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)

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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}

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## Comments

I found a small error and corrected it.

Thanks.