ODE solved using Euler's equation

This problem is getting messy fast and I'm not sure if I am doing it correctly. Any advice will be very appreciated.

y is a function of x

$\displaystyle x^2y'' + 7xy'+13y=0$; $\displaystyle y(-1) = 1$; $\displaystyle y'(-1)=3$

using $\displaystyle x = e^t$, $\displaystyle t = ln(x)$ and $\displaystyle y(x) = Y(t)$ this becomes

$\displaystyle Y'' +6Y'+13Y = 0$

The characteristic equation has roots $\displaystyle -3\pm{i2}$

so $\displaystyle Y = c_1e^{-3t}e^{i2t} + c_2e^{-3t}e^{-i2t}$

back substitution yields:

$\displaystyle y = (c_1+c_2)x^{-3}cos(2ln(x)) +j(c_1-c_2)x^{-3}sin(2ln(x))$

This is about where I ran out of steam. I had started to try and choose my constants to give me two separate equations which should be linearly independent which gives me the following:

$\displaystyle y = k_{1}x^{-3}cos(2ln(x))+k_{2}x^{-3}sin(2ln(x))$

but plugging in my initial conditions gets so messy taking the log of negative numbers giving me imaginary arguments to sinusoids which gives hyperbolic functions.

None of my other problems have been this tough and I'm guessing I'm doing something wrong. Thanks for any help.