# Problem solving ODE with Frobenius method and understand purpose of indicial equation

I am trying to solve $$y'' -xy'- y= 0, y'(0) = 0, y(0) =1$$ Using the Frobenius Method. I expand around 0 and take a solution of the form $$y(x) = \sum_{j=0}^{\infty} a_j x^{j+k}$$. Whilst doing so, I have realised that I do not understand at all why we need add this $$k$$ as the exponent. I still followed the method and found roots for the indicial equation: $$k=1, k=0$$. My second question is what does this mean? Do I now have 2 cases to consider ($$k=1$$ and $$k=0$$)? Then, I find a recurrence relation for $$k=0$$: $$a_{j+2} = \frac 1 {j+2} a_{j+1} + \frac 1{(j+1)(j+2)} a_j$$ I am a bit confused as to what I have to do now to find the solution(s) to this ODE. Any help would be great!

Frobenius Method is applicable if in your equation $$y''+a(x)y'+b(x)y=0$$ the coefficient $$a$$ has a singularity at zero not worse than $$1/x$$ and the coefficient $$b$$ not worse than $$1/x^2$$. The idea is to consider the equidimensional (Euler) equation which has precisely those singularities $$y''+\frac{A}{x}y'+\frac{B}{x^2}y=0, \quad a(x)\sim A/x;\, b(x)\sim B/x^2$$ as $$x\to 0$$, and can be solved exactly, with basic solutions $$y_1(x)=x^{k_1}$$ and $$y_2(x)=x^{k_2}$$ (this is where the indicial equation arises, when you try solutions of the form $$x^k$$ in Euler's equation).
Then, you expect that the solution of your original equation is a perturbation of the exact solution of Euler's equation. Frobenius' idea was to consider a multiplicative perturbation $$y_1(x)=x^{k_1}F_1(x); \quad y_2(x)=x^{k_2}F_2(x)$$ with analytic $$F_1$$ and $$F_2$$ and see if we can find a basic set of solutions in that form. It turns out to be possible, and the coefficients in the analytic expansions of $$F_1$$ and $$F_2$$ can be found recursively.
This is why you look for solutions in the form $$y_1(x)=x^{k}F(x)=\sum_{j=0}^\infty a_jx^{k+j}$$ for both solutions $$k_1$$ and $$k_2$$ of the indicial equation.