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I have the following differential equation: $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ I have tried to multiply by $e^{x^2/4}$ to obtain $$\frac{d}{dx}\left(e^{\frac{x^2}{4}}\frac{dy}{dx}\right)=ye^{\frac{x^2}{4}}$$ but I don't know how to proceed from here, or even if this is the best first step to take. I can't integrate this because of the right hand side, but because of the $x/2$ term multiplying the middle term of the differential equation, I thought that doing this multiplication would have been useful.

Any help would be appreciated.

John Doe
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3 Answers3

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Considering the equation $$\frac{d^2 y}{dx^2}+\frac{1}{2} x \frac{dy}{dx}-y=0.$$ multiplying instead by $e^{-\frac{x^2}{4}}$, you should arrive to $$\frac{d}{dx}\left(e^{-\frac{x^2}{4}}\frac{dy}{dx}\right)=y\,e^{-\frac{x^2}{4}}$$ the solution of which being related to Hermite_polynomials.

Quoting the Wilipedia page, "the probabilists' Hermite polynomials are solutions of the differential equation $$\frac{d}{dx}\left(e^{-\frac{x^2}{2}}\frac{du}{dx}\right)+\lambda\,e^{-\frac{x^2}{2}}u=0$$ where $\lambda$ is a constant.

So, applied to your case, the solution should be $$y=c_1 e^{-\frac{x^2}{4}} H_{-3}\left(\frac{x}{2}\right)+c_2 \left(\frac{x^2}{2}+1\right)$$

Edit

Expanded, the solution write $$y= c_1 \left(\sqrt{\pi } \left(x^2+2\right) \text{erfc}\left(\frac{x}{2}\right)-2 e^{-\frac{x^2}{4}} x\right)+ c_2 \left(x^2+2\right)$$

Claude Leibovici
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You have $${y}''+\frac{1}{2}x{y}'-y=0$$ Define $x=r,y=s\left( r \right){{r}^{2}}+2s\left( r \right)$

This yields $$\left( 2+{{r}^{2}} \right)s''+\frac{1}{2}\left( 10+{{r}^{2}} \right)rs'=0$$

Integration of this result should yeild error functions (erf(x)).

mathstackuser12
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  • I have attempted this integration, and obtained that $s'=Ae^{-\frac{x^2}{4}}(2+x^2)^2$ (correct?). You say the result would yield an error function so I attempted integration by parts on this, with the $e^{\cdot}$ term being integrated. However then the other term isn't very easy to deal with, so I am wondering how you did the integration of $s'$? – John Doe Apr 17 '17 at 12:45
  • Sorry that was meant to say $s'=Ae^{-\frac{x^2}{4}}(2+x^2)^{-2}$, but the problem with the integration remains. – John Doe Apr 17 '17 at 15:53
  • You'll need to introduce a factor of 2x into numerator and denominator then by parts twice – mathstackuser12 Apr 18 '17 at 05:54
  • Oh, ok got it, thanks :) – John Doe Apr 18 '17 at 12:25
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I would use the series solutions. Suppose: $y(x)=a_0+a_1x+a_2x^2+...+\,a_nx^n$.

$y=\displaystyle{\sum_{n=0}^\infty}a_nx^n\\ y'=\displaystyle{\sum_{n=1}^\infty}na_nx^{n-1} =\displaystyle{\sum_{n=0}^\infty}(n+1)a_{n+1}x^n\\ y''=\displaystyle{\sum_{n=2}^\infty}n(n-1)a_nx^{n-2} =\displaystyle{\sum_{n=0}^\infty}(n+2)(n+1)a_{n+2}x^n $

QFi
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