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.

In some degenerate cases, the indicial equation only has one real solution but you can generate a second independent solution by variation of parameters (logarithms appear, etc.) Hope this helps.


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