3.1 Example 1. homogeneous ode $2x^2{y}^{\prime \prime }\left(x\right)-x{y}^{\prime }\left(x\right)+(1-x^2)y\left(x\right)=0$

With expansion around $x=0$. This is a regular singular ODE. Let

$$y\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$

Therefore

$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above into the ode and simplifying gives

$$\begin{array}{rl}2x^2\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}-x\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}+(1-x^2)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}& =0\\ \sum _{n=0}^{\mathrm{\infty }}2(n+r)(n+r-1)a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}-x^2\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}& =0\\ \sum _{n=0}^{\mathrm{\infty }}2(n+r)(n+r-1)a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r+2}& =0\end{array}$$

The next step is to make all powers of $x$ to be $n+r$. This results in

$$\begin{array}{}\text{(1)}& \sum _{n=0}^{\mathrm{\infty }}2(n+r)(n+r-1)a_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}-\sum _{n=2}^{\mathrm{\infty }}{a}_{n-2}{x}^{n+r}=0\end{array}$$

The indicial equation is obtained from $n=0$

$$\begin{array}{rl}2(n+r)(n+r-1)a_n-(n+r)a_n+a_n& =0\\ a_n(2(n+r)(n+r-1)-(n+r)+1)& =0\\ 2(n+r)(n+r-1)-(n+r)+1& =0\\ 2(0+r)(0+r-1)-(0+r)+1& =0\\ 2r^2-3r+1& =0\end{array}$$

The roots are

$$\begin{array}{rl}{r}_1& =1\\ r_2& =\frac{1}{2}\end{array}$$

Since the roots differ by non integer, then the solutions are given by

$$\begin{array}{rl}{y}_1& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r_1}=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}\\ y_2& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r_2}=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+\frac{1}{2}}\end{array}$$

We start by finding $y_1$.  EQ. (1) gives for $r=1$

$$\sum _{n=0}^{\mathrm{\infty }}2(n+1)\left(n\right)a_n{x}^{n+1}-\sum _{n=0}^{\mathrm{\infty }}(n+1)a_n{x}^{n+1}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}-\sum _{n=2}^{\mathrm{\infty }}{a}_{n-2}{x}^{n+1}=0$$

For $n=1$ (we skip $n=0$ as that was used to find $r$) and we always let $a_0=1$ as it is arbitrary. Hence

$$\begin{array}{rl}2(n+1)\left(n\right)a_n{x}^{n+1}-(n+1)a_n{x}^{n+1}+a_n{x}^{n+1}& =0\\ 2(n+1)\left(n\right)a_n-(n+1)a_n+a_n& =0\\ 4a_1-2a_1+a_1& =0\\ 3a_1& =0\\ a_1& =0\end{array}$$

For $n\ge 2$, we have recursion relation. From it we can find that $a_2=\frac{1}{10},a_3=0,a_4=\frac{1}{360}$ and so on. Hence

$$\begin{array}{rl}{y}_1& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}\\ & =a_{0}x+a_1{x}^2+a_2{x}^3+a_3{x}^4+a_4{x}^5+\cdots \\ & =x+\frac{1}{10}{x}^3+\frac{1}{360}{x}^5+\cdots \end{array}$$

Now we do the same for $y_2$. EQ. (1) now becomes for $r_2=\frac{1}{2}$

$$\sum _{n=0}^{\mathrm{\infty }}2(n+\frac{1}{2})(n+\frac{1}{2}-1)a_n{x}^{n+\frac{1}{2}}-\sum _{n=0}^{\mathrm{\infty }}(n+\frac{1}{2})a_n{x}^{n+\frac{1}{2}}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+\frac{1}{2}}-\sum _{n=2}^{\mathrm{\infty }}{a}_{n-2}{x}^{n+\frac{1}{2}}=0$$

For $n=1$ (we skip $n=0$ as that was used to find $r$) and we always let $a_0=1$ as it is arbitrary. Hence

$$\begin{array}{rl}2(1+\frac{1}{2})(1+\frac{1}{2}-1)a_1-(1+\frac{1}{2})a_1+a_1& =0\\ a_1& =0\end{array}$$

For $n\ge 2$, we have recursion relation. From it we can find that $a_2=\frac{1}{6},a_3=0,a_4=\frac{1}{168}$ and so on. Hence

$$\begin{array}{rl}{y}_2& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+\frac{1}{2}}\\ & =x^{\frac{1}{2}}\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\\ & =x^{\frac{1}{2}}(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots )\\ & =x^{\frac{1}{2}}(1+\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\cdots )\end{array}$$

Hence the final solution is

$$\begin{array}{rl}{y}_h& =c_1{y}_1+c_2{y}_2\\ & =c_1(x+\frac{1}{10}{x}^3+\frac{1}{360}{x}^5+\cdots )+c_2\left(x^{\frac{1}{2}}(1+\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\cdots )\right)\end{array}$$