How can I solve $y'' + \left(4x-\tfrac{2}{x}\right)y' + 4x^2y= 3xe^{x^2}$?

Question

The DE is $$y'' + \left(4x-\frac{2}{x}\right)y' + 4x^2y= 3xe^{x^2}.$$

I've been told to use $t = x^2$ along with change of variable to solve it, but it's clear that's not possible due to the singularity on $x = 0$ on the term $\frac{2}{x}$. I tried using Frobenius method, but not luck with the solution. I would appreciate any help on how to solve it.

Thanks in advance.

Answer 1

A fundamental set of solutions of the homogeneous equation is $(\sqrt{2}\cos(\sqrt{2}x)x-\sin(\sqrt{2} x))\exp(-x^2)$ and $(\sqrt{2}\sin(\sqrt{2}x)x+\cos(\sqrt{2} x))\exp(-x^2)$. You can then use Variation of Parameters to get a solution of the non-homogeneous equation. However, this will involve integrals that can't be done in closed form.

Answer 2

$$y'' + \left(4x-\frac{2}{x}\right)y' + 4x^2y= 3xe^{x^2}$$ If this ODE is a classroom exercise, there is probably a typo. This is solvable, but of high level. The most likely the equation might be : $$y'' + \left(4x-\frac{1}{x}\right)y' + 4x^2y= 3xe^{-x^2}$$ SOLVING THE HOMOGENEOUS ODE : $$y'' + \left(4x-\frac{1}{x}\right)y' + 4x^2y=0$$ Let $\quad t=x^2\quad\implies\quad dt=2xdx\quad\implies\quad \frac{dt}{dx}=2x$

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=2x\frac{dy}{dt}$

$\frac{d^2y}{dx^2}=2x\frac{d^2y}{dt^2}\frac{dt}{dx}+2\frac{dy}{dt}=4x^2\frac{d^2y}{dt^2}+2\frac{dy}{dt}=4t\frac{d^2y}{dt^2}+2\frac{dy}{dt}$

$4t\frac{d^2y}{dt^2}+2\frac{dy}{dt} + \left(4x-\frac{1}{x}\right)2x\frac{dy}{dt} + 4x^2y=0$

$4t\frac{d^2y}{dt^2}+2\frac{dy}{dt} + \left(4x^2-1\right)2\frac{dy}{dt} + 4x^2y=0$

$4t\frac{d^2y}{dt^2}+2\frac{dy}{dt} + \left(4t-1\right)2\frac{dy}{dt} + 4ty=0$

After simplification : $$\frac{d^2y}{dt^2} + 2\frac{dy}{dt} + y=0$$ This ODE is easy to solve :$\quad y=c_1e^{-t}+c_2te^{-t}$ $$y=c_1e^{-x^2}+c_2x^2e^{-x^2}$$ SOLVING THE INHOMOGENEOUS ODE : $$y'' + \left(4x-\frac{1}{x}\right)y' + 4x^2y= 3xe^{-x^2}$$

The form of the equation and above solutions suggest the change $\quad y(x)=u(x)e^{-x^2}$

$y'= (u'-2xu)e^{-x^2}$

$y''= (u''-4xu'-2u+4x^2u)e^{-x^2}$

$(u''-4xu'-2u+4x^2u)e^{-x^2} + \left(4x-\frac{1}{x}\right)(u'-2xu)e^{-x^2} + 4x^2ue^{-x^2}= 3xe^{x^2}$

$(u''-4xu'-2u+4x^2u) + \left(4x-\frac{1}{x}\right)(u'-2xu) + 4x^2u= 3x$

$u''-4xu'-2u+4x^2u + 4xu'-8x^2u - \frac{1}{x}u'+2u + 4x^2u= 3x$

After simplification : $$u'' - \frac{1}{x}u' = 3x$$

No need to solve for the general solution. Only one particular solution is sufficient.

Obviously $u'=3x^2$ , then $u=x^3$ is a particular solution. Thus, the general solution of the inhomogeneous ODE is : $$y=c_1e^{-x^2}+c_2x^2e^{-x^2}+x^3e^{-x^2}$$

NOTE :

If the above supposition was not fully true, another supposition might be : $$y'' + \left(4x-\frac{1}{x}\right)y' + 4x^2y= 3xe^{x^2}$$ It is solvable with the same method, but the final computation of a particular solution would be a little more complicated, involving the special function erfi : http://mathworld.wolfram.com/Erfi.html