Solving DE with substitution given the initial values

Question

I have to solve the following differential equation,

$y''-xy'-y=0$, while $y(0)=0, y'(0)=1$, we were suggested to use the transofrmation, $y(x) = h(x)e^{\frac{x^2}{2}}$

After substitution in the DE I end up with: $h''+xh'=0$, from here an easy solution would be $h(x) = c$, where $c$ is a constant.

But given that $y(0)=0$, we get $c=0$, which is not something we want since we are told $y'(0)=1$.

Is there a different valid solution? We were told that these is very easy to solve, which is, but the initial values create the above problem. I am thinking that maybe the initial values were mixed up, while they should be $y(0)=1, y'(0)=0$.

Answer 1

Note $$ h''+xh'=0\Longrightarrow\frac{h''}{h'}=-x\Longrightarrow (\ln h')'=-x\Longrightarrow h'=c_1e^{-x^2/2} $$ and hence $$ h=c_1\int_0^xe^{-t^2/2}dt+c_2. $$

Answer 2

$$y''-xy΄-y=0$$ $$y''-(xy)'=0$$ $$y'-xy=C_1$$ $$y'(0)=1 \implies C_1=1$$ $$(ye^{-x^2/2})'=e^{-x^2/2}$$ $$ye^{-x^2/2}=\int_0^x e^{-s^2/2}ds$$ $$y(x)=e^{x^2/2}\int_0^x e^{-s^2/2}ds$$ (This is a valid solution since we have $y(0)=0$)