# Frobenius Method Constant Coefficient

Need help solving this differential equation but using the power series method.

$y'' -5y'+4y=0$

Since there are no external $x$ terms there three different powers of $x$ and I am not sure how to get a recurrence relation.

Thanks

• There are infinitely many solutions. You may choose $y(0)$ and $y'(0)$ to be anything you want.(Even both $0$, although that does yield only the constant solution $y=0$) – DanielWainfleet Oct 5 '15 at 4:33

Since you want series, start with $$y=\sum_{i=0}^\infty a_ix^i$$ $$y'=\sum_{i=0}^\infty ia_ix^{i-1}$$ $$y''=\sum_{i=0}^\infty i(i-1)a_ix^{i-2}$$ So, the differential equation write $$\sum_{i=0}^\infty i(i-1)a_ix^{i-2}-5\sum_{i=0}^\infty ia_ix^{i-1}+4\sum_{i=0}^\infty a_ix^i=0$$ Now, consider the term corresponding to $x^n$. You then have $$(n+2)(n+1)a_{n+2}x^n-5(n+1)a_{n+1}x^n+4a_nx^n=0$$
I am sure that you can take from here (remembering that $a_0$ and $a_1$ are undefined since there are no initial conditions given).