Frobenius method to solve this equation: $$xy''+(1-x)y'-y=0$$
I am trying to find the solution for the DE but I'm stuck at the step where to find the equation to find the terms of the sequence.Attached is my working.Therefore,I would like to ask that how can I get the equation to calculate the terms?
1.Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$. Plugging in the appropriate series expansions, we get that $$\sum_{n=2}^{\infty}n(n-1)a_n x^{n-1}+(1-x)\sum_{n=1}^{\infty}na_nx^{n-1}-\sum_{n=0}^{\infty}a_n x^n=0.$$ Hence, we conclude that $$a_{n+1}=\frac{a_n}{n+1},\quad\forall\ n\in\mathbb{N} \implies a_n=\frac{a_0}{n!}.$$
Thus we get a solution $$y_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x,$$ and it is easy to check this solution.
2.Let $y(x)=y_1(x)v(x)$, put it in the equation, then we have $$xv''+(1+x)v'=0.$$ Denote $u(x)=v'(x)$ and assume $u\neq0$. Solve the 1st ODE $$\frac{u'}{u}=-(1+\frac{1}{x})\implies u(x)=C_1\frac{e^{-x}}{x},\quad v(x)=C_1\int \frac{e^{-x}}{x}dx+C_2.$$ Thus all solutions are $$y(x)=e^x(C_1\int \frac{e^{-x}}{x}dx+C_2).$$