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So, as a homework question, I am trying to solve $y''+xy'+y=0$.

I checked that this is exact and gives $$(y'+xy)'=0$$ $$y'+xy = C_1$$ Using integrating factor $e^{\int xdx} = e^{x^2/2}$ : $$(ye^{x^2/2})' = C_1 e^{x^2/2}$$ At this point, integrating $e^{x^2/2}$ is needed which can't be done (without using the error function).

I did go ahead to solve it and get $$y=C_1e^{-x^2/2}\int e^{x^2/2}dx + C_2e^{-x^2/2}$$ which doesn't seem to be the solution when subsituted back into the original equation.

Am I doing something wrong here?

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  • $\begingroup$ Why do you think that its wrong ? Check again the subsitution back into the original equation. $\endgroup$ – JJacquelin Feb 21 '14 at 16:14
  • $\begingroup$ @JJacquelin Turns out I did something wrong when differentiating. Thanks for the quick response. $\endgroup$ – TwiNight Feb 21 '14 at 16:34
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It is correct and I think it could be book keeping error that prevents you getting back to your ode.

$$ y = C_{1}\mathrm{e}^{-\frac{x^{2}}{2}}\int \mathrm{e}^{\frac{x^{2}}{2}}dx + C_{2}\mathrm{e}^{-\frac{x^{2}}{2}}= C_{1}\mathrm{e}^{-\frac{x^{2}}{2}}I +C_{2}\mathrm{e}^{-\frac{x^{2}}{2}} $$ if we denote the error integral as $I$ for simplicity we find $$ y^{'} = -xC_{1}\mathrm{e}^{-\frac{x^{2}}{2}}I + C_{1}\mathrm{e}^{-\frac{x^{2}}{2}}\left(\mathrm{e}^{\frac{x^{2}}{2}}\right) - xC_{2}\mathrm{e}^{-\frac{x^{2}}{2}} $$ here the derivative of the error function is simple the $\mathrm{e}^{\frac{x^{2}}{2}}$ as you know. Simplifying yields $$ y^{'} = -x\left[C_{1}\mathrm{e}^{-\frac{x^{2}}{2}}I +C_{2}\mathrm{e}^{-\frac{x^{2}}{2}}\right] + C_{1} = -xy + C_{1} $$ now using that we can take the derivative $$ y^{''} = -y - xy^{'} $$ and re-arranging gives back your original equation $$ y^{''} + xy^{'}+y = 0 $$

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