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I have a second order ordinary differential equation of the form :

$ a \frac{d^2p(x)}{dx^2} + x \frac{dp(x)}{dx} + p(x) = 0 $

Can anyone tell me how I can solve it?

Thanks in advance .

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    $\begingroup$ Any tries or thoughts yourself? How did you meet the problem? There is one rather simple solution, and one non-elementary. $\endgroup$ – mickep Jun 12 '16 at 11:06
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    $\begingroup$ @mickep I was trying to solve the Fokker-Plank equation for linear harmonic oscillator, in the region where the probability density is time independent, that's where I met with this kind of a ODE, and I had no idea about how to solve it. $\endgroup$ – kanayamalakar Jun 13 '16 at 5:53
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Assuming that $a$ is a constant:

$$ap''(x)+xp'(x)+p(x)=0\Longleftrightarrow$$


Substitute $\frac{\text{d}}{\text{d}x}\left(x\right)=1$:


$$ap''(x)+xp'(x)+\frac{\text{d}}{\text{d}x}\left(x\right)p(x)=0\Longleftrightarrow$$


Apply the reverse product rule:


$$\frac{\text{d}}{\text{d}x}\left(ap'(x)\right)+\frac{\text{d}}{\text{d}x}\left(xp(x)\right)=0\Longleftrightarrow$$ $$\frac{\text{d}}{\text{d}x}\left(ap'(x)+xp(x)\right)=0\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(ap'(x)+xp(x)\right)\space\text{d}x=\int0\space\text{d}x\Longleftrightarrow$$ $$ap'(x)+xp(x)=\text{C}\Longleftrightarrow$$ $$p'(x)+\frac{xp(x)}{a}=\text{C}\Longleftrightarrow$$


Let $v(x)=\exp\left[\int\frac{x}{a}\space\text{d}x\right]=e^{\frac{x^2}{2a}}$.

Multiply both sides by $v(x)$:


$$e^{\frac{x^2}{2a}}p'(x)+\frac{xe^{\frac{x^2}{2a}}p(x)}{a}=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$


Substitute $\frac{xe^{\frac{x^2}{2a}}}{a}=\frac{\text{d}}{\text{d}x}\left(e^{\frac{x^2}{2a}}\right)$:


$$e^{\frac{x^2}{2a}}p'(x)+\frac{\text{d}}{\text{d}x}\left(e^{\frac{x^2}{2a}}\right)p(x)=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$


Apply the reverse product rule:


$$\frac{\text{d}}{\text{d}x}\left(p(x)e^{\frac{x^2}{2a}}\right)=\text{C}e^{\frac{x^2}{2a}}\Longleftrightarrow$$ $$\int\frac{\text{d}}{\text{d}x}\left(p(x)e^{\frac{x^2}{2a}}\right)\space\text{d}x=\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x\Longleftrightarrow$$ $$p(x)e^{\frac{x^2}{2a}}=\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x\Longleftrightarrow$$ $$p(x)=\frac{\int\text{C}e^{\frac{x^2}{2a}}\space\text{d}x}{e^{\frac{x^2}{2a}}}$$

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    $\begingroup$ In the last integration you should have another constant appearing. $\endgroup$ – mickep Jun 12 '16 at 11:40
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This can also be solved as an Eigenvalue problem. With some manipulation, you can turn it into Hermite's Differential Equation having Hermite polynomials as solutions. One benefit, this allows the use of Rodrique's Formulas to generate new solutions.

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