2nd order differential equations

Question

Find the general solution to the non-homogeneous differential equation.

y'' − 5y' + 4y = 3e^x

the homogenous part would be y(x) = C1e^x+C2e^4x

for yp would i guess Ae^x or Axe^x

Answer 1

You're right on the homogeneous part. e^x is already included in the homogenous part, so let's try your second guess yp = Axe^x. To find A, we plug yp into the differential equation. Since yp' = A(xe^x+e^x),

yp'' = A(xe^x+2e^x). Plugging into yp''-5yp' +4yp = 3e^x, we get

A(xe^x+2e^x)-5A(xe^x+e^x)+4Axe^x = 3e^x,

and grouping terms we get

(A-5A+4A)xe^x+(2A-5A)e^x = 3e^x

-3Ae^x = 3e^x.

so A must be -1.

Thus the general solution is yp plus the homogenous solution.