Solve differential equation $\left((1-x^2)\frac{d^2}{dx^2}-2x\frac{d}{dx}-\frac{n^2}{1-x^2}\right)y(x)=0$

Question

I have to solve the differential equation $$\left((1-x^2)\frac{d^2}{dx^2}-2x\frac{d}{dx}-\frac{n^2}{1-x^2}\right)y(x)=0$$ In order to solve above differential equation I have substitution i.e. if we let $$z=\ln\frac{1+x}{1-x}$$

This substitution can solve the above differential equation,but I don't know how we suppose this substitution, and what is basic concept behind this substitution.

Is there any best method to solve this differential equation (any other substitution), if I face any other such type of differential equation?

Answer 1

If it helps, your equation is of Sturm-Liouville form $\frac{d}{dx}[(1-x^2)\frac{dy}{dx}] - \frac{n^2}{1-x^2}y = 0$. This can be seen by applying the integrating factor approach to the first two terms of the equations, and seeing that $(1-x^2)$ is exactly the integrating factor. There is an entire theory devoted to solving exactly this kind of equations.

https://en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory

Sorry, I've never seen a substitution for the parameter variable used for an ODE solution. Surely, to see this clearly the equation needs to be transformed, and that is already half of the solution. I suspect this is one of those olympiad problems, where somebody discovered this approach by accident, without using it within any formal solution protocol, and then used to troll poor students

Answer 2

If you assume the form $$ y(x) = \sum_{k=0}^\infty a_k x^k $$ and assume $a_0=0$, and $a_1=q$ for some constant $q$. Then the solution follows that $a_{2n}=0$ and otherwise $$ a_k = \frac{q P_{k-1}(n)}{k!} $$ with $$ P_0(n)=1\\ P_2(n)=(2+n^2)\\ P_4(n)=(24+20n^2+n^4)\\ P_6(n)=(720+784n^2+70n^4+n^6)\\ \cdots $$ the coeffieicients of the temrs in the $P_k(n)$ seem to be wholly descibed by the table A111594 in the OEIS. These coefficients are made by even powers on the exponential generating functions $$ \frac{\tanh^{-2k-1}(x)}{(2k+1)!} $$