Power series method to solve this ODE: what did I do wrong? Problem: Use the power series method to solve the ODE $$ 2(x - 1) y' = 3y. $$
Attempt: I solved this with power series method, and then compared with the technique of separation of variables, and I'm not getting the same answer. Here is what I did:
We look for a power series around $x = 0$. Hence we look for a solution of the form $y(x) = \sum_{n=0}^{\infty} a_n x^n$. Then $y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} $, and if we plug this in the ODE we get \begin{align*} 2(x-1) \sum_{n=1}^{\infty} n a_n x^{n-1} - 3 \sum_{n=0}^{\infty} a_n x^n = 0 
\end{align*} or, after distributing and equating powers: \begin{align*} 2 \sum_{n=1}^{\infty} n a_n x^n - 2 \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n - 3 \sum_{n=0}^{\infty} a_n x^n = 0 \end{align*} Hence by the identity theorem we get \begin{align*} \begin{cases} -2 a_1 - 3 a_0 &= 0 \qquad (n = 0) \\ 2n a_n - 2 (n+1) a_{n+1} - 3 a_n &= 0 \qquad (n \geq 1) \end{cases} \end{align*} From the first equation, we have $a_1 = - \frac{3}{2} a_0$, and from the second $$ a_{n+1} = a_n \frac{(2n-3)}{2(n+1)} $$  Since $$ a_2 = \frac{(-3)(-3)}{2 \cdot 2} a_0 $$ and $$ a_3 = \frac{(-3)(-3)(-3)}{2 \cdot 2 \cdot 2} a_0 $$ the general pattern is $$ a_n = (-1)^n (\frac{3}{2})^n a_0 $$ Hence the solution is given as $$ y(x) = a_0 \sum_{n=0}^{\infty} (- \frac{3}{2})^n x^n $$ and the radius of convergence is $R = 2 / 3$  because $\lim_{n \to \infty} \big( ( 3/2)^n \big)^{1/n}  = 3/2 $ by Cauchy's test. I rewrote the solution as $$ y(x) = a_0 \bigg(\frac{1}{1 + \frac{3}{2} x} \bigg) = a_0 \big( \frac{2}{2 + 3x} \big). $$
But If I solve the ODE by seperation of variables, I get a different answer which is also what Maple gives me: $y(x) = C (x-1)^{3/2}$. So what did I do wrong here?

There is a mistake in your computation of the coefficients $a_n$ (see below) :
One can reconize the binomial coefficients
\begin{pmatrix}
  \frac{3}{2}  \\
  n
   \end{pmatrix}
wich allows to link with $(1-x)^{3/2}$

