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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!

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  • $\begingroup$ $u'=Cxe^{x^2/2}, u''=C(1+x^2)e^{x^2/2}$. $\endgroup$
    – Gonçalo
    Commented May 13 at 21:46
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    $\begingroup$ I have corrected my mistake, thanks fot that! $\endgroup$
    – good12
    Commented May 13 at 22:25
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    $\begingroup$ Hint. Compare the equation $z''+z(-x^2-1)=0$ with $u''=C(1+x^2)e^{x^2/2}=(1+x^2)u$. $\endgroup$
    – Gonçalo
    Commented May 13 at 23:16
  • $\begingroup$ I have added an attempt and corrected my mistake. Any help would be appreciated! $\endgroup$
    – good12
    Commented May 14 at 12:26

1 Answer 1

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Your work so far looks correct.

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 $v'$.

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  • $\begingroup$ I have added an attempt, because we have been doing this things differently. $\endgroup$
    – good12
    Commented May 14 at 8:50
  • $\begingroup$ 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 $\endgroup$ Commented May 14 at 16:13

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