Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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…;, 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 – doraemonpaul Oct 14 '12 at 22:19

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.