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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

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  • $\begingroup$ 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. $\endgroup$ – user147263 Nov 3 '14 at 6:00
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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.

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

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