Solving the IVP $8y''+26y=0$, $y(0)=2$, and $y'(0)=7$ Im supposed to find $y$ as a function of $t$ given the equation: 

$$8y''+26y=0$$

Initial values are $y(0)=2$ and $y'(0)=7$. 

I found the function of $y$ is $y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$. Then I take the derivative, $y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$. Then using the initial values i'd solve for both arbitrary constants $c_1$ and $c_2$ and I ended up getting $2$ and $\frac{14}{\sqrt{13}}$. However this answer isn't correct. There's gotta be a mistake somewhere but I've been struggling to find it.
The DE's characteristic equation has two distinct complex roots: $+\frac{\sqrt{13}}{2}i$ and $-\frac{\sqrt{13}}{2}i$, so the general solution (indefinite integral) is:

$$y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$$

First initial condition, $y(0)=2$:

$$2=c_1\cos (\frac{\sqrt{13}}{2}0)+c_2\sin(\frac{\sqrt{13}}{2}0) \implies c_1=2$$

Second initial condition, $y'(0)=7$:

$$y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$$

$$7=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}0)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}0) \implies c_2=\frac{14}{\sqrt{13}}$$

Your solution is correct.
With free Computer Algebra System Maxima 5.42.1: