Introduce $y(x)=u(x)z(x)$ into the equation $y''-2xy'-2y=0$ so that there won't be a term with $z'$ in the new equation. Find all solutions of such equations and also explore the possibility when $z'=0$.
Attempt: From $y=uz$, we have $y'=u'z+uz'$ and $y''=u''z+2u'z'+uz''$. Substituting this into the original equation, we get $uz''+z'(2u'-2xu)+z(u''-2xu'-2u)=0$, which leads to $2u'-2xu=0$ or $du/u=xdx$, implying $u=Ce^{(x^2/2)}$. Then I calculated $u'=Cxe^{(x^2/2)}$ and $u''=C(1+x^2)e^{(x^2/2)}$, and when I substituted these into the equation involving $z$, $z'$, and $z''$, I obtained $z''+z(-x^2-1)=0$ (since I divided by $Ce^{(x^2/2)}$). Are all steps up to this point correct? Any idea on how to solve this differential equation? I know it's a second-order linear differential equation and I need to use the Wronskian determinant, but usually we were given one solution. Also, I don't know how to find solutions when $z'=0$.