General solution to cauchy-euler DE when t < 0

Question

The question is:

Find a general solution for $$y''(t) - \frac 1t y'(t) + \frac {5}{t^2}y(t) = 0$$ for t < 0.

I multiplied through by $t^2$ to get it into the proper form, found the roots $$1 \pm 2i$$ and obtained the general solution $$y = (-t)(c_1cos(2ln(-t))) + c_2sin(2 ln(-t)))$$ However, the solution in my book considers the $t$ on the outside to be positive while keeping the $t$'s in the ln expressions to be negative. I saw a similar question elsewhere, where the solution was derived by substituting u = -t, solving the DE regularly, and then substituting back in a -t for every u in the equation, which made sense to me. Is there some fine print I'm missing?

Answer 1

The general solution is $$y(t)=C_1t^{1+2i}+C_2t^{1-2i}$$ $t^{2i}=(t^2)^i=e^{i\ln(t^2)}=\cos\left(\ln(t^2)\right)+i\sin\left(\ln(t^2)\right)$

This trick avoid the problem of sign. Otherwise, one have to determine the branche of the function in the complex plane depending on the sign of $t$. $$y(t)=c_1t\cos\left(\ln(t^2)\right)+c_2t\sin\left(\ln(t^2)\right)$$ Note that you can use $\ln(t^2)=2\ln|t|$