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)
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}
Comments
I found a small error and corrected it.
Thanks.