Differential equation- Solutions to initial condition problem

Question

The equation given is $$x(t)=C_{1}e^{-t}+C_{2}e^{2t}$$ $$x^{''}-x^{'}-2x=0$$ The first and second derivatives are given respectfully $$-C_{1}e^{-t}+2C_{2}e^{2t}$$ $$C_{1}e^{-t}+4C_{2}e^{2t}$$ I have verified that the problem does in fact equal zero. The given conditions are: $$x(0)=10$$ $$x^{'}(0)=8$$ After substitution and evaluation of the original and first prime equations I got $C_2=\frac{8}{3}$ and $C_1=\frac{16}{3}$ and having no one around to check if I'm correct,I was hoping that someone could tell me if I'm right or if I made a small mistake.

Answer 1

$$ \begin{cases} c_1+c_2=10 \\ -c_1+2c_2=8 \\ \end{cases} $$ As Thomas commented the sum of the two equations gives: $$3c_2=18 \implies c_2=6$$ and $$c_1=10-c_2=4$$ You made a mistake somewhere. Can you post your solution ? So we can check it ?