$$y^{''} = k^2y,y(0)=A, y^{'}(0) = B$$
I used the characteristic equation and found that $r=k,-k$.
Then the general solution is $y(x) = C_1e^{kx}+C_2e^{-kx}$.
And $y^{'}(x) = C_1ke^{kx}-C_2ke^{-kx}$
Use the initial condition I got $A=C_1+C_2$ and $B=k(C_1-C_2)$
How do I express $C_1$ and $C_2$ in terms of $A$ and $B$? Did I do something wrong in my calculation?