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I'm trying to solve this ODE:

$$y''-2xy'+3y=x^3 $$

With the conditions:

$$\lim_{x\to\pm\infty}e^{-x^2/2}y(x)=\lim_{x\to \pm\infty}e^{-x^2/2}y'(x)=0$$

The homogeneous part is Hermite's equation for noninteger $n$.

I tried multiplying by the exponential and take the limit:

$$\lim_{x\to\pm\infty} \left( e^{-x^2/2}y'' - 2e^{-x^2/2}xy'= e^{-x^2/2}x^3\right)$$

But I think this leads nowhere. I was told that I should use the generating function: $g(x,t)=e^{-x^2+2tx}$, which I don't see how could it be helpful.

So, where's the trick?

I would appreciate just some hints.

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Hint: Could a polynomial be a solution? Also, what do the extra conditions say about the homogenous part of the solution?

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  • $\begingroup$ I'll try to think more about it, but the boundary conditions don't seem of much use in the homogeneous case, because the exponential always decays faster than $H_{3/2}$ and the confluent hypergeometric series. $\endgroup$ – jinawee May 16 '14 at 22:16
  • $\begingroup$ In which case the solution wouldn't be unique. But anyway you can find a particular solution using the method of undetermined coefficients: plug in an arbitrary cubic and solve for the coefficients. $\endgroup$ – John M May 16 '14 at 22:26
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Hint:

Let $y=u+ax^3+bx^2+cx+d$ ,

Then $y'=u'+3ax^2+2bx+c$

$y''=u''+6ax+2b$

$\therefore u''+6ax+2b-2x(u'+3ax^2+2bx+c)+3(u+ax^3+bx^2+cx+d)=x^3$

$u''+6ax+2b-2xu'-6ax^3-4bx^2-2cx+3u+3ax^3+3bx^2+3cx+3d=x^3$

$u''-2xu'+3u-(3a+1)x^3-bx^2+(6a+c)x+2b+3d=0$

$\therefore$ By taking $a=-\dfrac{1}{3}$ , $b=0$ , $c=2$ and $d=0$ , we have $u''-2xu'+3u=0$

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  • $\begingroup$ I want to know how the generating function is supposed to solve the equation. Do you have any idea? $\endgroup$ – abnry May 16 '14 at 22:48

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