how to find the general series solution of an ODE

Question

The questions states to find the general series solution about the point is $x=0$ and the equation is $(x^2-1)y''-2xy'+2y=0$. State the domain of validity.

I know that to begin we assume the general power series solution.

Firstly having trouble understanding how to determine the domain of validity. I arrived at the conclusion that $|x|<1$ but am uncertain on how exactly that is right.

The second issue i am having is when i arrive at the point where all the summations have the same power moving forward and finding the coefficients are a bit confusing.

Answer 1

When dividing the coefficient of $y'$ and the coefficient of $y$ by the coefficient of $y''$ we get $$ P(x) = \frac{2x}{x^2 -1} \quad \text{and} \quad Q(x)=\frac{2}{x^2 -1},$$ both of which exist for values of $|x| >1$.

The ODE you're trying to solve is very well known in the literature as a modified version of the Legendre ODE. Here I provide a solution for the form $$ y(x) = (1-x^2)y'' -2xy' + \alpha(\alpha+1),$$ which should help you to solve the given form of your ODE.

\begin{align*} (1-x^{2}) \sum_{n=0}^{\infty} n(n-1)a_{n}x^{n-2} -2x\sum_{n=0}^{\infty} n a_{n}x^{n-1} + \alpha(\alpha+1)\sum_{n=0}^{\infty} a_{n}x^{n} &= 0\\ \sum_{n=0}^{\infty} n(n-1)a_{n}x^{n-2} - \sum_{n=0}^{\infty}n(n-1)a_{n}x^{n} - 2\sum_{n=0}^{\infty} n a_{n}x^{n} + \alpha(\alpha+1)\sum_{n=0}^{\infty}a_{n}x^{n} &= 0\\ \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n} \sum_{n=0}^{\infty}n(n-1)a_{n}x^{n} - 2\sum_{n=0}^{\infty} n a_{n}x^{n} + \alpha(\alpha+1)\sum_{n=0}^{\infty}a_{n}x^{n} &= 0 \\ \implies \sum_{n=0}^{\infty} \left((n+2)(n+1)a_{n+2} - n(n-1)a_{n} -2na_{n} + \alpha(\alpha+1)a_{n}\right)x^{n} &= 0 \end{align*} \begin{align*} \implies a_{n+2} &= \frac{n(n-1)a_{n} + 2na_{n} - \alpha(\alpha+1)a_{n}}{(n+2)(n+1)} \\ &= \frac{a_{n}((n-\alpha)(n+\alpha+1))}{(n+2)(n+1)} \quad n \ge 1. \end{align*} \begin{align*} a_{2} &= \frac{a_{0}(-\alpha)(\alpha+1)}{2}\\ a_{4} &= \frac{a_{0}(\alpha+1)(\alpha-2)(\alpha+3)}{4} \\ a_{6} &= \frac{a_{0}-(\alpha+1)(\alpha-2)(\alpha+3)(\alpha-4)(\alpha+5)}{6!} \end{align*}
\begin{align*} a_{3} &= \frac{-a_{1}(\alpha-1)(\alpha+2)}{3!}\\ a_{5} &= \frac{a_{1}(\alpha-1)(\alpha+2)(\alpha-3)(\alpha+4)}{8} \\ a_{7} &= \frac{-a_{1}(\alpha-1)(\alpha+2)(\alpha-3)(\alpha+4)(\alpha-5)(\alpha+6)}{7!} \end{align*} This leads to the recursion relationship \begin{align*} &\textbf{EVEN} : a_{n} = a_{2k} = \frac{(-1)^{k}a_{0}\alpha (\alpha+2k-1)(\alpha+2k-3)\dots(\alpha+1)(\alpha-2)\dots(\alpha-2k+2)}{(2k)!} , \ k\ge 1\\ &\textbf{ODD} : a_{n} = a_{2k+1} = \frac{(-1)^{k} a_{1}(\alpha+2k)(\alpha+2k-2)\dots(\alpha+2)(\alpha-1)\dots(\alpha-2k+1)}{(2k+1)!} \ \ ,k \ge 1. \end{align*} Therefore we write the series solution as \begin{gather*} y(x) = y_{0} y_1(x) + y_{0}' y_2(x),\\ \text{where } \quad y_{1}(x) = 1 + \sum_{k=1}^{\infty}\frac{(-1)^{k}a_{0}\alpha (\alpha+2k-1)(\alpha+2n-3)\dots(\alpha+1)(\alpha-2)\dots(\alpha-2k+2)}{(2k)!} \\ \text{and} \quad y_{2}(x) = x + \sum_{k=0}^{\infty} \frac{(-1)^{k} a_{1}(\alpha+2k)(\alpha+2k-2)\dots(\alpha+2)(\alpha-1)\dots(\alpha-2k+1)}{(2k+1)!} \end{gather*}

At a first glance your ODE should have a slightly easier solution ,but very similar to the one which I'm providing.

Answer 2

Legendre equation is given as: $$(1-x^2)u''-2xu'+\nu(\nu+1)u=0~~~(1)$$ Whose standard solution is given as $$u(x)=C_1 P_{\nu}(x)+ C_2Q_\nu(x)]$$ In he given case the ODE is $$(1-x^2)y''+2xy'-2y=0~~~~(2)$$ In (1) if we change $x$ to $-x$ the solution of (2) can be written as $$y(x)=C_1 P_{\nu}(-x)+C_2 Q_{\nu}(-x)$$ where $$\nu^2+ \nu=- \implies \nu=\frac{-1\pm i\sqrt{7}}{2}.$$