I want to solve $$x(2-x) y'' +(1+x)y' -\frac{3}{x} y = 0$$ with singular-regular point $x_0 =0$, the general solution by Frobenius is of the form $y = x^r \sum \limits_{n=0}^{\infty} a_n x^n$. i found that $r = -1, \frac{3}{2}$. substituting the value $r=-1$ and the standard value $a_0 =1$ gave me that $$a_{n+1} = \frac{n^2-4n+3}{2n^2-n-3}a_n$$ which means that $a_1 = a_2 = \cdots =0$
So in the end i have one solution which is $y = \frac{1}{x}(1+0 x^1 +0 x^2+\cdots) = \frac{1}{x}$ but the correct solution is $y = \frac{1}{x}-1$, what went wrong in my solution ?
You have
$$ a_{n + 1} = a_n\frac{n^2 - 4n + 3}{2n^2 - n - 3} $$
Using $a_0 = \color{blue}{1}$ you get
$$ a_1 = 1 \frac{3}{(-3)} = \color{red}{-1} $$
and
$$ a_2 = 0 = a_3 = \cdots $$
So one of the solutions is
$$ y_1(x) = x^{-1}\left(a_0 + a_1 x + a_2 x^2 + \cdots \right) = x^{-1}\left[\color{blue}{1} + (\color{red}{-1})x\right] = -1 + 1/x $$