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

With expansion around $x=0$. This is a regular singular ODE. This is same ode solved in example 1, but now with non-zero on the right side. Hence solution is given by

$$y=y_h+y_p$$

Where we found $y_h$ to be

$$y_h=c_1(x+\frac{1}{10}{x}^3+\frac{1}{360}{x}^5+\cdots )+c_2\left(x^{\frac{1}{2}}(1+\frac{1}{6}{x}^2+\frac{1}{168}{x}^4+\cdots )\right)$$

To find $y_p$, we see there are two terms on the right side. we find $y_{p_1}$ that corresponds to $1$ on the right side and then $y_{p_2}$ that corresponds to $x$ on the right side. We will find that $y_{p_2}$ does not exist. Hence no solution exist using series. Let us now find $y_{p_1}$. The ode now is

$$2x^2{y}^{\prime \prime }\left(x\right)-x{y}^{\prime }\left(x\right)+(1-x^2)y\left(x\right)=1$$

Let

$$y_{p_1}=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+r}$$

Using same expansion as in example 1, but now using $c_n$ instead of $a_n$ gives EQ. (1) from example 1 as

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

For $n=0$

$$\begin{array}{rl}2r(r-1)c_0{x}^r-r{c}_0{x}^r+c_0{x}^r& =x^0\\ (2r(r-1)c_0-r{c}_0+c_0)x^r& =x^0\\ (2r(r-1)-r+1)c_0{x}^r& =x^0\\ (2r^2-3r+1)c_0{x}^r& =x^0\end{array}$$

We see that for balance, then $r=0$. Which implies $(2r^2-3r+1)c_0=1$ or $c_0=1.$ Hence

$$y_{p_1}=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n$$

Now that we used $n=0$ to find $r$ by matching against the RHS which is $1$, from now on, for all $n>0$ we will use (1A) to solve for all other $c_n$. From now on, we just need to solve with RHS zero, since there can be no more matches for any $x^n$ on the right.

Then (1A) becomes (using $r=0$)

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

For $n=1$ the above gives

$$-c_{1}x+c_{1}x=0$$

$c_1$ is arbitrary Let $c_1=0$. For $n=2$ EQ. (2) gives

$$\begin{array}{rl}4c_2{x}^2-2c_2{x}^2+c_2{x}^2-c_0{x}^2& =0\\ (4c_2-2c_2+c_2-c_0)x^2& =0\\ (3c_2-1)x^2& =0\end{array}$$

Hence $3c_2-1=0$ since there is no $x^2$ term on the right side. Hence $c_2=\frac{1}{3}$. For $n=3$ EQ. (2) gives

$$\begin{array}{rl}(12c_3-3c_3+c_3-c_1)x^3& =0\\ \left(10c_3\right)x^3& =0\end{array}$$

Hence $c_3=0$. For $n=4$ EQ. (2) gives

$$\begin{array}{rl}24c_4{x}^4-4c_4{x}^4+c_4{x}^4-c_2{x}^4& =0\\ (21c_4-\frac{1}{3})x^4& =0\end{array}$$

Hence $21c_4-\frac{1}{3}=0$ or $c_4=\frac{1}{63}$ and so on. We find that

$$\begin{array}{rl}{y}_{p_1}& =\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n\\ & =c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\cdots \\ & =1+\frac{1}{3}{x}^2+\frac{1}{63}{x}^4+\cdots \end{array}$$

Now we find $y_{p_2}$. EQ. (1A) now is

$$\sum _{n=0}^{\mathrm{\infty }}2(n+r)(n+r-1)c_n{x}^{n+r}-\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+r}-\sum _{n=2}^{\mathrm{\infty }}{c}_{n-2}{x}^{n+r}=x$$

For $n=0$

$$\begin{array}{rl}2(n+r)(n+r-1)c_n{x}^{n+r}-(n+r)c_n{x}^{n+r}+c_n{x}^{n+r}& =x\\ 2r(r-1)c_0{x}^r-r{c}_0{x}^r+c_0{x}^r& =x\\ (2r(r-1)-r+1)c_0{x}^r& =x\\ (2r^2-3r+1)c_0{x}^r& =x\end{array}$$

For balance we need $r=1$. This results in

$$\begin{array}{rl}(2r^2-3r+1)c_0& =1\\ (2-3+1)c_0& =1\\ 0c_0& =1\end{array}$$

Not possible. We see why there is no series solution. It is not possible to solve for $y_{p2}$.