Homogeneous solution: y′′+y′−6y=0 yields the characteristic equation r2+r−6=0 ⇒ r=−3,2 So homogeneous part is yc=C1e−3t+C2e2t. Non-homogeneous solution: y′′+y′−6y=12e3t+12e−2t Use the method of undetermined coefficients. Suppose yp=Ae3t+Be−2t is a solution, so you have yp=Ae3t+Be−2ty′p=3Ae3t−2Be−2ty′p=9Ae3t+4Be−2t Substitute into the original equation: (9Ae3t+4Be−2t)+(3Ae3t−2Be−2t)−6(Ae3t+Be−2t)6Ae3t−4Be−2t=12e3t+12e−2t=12e3t+12e−2t This tells you that A=2 and B=−3, and so your non-homogeneous part is yp=2e3t−3e−2t. Your final solution would be the sum of the non/homogeneous parts, or y=yc+yp