# 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.

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 $$z'=0$$.

Added attempt: Indeed, one solution is $$z = e^{\frac{x^2}{2}}$$. Then, I used the Wronskian determinant and obtained $$W(x) = e^{\frac{x^2}{2}} z_2' - x e^{\frac{x^2}{2}} z_2$$. On the other hand, we have $$e^{-\int \frac{b(x)}{a(x)} \ dx} = e^0 = 1$$. Therefore, $$z_2' - x z_2 = e^{-x^2/2}$$. From this, we see that $$z(x) = C e^{\frac{x^2}{2}} + e^{\frac{x^2}{2}} \int e^{-x^2} \ dx$$, but now I don't know how to compute this integral ...

Are even my last few steps correct or have I made a mistake somewhere? Any help would be appreciated!

• $u'=Cxe^{x^2/2}, u''=C(1+x^2)e^{x^2/2}$. Commented May 13 at 21:46
• I have corrected my mistake, thanks fot that! Commented May 13 at 22:25
• Hint. Compare the equation $z''+z(-x^2-1)=0$ with $u''=C(1+x^2)e^{x^2/2}=(1+x^2)u$. Commented May 13 at 23:16
• I have added an attempt and corrected my mistake. Any help would be appreciated! Commented May 14 at 12:26

We've already seen that $$u'' = C (1 + x^2) \exp \left(\frac{x^2}2\right) = (1 + x^2) u$$, so, like Gonçalo pointed out in the comments, $$z(x) = \exp \left(\frac{x^2}2\right)$$ is a solution. Substituting the ansatz $$z(x) = \exp \left(\frac{x^2}2\right) v(x)$$ in the equation in $$z$$ (or just as well $$y(x) = e^{x^2} v(x)$$ in the equation in $$y$$) then gives $$v''(x) + 2 x v'(x) ,$$ which in turn is first-order and separable in $$w(x) := v'(x)$$: $$w'(x) + 2 x w(x) = 0.$$
Remark With this computation in hand, we can see that it's more efficient to change variables first via $$y(x) = e^{x^2} v(x)$$, which directly transforms the original equation in $$y$$ to the equation in $$v'$$.
• Your additional work looks good to me, though I'd write the integral as definite: $\int^x \exp(-t^2) \,dt$. This integral has no elementary antiderivative, and we usually express it in terms of the error function: en.wikipedia.org/wiki/Error_function Commented May 14 at 16:13