Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\exp(kx^2)=0
$
$2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2=0\iff k=\frac{1}{2}$
Plug in:
$z''+z'(2x-x^{-1})=0$
$z'=:u$
$u'+u(2x-x^{-1})=0$
$\frac{\mathrm{d}u}{u}=-(2x-x^{-1})\mathrm{d}x$
$\ln|u|=-(x^2-\ln|x|)+c_1$
$u=\frac{c_1x}{e^{x^2}}$
Substitute back:
$z'=c_1xe^{-x^2}$
$z=-\tfrac{1}{2}c_1e^{-x^2}+c_2$
$y(x)=(-\tfrac{1}{2}c_1e^{-x^2}+c_2)e^{\frac{1}{2}x^2}$
$y(x)=c_2e^{\frac{1}{2}x^2}-\tfrac{1}{2}c_1e^{-\frac{1}{2}x^2}$
Right? Is there another way solving this ode? Thanks!
edit: @Lutz Lehmann: Sorry there was a square missing. Fixed it.
 A: We can really smoke this with the substitution $u=x^2$:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=2x\frac{dy}{du}$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(2x\frac{dy}{du}\right)=2\frac{dy}{du}+4x^2\frac{d^2y}{du^2}$$
So
$$x\frac{d^2y}{dx^2}-\frac{dy}{dx}-x^3y=4x^3\frac{d^2y}{du^2}-x^3y=4x^3\left(\frac{d^2y}{du^2}-\frac14y\right)=0$$
With solution
$$y=c_1e^{\frac12u}+c_2e^{-\frac12u}=c_1e^{\frac12x^2}+c_2e^{-\frac12x^2}$$
A: Another way to solve it maybe:
$$xy''-y'-x^3y=0$$
Consider $x'$, then the equation becomes:
$$-x''x-x'^2=(x'x)^3y$$
$$-(x'x)'=(x'x)^3y$$
This differential equation is separable:
$$\dfrac {d(x'x)}{(x'x)^3}=-ydy$$
$$\dfrac {1}{(x'x)^2}=y^2+K$$
$$2x'x=\pm \dfrac {2}  { \sqrt {y^2+K}}$$
$$(x^2)'= \pm \dfrac {2}  { \sqrt {y^2+K}}$$
Integrate : 
$$x^2+C=\pm 2 \int \dfrac {dy}  { \sqrt {y^2+K}}$$
The last integral depends on the value of the constant K.
A: Dividing by $x^2$, the equation is
$$\left(\frac{y'}x\right)'-xy=0$$
or
$$\frac{y'}x\left(\frac{y'}x\right)'-yy'=0.$$
After integration,
$$\left(\frac{y'}x\right)^2-y^2=c,$$
which is separable.
$$\frac{y'}{\sqrt{y^2+c}}=\pm x$$
gives 
$$\text{arsinh}\frac y{\sqrt{|c|}}=c'\pm\frac{x^2}2$$ or 
$$\text{arcosh}\frac y{\sqrt{|c|}}=c'\pm\frac{x^2}2$$ depending on the sign of $c$.

Final form:
We can rewrite as
$$y=\frac{\sqrt{|c|}}2\left(e^{c'+x^2/2}\pm e^{-c'-x^2/2}\right)$$
or
$$y=c_+e^{x^2/2}+c_-e^{-x^2/2}.$$
