I am finding this ode can't be solved by power series. Maple can't solve it either. To find why, I solved it by hand.
Here is my hand solution. Did I do something wrong, or is this all correct, and is the following why Maple can't also solve it?
\begin{align*} x^{2}y^{\prime\prime}+xy^{\prime}+xy & =1\\ y^{\prime\prime}+p\left( x\right) y^{\prime}+q\left( x\right) y & =1 \end{align*}
Expansion around $x=0$. This is regular singular point. Hence Frobenius is needed. Therefore $p=\frac{x}{x^{2}}=\frac{1}{x},q=\frac{x}{x^{2}}=\frac{1} {x}$ and $$ p_{0}=\lim_{x\rightarrow0}xp\left( x\right) =\lim_{x\rightarrow0}x\frac {1}{x}=1 $$ And $$ q_{0}=\lim_{x\rightarrow0}x^{2}q\left( x\right) =\lim_{x\rightarrow0} x^{2}\frac{1}{x}=0 $$ Hence the indicial equation is \begin{align*} r\left( r-1\right) +p_{0}r+q_{0} & =0\\ r\left( r-1\right) +r & =0\\ r^{2} & =0 \end{align*} Hence $r_{1}=0,r_{2}=0$. Since roots are repeated, then
$$ y=c_{1}y_{1}+c_{2}y_{2} $$
Where
$$ y_{1}=x^{r_{1}}\sum_{n=0}^{\infty}a_{n}x^{n}% $$
And
$$ y_{2}=y_{1}\ln\left( x\right) +x^{r_{2}}\sum_{n=1}^{\infty}b_{n}x^{n} $$
Now let
\begin{align} y & =\sum_{n=0}^{\infty}a_{n}x^{n+r}\nonumber\\ y^{\prime} & =\sum_{n=0}^{\infty}\left( n+r\right) a_{n}x^{n+r-1} \nonumber\\ y^{\prime\prime} & =\sum_{n=0}^{\infty}\left( n+r\right) \left( n+r-1\right) a_{n}x^{n+r-2}\tag{1} \end{align} For $r=0$ the above becomes \begin{align} y_{1} & =\sum_{n=0}^{\infty}a_{n}x^{n}\nonumber\\ y_{1}^{\prime} & =\sum_{n=0}^{\infty}na_{n}x^{n-1}=\sum_{n=1}^{\infty} na_{n}x^{n-1}\nonumber\\ y_{1}^{\prime\prime} & =\sum_{n=1}^{\infty}n\left( n-1\right) a_{n} x^{n-2}=\sum_{n=2}^{\infty}n\left( n-1\right) a_{n}x^{n-2}\tag{2} \end{align} Solve for $y_{1}$ first. Hence the ode becomes \begin{align*} x^{2}y_{1}^{\prime\prime}+xy_{1}^{\prime}+xy_{1} & =1\\ x^{2}\sum_{n=2}^{\infty}n\left( n-1\right) a_{n}x^{n-2}+x\sum_{n=1}^{\infty }na_{n}x^{n-1}+x\sum_{n=0}^{\infty}a_{n}x^{n} & =1\\ \sum_{n=2}^{\infty}n\left( n-1\right) a_{n}x^{n}+\sum_{n=1}^{\infty} na_{n}x^{n}+\sum_{n=0}^{\infty}a_{n}x^{n+1} & =1\\ \sum_{n=2}^{\infty}n\left( n-1\right) a_{n}x^{n}+\sum_{n=1}^{\infty} na_{n}x^{n}+\sum_{n=1}^{\infty}a_{n-1}x^{n} & =1 \end{align*}
Since sums starts at $n=1$ then not possible to match the RHS which is $x^{0}$ as there is no $x^{0}$ in the LHS. Hence no solution exist.
