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
