Solve $xy''+y'-xy=0$ using Frobenius method 
Using Frobenius method solve $$xy''+y'-xy=0$$

Comparing the given equation with $y''+P(x)y'+Q(x)y=0$ we have,
$$P(x)=\frac{1}{x}\qquad Q(x)=-1$$
Here $x=0$ is a singular point of the given differential equation. Now
$$
\begin{aligned}
 \lim_{x\rightarrow 0}(x-0)P(x)&=1\\
 \lim_{x\rightarrow 0}(x-0)^2Q(x)&=0
\end{aligned}
$$
Hence $x=0$ is a regular singular point. The indicial equation is,
$$
\begin{aligned}
r(r-1)+pr+q&=0\\
r^2-r+r&=0\\
r=0
\end{aligned}
$$
$$
\begin{aligned}
 y'&=\sum_{k=0}^\infty a_k(k+r)x^{k+r-1}\\
 y''&=\sum_{k=0}^\infty a_k(k+r)(k+r-1)x^{k+r-2}
\end{aligned}
$$
Substituting these on the DE,
$$
\begin{aligned}
 \sum_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-1}+\sum_{k=0}^\infty(k+r)x^{k+r-1}-\sum_{k=0}^\infty a_kx^{k+r+1}&=0\\
 \sum_{k=0}^\infty ((k+r)(k+r-1)+(k+r))a_k x^{(k+r-1)} +\sum_{k=2}^\infty a_{k-2}x^{(k+r-1)} &=0\\
 r^2a_0x^{r-1}+(r+1)^2a_1x^r+\sum_{k=2}^\infty [(n+r)^2a_n-a_{n-2}]x^{n+r-1} &=0
\end{aligned}
$$
$$r^2a_0=0\qquad (r+1)^2a_1=0\implies a_1=0$$
$$
\begin{aligned}
(n+r)^2a_n-a_{n-2} &=0\\
an&=\frac{a_{n-2}}{(n+r)^2}\\
{a_{n+2}}&=\frac{a_n}{(n+2+r)^2}\\
&=\frac{a_n}{(n+2)^2}\quad\text{ Using }r=0
\end{aligned}
$$
$a_3=a_5=\cdots=a_{2n+1}=0$ because $a_1=0$. Now,
$$
\begin{aligned}
a_2&=\frac{a_0}{2^2}\\
a_4&=\frac{a_2}{4^2}=\frac{a_0}{4^22^2}=\frac{a_0}{2^{2.2}(2.1)^2}\\
a_6&=\frac{a_4}{6^2}=\frac{a_0}{6^24^22^2}=\frac{a_0}{2^{2.3}(3.2.1)^2}\\
&\vdots\\
a_{2n}&=\frac{a_0}{2^{2n}(n!)^2}
\end{aligned}
$$
One solution of the DE is,
$$
y(x)=x^0(\sum_{k=0}^\infty a_kx^k)=a_0(1+\frac{1}{2^2}x^2+\cdots+\frac{1}{2^{2n}n!}x^{2n}+\cdots)
$$
what about another solution? How to get that one?

The indicial equation has double root $r=0$.  So the second solution will involve $\log x$.
$$
y_1(x)=\,  \left(1+{\frac{1}{4}}{x}^{2}+{\frac
{1}{64}}{x}^{4}+\dots\right),
\\
y_2(x) =
\log(x) \left(1+{\frac{1}{4}}{x}^{2}+{\frac{1
}{64}}{x}^{4}+\dots \right) +\left(-{\frac{1}{4}}{x}^
{2}-{\frac{3}{128}}{x}^{4}+\dots\right) 
$$

Maple code

