Power series differential check $(x-1)y'' + y' = 0$ I'm a bit stuck on how to solve this:
$$(x-1)y'' + y' = 0 $$
so assuming y is a solution in this form:
$$\sum_{n=0}^\infty C_nx^n$$
$$\sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$
subbing in and distributing:
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty n(n-1)C_nx^{n-2} + \sum_{n=1}^\infty nC_nx^{n-1}$$
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=1}^\infty (n+1)(n)C_{n+1}x^{n-1} \sum_{n=1}^\infty nC_nx^{n-1}$$
start at n=2
$$\sum_{n=2}^\infty n(n-1)C_nx^{n-1} - \sum_{n=2}^\infty (n+1)(n)C_{n+1}x^{n-1} + \sum_{n=2}^\infty nC_nx^{n-1} - 2C_2 + C_1$$
so $-2C_2 + C_1 = 0$
and $$(n(n-1)C_n - (n+1)(n)C_{n + 1} + nC_n = 0$$
and $$(n^2C_n - (n+1)(n)C_{n + 1} = 0$$
and $c_{n+1} = \frac{nC_n}{n+1}$ for $n \ge 2$
so writing out the first few terms:
$c_0 = c_0$ and $c_1 = c_1$ and $c_2 = \frac{c_1}{2}$ and $c_3 = \frac{2c_2}{3}$ and $c_4 = \frac{3c_3}{4} = \frac{3 \cdot 2 \cdot c_1}{4!}$  and so $$c_n = \frac{c_1}{n}$$
so can I write that $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$ 

Calling $Y = y' = \sum_{k=0}^{\infty}a_k x^k$ we have in $(x-1)Y'+Y = 0$
$$
(x-1)\sum_{k=1}^{\infty}k a_k x^{k-1}+\sum_{k=0}^{\infty}a_k x^k = a_0-a_1+\sum_{k=1}^{\infty}(k+1)a_k x^k-\sum_{k=1}^{\infty}(k+1)a_{k+1} x^k = 0
$$
hence $a_k - a_{k+1} = 0$ or $Y = a_0\sum_{k=0}^{\infty}x^k$ but $Y = y'$ so finally
$$
y = \int_0^x Y dx = a_0\sum_{k=1}^{\infty}\frac{x^k}{k}+ C_0
$$
NOTE
$$
\sum_{k=1}^{\infty}k a_k x^{k-1} = a_1+\sum_{k=1}^{\infty}(k+1) a_{k+1} x^k
$$
