The differential equation
\begin{align}
y^{''} - x y^{'} + y = 0
\end{align}
can be solved via a power series of the form
\begin{align}
y(x) = \sum_{k=0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \cdots .
\end{align}
It is fairly evident that
\begin{align}
\sum_{k=0}^{\infty} k(k-1) a_{k} x^{k} = \sum_{k=0}^{\infty} (k-1) a_{k} x^{k}
\end{align}
which yields the equation for the coefficients
\begin{align}
a_{k+2} = \frac{ (k-1) a_{k} }{ (k+1) (k+2) }.
\end{align}
It is discovered that $a_{3} = 0 \cdot a_{1}$. Since, for $k$ being odd, say $k \rightarrow 2k+1$,
\begin{align}
a_{2k+3} = \frac{k a_{2k+1} }{(k+1)(2k+3)}
\end{align}
it is clear that all the odd coefficients depend of $a_{3}$ for $k \geq 1$ and leads to $a_{2k+1} = 0$ for $k \geq 1$. The even $k$ values are
\begin{align}
a_{2} &= - \frac{a_{0}}{2!} \\
a_{4} &= - \frac{a_{0}}{4!} \\
a_{6} &= - \frac{(1 \cdot 3) a_{0}}{6!} \\
a_{8} &= - \frac{(1\cdot 3 \cdot 5)a_{0}}{8!}
\end{align}
which has the general form
\begin{align}
a_{2k} = - \frac{a_{0}}{2^{k} k! (2k-1)}.
\end{align}
The series for $y(x)$ now be seen in the form
\begin{align}
y(x) = a_{0} + a_{1} x - a_{0} \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)}.
\end{align}
The power series discovered can be evaluated as follows. Consider
\begin{align}
\partial_{x} \left( \sum_{k=1}^{\infty} \frac{ x^{2n-1} }{2^{k} k! (2k-1)} \right) &= \sum_{k=1}^{\infty} \frac{ x^{2n-2} }{2^{k} k!} = \frac{1}{x^{2}}( e^{x^{2}/2} -1).
\end{align}
Integrating both sides
\begin{align}
\sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} &= x \int^{x} \frac{e^{u^{2}/2} -1}{u^{2}} du = x \left[ \sqrt{\frac{\pi}{2} } erfi\left( \frac{x}{\sqrt{2}} \right) - \frac{e^{x^{2}/2}}{x} + \frac{1}{x} \right] \\
&= \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} + 1.
\end{align}
With this series the general solution of $y(x)$ can be sen by
\begin{align}
y(x) &= a_{0} + a_{1} x - \sum_{k=1}^{\infty} \frac{ x^{2n} }{2^{k} k! (2k-1)} \\
&= a_{1} x - a_{0} \left[ \sqrt{\frac{\pi}{2} } \cdot x \cdot erfi\left( \frac{x}{\sqrt{2}} \right) - e^{x^{2}/2} \right],
\end{align}
where $erfi(x)$ is the imaginary error function (erfi(x) = -i erf(ix)).
An elementary (and not particularly smart) approach could be:
Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$
Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.
You know that $y'(0)=f(0,y_0)$.
Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$
Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.
Keep deriving
$$y^{(3)}(x)=\\=
\dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x))
$$
Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.
The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.
Best Answer
A power series solution can be easily obtained. Making $y=\sum_{k=0}^n a_k t^k$ and substituting into the ODE we have
$$ \left(\sum_{k=0}^n a_k t^k\right)'-t\left(\sum_{k=0}^n a_k t^k\right)-t^2=0 $$
after grouping for powers of $t$ we have
$$ \left\{ \begin{array}{rcl} a_1&=&0 \\ 2 a_2-a_0&=&0 \\ 3 a_3-a_1 &=&1 \\ 4 a_4-a_2&=&0 \\ 5 a_5-a_3&=&0 \\ 6 a_6-a_4&=&0 \\ 7 a_7-a_5&=&0 \\ 8 a_8-a_6&=&0 \\ 9 a_9-a_7&=&0 \\ 10 a_{10}-a_8&=&0 \\ \vdots & \vdots & \vdots \end{array} \right. $$
solving for $n=10$ we obtain
$$ y_{10} = a_0\left(1+\frac{t^2}{2}+\frac{t^4}{8}+\frac{t^6}{48}+\frac{t^8}{384}+\frac{t^{10}}{3840}\right)+\frac{t^3}{3}+\frac{t^5}{15}+\frac{t^7}{105}+\frac{t^9}{945}+O\left(t^{11}\right) $$
Follows a plot showing in blue the closed solution for $y(0)=1$ as well as in red the series approximation for $a_0=1$.