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 $$y = c_0 + c_1 + \sum_{n=2}^\infty \frac{x^n}{n}$$
Best Answer
You initially wrote that,
$$ y = \sum_{n=0}^\infty c_n x^n,$$
and then you determined that for $n \geq 1$ we have $c_n = c_1/n$.
$$ y = c_0 + \sum_{n=1}^\infty c_n x^n,$$
$$ y = c_0 + \sum_{n=1}^\infty \frac{c_1}{n} x^n,$$
$$ y = c_0 + c_1 \sum_{n=1}^\infty \frac{x^n}{n},$$
which is consistent with the answer given by @peter in the comments.