Answer
$\{e^{-3x}\cos 4x, e^{-3x}\sin 4x\}$ is a basis for the solution space.
Work Step by Step
Solve the characteristic equation for the differential equation. $$r^2+6r+25=0$$
Factor and solve for the roots. $$r_1=-3-4i, r_2=-3+4i$$
The general equation is equal to $y=C_1e^{-3x}cos 4x+C_2e^{-3x}\sin 4x$
Therefore, $\{e^{-3x}\cos 4x, e^{-3x}\sin 4x\}$ is a basis for the solution space.