Answer
$y(x)=C_1e^{-2x}\cos x+C_2 e^{4x}$
Work Step by Step
Solve the characteristic equation for the differential equation. $$(D-4)(D+2)y=0$$
Factor and solve for the roots. $$(r-4)(r+2)=0$$ $r_1=-2; r_1=4$ as roots.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-2x}$ and $y_2= e^{4x}$
Therefore, the general equation is equal to $y(x)=C_1e^{-2x}\cos x+C_2 e^{4x}$
