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

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

• Notice the pattern: the powers of $x$ in coefficients are the same as the order of derivative. This means $y=x^a$ should work: you have a Cauchy–Euler equation. – user147263 Nov 3 '14 at 6:00

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.
• Thanks!! I think I confused myself when doing the second attemt, since I did get $\alpha = \frac{1\pm \sqrt{13}}{2}$ – lll Nov 3 '14 at 6:05