How solve inhomogeneous ODE with boundary conditions?

Question

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.

Answer 1

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

Answer 2

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 $u''-2xu'+3u=0$