Find the series solution for the ODE $x^2y''(x)-3y(x) = 0$

Question

Find the series solution for the ODE $x^2y''(x)-3y(x) = 0$

I assume $y(x) = \sum{a_nx^n}$ then substitute in the equation and get

$$\sum_{n=0}^{\infty} ({a_nn(n-1) - 3a_n)x^n}=0$$

When I equate the coefficients I get $a_i = 0$ for all i

Then I tried to put $y(x) =x^\alpha \sum{a_nx^n}$ but I still can't get a solution.

Can someone help? Thanks

Answer 1

Your second attempt should have led you to a solution. If you try $ y (x)=x^\alpha $, you get $\alpha^2-\alpha-3=0$, i.e. $\alpha=\frac {1\pm\sqrt {13}} 2$. This gives you two linearly independent solutions.