Find the general solution of y"'-y"-y'+y=2e^(-t)+3.

y=c_{1}e^{t}+c_{2}te^{t}+c_{3}e^{-t}+(1/2)te^{-t}+3 (This is the answer)

Find the general solution of y"'-y"-y'+y=2e^(-t)+3.

y=c_{1}e^{t}+c_{2}te^{t}+c_{3}e^{-t}+(1/2)te^{-t}+3 (This is the answer)

Tutors, please sign in to answer this question.

Plugging y = e^{rt} into the differential equation, you have

r^{3} - r^{2} - r + 1 = r^{2}(r-1) - (r-1) = (r+1)(r-1)^{2}

r = -1, and r = 1 (double roots)

So, the homogeneous solution is

y_{h} = c_{1}e^{t}+c_{2}te^{t}+c_{3}e^{-t
}

Since r = -1 matches the particular solution too, the particular solution should have a form: y* = Cte^{-t
}+ 3.

y*' = Ce^{-t} (-t+1)

y*'' = Ce^{-t} (t-2)

y*''' = Ce^{-t} (-t+3)

So, from y*"'-y*"-y*'+y*=2e^{-t}+3 we have,

Ce^{-t} (-t+3-t+2+t-1+t) = 2e^{-t}

C = 1/2

Answer: y = y_{h} + y* = c_{1}e^{t}+c_{2}te^{t}+c_{3}e^{-t}+(1/2)te^{-t}+3 {edited}

Steven L.

Cornell Alum - Enthusiast in Math and Science (Regents, SAT/ACT, AP)

Fresh Meadows, NY

4.9
(146 ratings)

Nikunj P.

Math and Science Specialist, Over 14 Years of Educational Experience

Scarsdale, NY

4.9
(178 ratings)

Kimberly P.

Get Ready For Next Year With a Principal-Recommended Tutor!

Rego Park, NY

4.8
(28 ratings)

## Comments

I found a small error and corrected it.

Thanks.