Power series in a differential equation I have a differential equation that I'm having trouble wrapping my head around. 
Use a power series to solve
$$
(1-x)y'(x) + y(x) = 0
$$
with the initial condition that $y(\frac{1}{2}) = 17.$
I've tried setting up the power series as $\sum_{k=0}^\infty a_k x^k$ and differentiating from there, but I'm not managing to land anywhere useful. Some pointers as to where to go from here would be appreciated. 
 A: Let's instead use the series
$$y=\sum_{k\ge0}a_kz^k\implies y'=\sum_{k\ge0}(k+1)a_{k+1}z^k$$
where $z=x-\frac12$. This is so we can make use of the given initial value; if $x=\frac12$, then all but the constant terms in the series vanish, so $y\left(\frac12\right)=a_0=17$. The ODE becomes
$$\left(\frac12-z\right)y'+y=0$$
Plug in the series:
$$\begin{align*}
0&=\left(\frac12-z\right)y'+y\\[1ex]
&=\frac12\sum_{k\ge0}(k+1)a_{k+1}z^k-\sum_{k\ge0}(k+1)a_{k+1}z^{k+1}+\sum_{k\ge0}a_kz^k\\[1ex]
&=\frac12\left(a_1+\sum_{k\ge1}(k+1)a_{k+1}z^k\right)-\sum_{k\ge1}ka_kz^k+\left(a_0+\sum_{k\ge1}a_kz^k\right)\\[1ex]
&=\frac{a_1}2+a_0+\sum_{k\ge1}\left(\frac12(k+1)a_{k+1}-(k-1)a_k\right)z^k
\end{align*}$$
It follows that
$$\frac{a_1}2+a_0=0\implies a_1=-34$$
The remaining coefficients are governed by the recurrence relation,
$$\frac12(k+1)a_{k+1}-(k-1)a_k=0\implies a_{k+1}=\frac{2(k-1)}{k+1}a_k$$
Notice that when $k=1$, we have $a_2=0$, and this forces $a_n=0$ for all $n\ge2$.
Hence

 $$y(z)=17-34z\iff y(x)=34-34x$$