Is the above the reason why this ode can't be solved using power series? If you think this can be solved using power series, then how?
restart;
ode:=x^2*diff(y(x),x$2)+x*diff(y(x),x)+x*y(x)=1;
sol:=dsolve(ode,y(x),'series');
Maple does solve $x^2 y'' + x y' + x y = 1$, obtaining $$ y \! \left(x \right) = c_1 J_{0}\! \left(2 \sqrt{x}\right) +c_2 Y_{0}\! \left(2 \sqrt{x}\right) -\left(-J_{0}\! \left(2 \sqrt{x}\right) \mathit{MeijerG} \! \left(\left[\left[\;\right], \left[-\frac{1}{2}, 1\right]\right], \left[\left[0, 0, 0\right], \left[-\frac{1}{2}\right]\right], x\right)+Y_{0}\! \left(2 \sqrt{x}\right) \left(x\; {}_{2}^{}{{{F_{3}^{}}}}\! \left(1,1;2,2,2;-x \right)-2 \gamma -\ln \! \left(x \right)\right)\right) \pi $$ However, it can't expand this MeijerG function in a series about $x=0$.
I suspect there is no series solution for this inhomogeneous differential equation.
EDIT: Sorry, I take that back. You can use the method of Variation of Parameters to get a particular solution $y_p = u_1 y_1 + u_2 y_2$ where $$ \eqalign{y_1 &= Y_0(2 \sqrt{x}) \cr y_2 &= J_0(2 \sqrt{x}) \cr u_1 &= - \int \frac{y_2\; dx}{x^2 W(y_1,y_2)}\cr u_2 &= \int \frac{y_1 \; dx}{x^2 W(y_1, y_2)}\cr W(y_1,y_2) &= y_1 y_2' - y_1' y_2}$$ Maple will find series expansions for $u_1$ and $u_2$: $$\eqalign{ u_1 &= \pi \ln \! \left(x \right)-\pi x +\frac{1}{8} \pi \,x^{2}-\frac{1}{108} \pi \,x^{3}+\frac{1}{2304} \pi \,x^{4}-\frac{1}{72000} \pi \,x^{5}+\mathrm{O}\! \left(x^{6}\right)\cr u_2 &= -\frac{\ln \! \left(x \right)^{2}}{2}-2 \ln \! \left(x \right) \gamma +\left(2 \gamma +\ln \! \left(x \right)-3\right) x +\left(-\frac{\ln \! \left(x \right)}{8}+\frac{7}{16}-\frac{\gamma}{4}\right) x^{2}+\left(\frac{\ln \! \left(x \right)}{108}-\frac{1}{27}+\frac{\gamma}{54}\right) x^{3}+\left(-\frac{\ln \! \left(x \right)}{2304}+\frac{53}{27648}-\frac{\gamma}{1152}\right) x^{4}+\left(\frac{\ln \! \left(x \right)}{72000}-\frac{143}{2160000}+\frac{\gamma}{36000}\right) x^{5}+\mathrm{O}\! \left(x^{6}\right)\cr}$$ so that $$ y_p = \frac{\ln \! \left(x \right)^{2}}{2}+\left(-\frac{\ln \! \left(x \right)^{2}}{2}+2 \ln \! \left(x \right)-3\right) x +\left(\frac{\ln \! \left(x \right)^{2}}{8}-\frac{3 \ln \! \left(x \right)}{4}+\frac{23}{16}\right) x^{2}+\left(-\frac{\ln \! \left(x \right)^{2}}{72}+\frac{11 \ln \! \left(x \right)}{108}-\frac{97}{432}\right) x^{3}+\left(\frac{\ln \! \left(x \right)^{2}}{1152}-\frac{25 \ln \! \left(x \right)}{3456}+\frac{485}{27648}\right) x^{4}+\left(-\frac{\ln \! \left(x \right)^{2}}{28800}+\frac{137 \ln \! \left(x \right)}{432000}-\frac{14269}{17280000}\right) x^{5}+\mathrm{O}\! \left(x^{6}\right) $$ For some reason, Maple won't compute a series for $u_1 y_1$ directly, but you can find the series for $u_1$ and for $y_1$ separately and then multiply them together, similarly for $u_2 y_2$.
EDIT: Actually, of course, the $O(x^6)$ should be $O(\ln(x)^2 x^6)$.