Solve the following ODE: $$ f''(y) - yf'(y) - f(y) = 0 $$
I think I should start with the fact that $\frac{d}{dy}(-y)=-1$. So the equation is: $$ \frac{d^2f(y)}{dy^2}-y\frac{df(y)}{dy}+\frac{d}{dy}(-y)f(y)=0 $$ Then I should use reverse product rule solution. And then I am lost. I would like to have steps how to solve this equation till the end.
I know the answer is $$f(y) = \frac{π}{2} c_1 e^{x^2/2} {\rm erf}\left(\frac{x}{\sqrt2}\right) + c_2 e^{x^2/2}$$ where $${\rm erf}\left(\frac{x}{\sqrt2}\right)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y} e^{-z^2/2} \,dz .$$
So we want to solve the ODE $$y''-xy'-y=0\Longleftrightarrow y''+\left(-xy'-1\cdot y\right)=0$$ We apply the reverse product rule and factor out the derivative via linearity $$(y')'+\left(-xy\right)'=0\Longleftrightarrow \left(y'-xy\right)'=0$$ $$\int\left(y'-xy\right)'\text{ d}x=\int 0\text{ d}x$$ $$\implies y'-xy=A$$ for some constant $A$.
This is just a linear first order ODE now with a simple integrating factor of $$e^{\int-x\text{ d}x} = e^{-x^2\over2}$$
I believe you can take it from here.