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As a second part of my problem I end up with the differential equation looking like: $$ \frac{d^2 y}{dx^2} + \frac{1}{x}\frac{dy}{dx} - \frac{a}{x^2}y - \frac{c}{x}y + b x e^{-x^2/p^2}y - d e^{-x^2/p^2}y = 0. $$ It is more complex that my previous question. Can someone suggestion a solution method for this?

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As before, what's the domain of $x$? –  Pragabhava Oct 14 '12 at 17:48
    
It has to be >0. and all coefficients a-d are non-zero. –  nagendra Oct 14 '12 at 17:51
    
The first suggestion would be to write it in a easy to look form, i.e. $$ \frac{d^2 y}{d x^2} + \frac{1}{x} \frac{d y}{d x} + \left(-\frac{a}{x^2} - \frac{c}{x} + bxe^{-x^2/p^2}-de^{-x^2/p^2}\right) y = 0$$ –  Pragabhava Oct 14 '12 at 17:54
    
i was wondering, if the nature of the equation is to blow up due to a singularity at x=0, would be incorrect to change x to x + \delta where \delta<<1 and then proceed with the solution? (sorry, I couldn't get latex to do a greek delta for me ) –  drN Oct 14 '12 at 18:10
    
After transfering the ODE of the form $p(x)\dfrac{d^2y}{dx^2}+q(x)\dfrac{dy}{dx}+r(x)y=0$ to the ODE of the form $\dfrac{d^2z}{dx^2}+f(x)z=0$ by considering the method in mathworld.wolfram.com/…;, eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf claims that $\dfrac{d^2z}{dx^2}+f(x)z=0$ have method to solve generally for general $f(x)$ . But how is the reliability of eqworld.ipmnet.ru/en/methods/methods-ode/Khorasani2003.pdf? –  doraemonpaul Oct 14 '12 at 22:19
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1 Answer 1

The added complication makes closed form solutions even less likely, but you still have $x=0$ as a regular singular point with indicial roots $\pm \sqrt{a}$, and corresponding series solutions.

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