# Frobenius series solution

The equation $x^{2}y''+xy'+5y=0$ is equidimensional and has the general solution $y(x) = c_{1}\cos(\sqrt{5}\log x) + c_{2}\sin(\sqrt{5}\log x)$. But this differential equation also has a regular singular point around $x = 0$ and hence we can use the Frobenius method to find $a_{n}$ such that $y(x) = \sum_{n = 0}^{\infty} a_{n}x^{n + s}$. But wouldn't this imply that $\sin(\sqrt{5}\log x)$ has a series expansion around $x = 0$? Also when I use the indicial equation to find $s$, I get $s = \pm i\sqrt{5}$ which doesn't seem right?

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Remember $\sin(u)=(e^{iu}-e^{-iu})/(2i)$, so of course substituting $u=\sqrt{5}\log x$ you'll get that the function is a difference of two powers of $x$, namely the powers $s=\pm i\sqrt{5}$ as you solved for.