Answer
$y(x)=C_1e^{-5x}+C_2e^{x}+C_3e^{2x}\cos x+C_4e^{2x} \sin x$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$r^4-16r^2+40r-25=0$$
Factor and solve for the roots. $$(r-1)(r+5)(r^2-4r+5)=0$$
Roots are: $r_1=-5, r_2=1,r_3=2-i, r_4=2+i$
This implies that 4 are two independent solutions to the differential equation.
Therefore, the general equation is equal to $y(x)=C_1e^{-5x}+C_2e^{x}+C_3e^{2x}\cos x+C_4e^{2x} \sin x$
