} = (\frac{2}{x} + x)\, dx $$
on integration you get $$w = Cx^2e^{x^2/2}.$$
A: Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du}=4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}$
$\therefore x^2\left(4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}\right)+(-2x-x^3)2x\dfrac{dy}{du}+5y=0$
$4x^4\dfrac{d^2y}{du^2}-2x^2(x^2+1)\dfrac{dy}{du}+5y=0$
$4u^2\dfrac{d^2y}{du^2}-2u(u+1)\dfrac{dy}{du}+5y=0$
