Problem with 2nd order differential equation I have the following differential equation: $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ I have tried to multiply by $e^{x^2/4}$ to obtain $$\frac{d}{dx}\left(e^{\frac{x^2}{4}}\frac{dy}{dx}\right)=ye^{\frac{x^2}{4}}$$ but I don't know how to proceed from here, or even if this is the best first step to take. I can't integrate this because of the right hand side, but because of the $x/2$ term multiplying the middle term of the differential equation, I thought that doing this multiplication would have been useful.
Any help would be appreciated.
 A: Considering the equation $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ multiplying instead by $e^{-\frac{x^2}{4}}$, you should arrive to $$\frac{d}{dx}\left(e^{-\frac{x^2}{4}}\frac{dy}{dx}\right)=y\,e^{-\frac{x^2}{4}}$$ the solution of which being related to Hermite_polynomials.
Quoting the Wilipedia page, "the probabilists' Hermite polynomials are solutions of the differential equation $$\frac{d}{dx}\left(e^{-\frac{x^2}{2}}\frac{du}{dx}\right)+\lambda\,e^{-\frac{x^2}{2}}u=0$$ where $\lambda$ is a constant. 
So, applied to your case, the solution should be $$y=c_1 e^{-\frac{x^2}{4}} H_{-3}\left(\frac{x}{2}\right)+c_2
   \left(\frac{x^2}{2}+1\right)$$
Edit
Expanded, the solution write $$y= c_1 \left(\sqrt{\pi } \left(x^2+2\right)
   \text{erfc}\left(\frac{x}{2}\right)-2 e^{-\frac{x^2}{4}} x\right)+ c_2 \left(x^2+2\right)$$
A: You have
$${y}''+\frac{1}{2}x{y}'-y=0$$
Define $x=r,y=s\left( r \right){{r}^{2}}+2s\left( r \right)$
This yields
$$\left( 2+{{r}^{2}} \right)s''+\frac{1}{2}\left( 10+{{r}^{2}} \right)rs'=0$$
Integration of this result should yeild error functions (erf(x)).
A: I would use the series solutions. Suppose:
$y(x)=a_0+a_1x+a_2x^2+...+\,a_nx^n$. 
$y=\displaystyle{\sum_{n=0}^\infty}a_nx^n\\
y'=\displaystyle{\sum_{n=1}^\infty}na_nx^{n-1}
  =\displaystyle{\sum_{n=0}^\infty}(n+1)a_{n+1}x^n\\
y''=\displaystyle{\sum_{n=2}^\infty}n(n-1)a_nx^{n-2}
   =\displaystyle{\sum_{n=0}^\infty}(n+2)(n+1)a_{n+2}x^n
$
