Find the general solution?

Question

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

y=c₁e^t+c₂te^t+c₃e^-t+(1/2)te^-t+3  (This is the answer)

 

Answer 1

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

r³ - r² - r + 1 = r²(r-1) - (r-1) = (r+1)(r-1)²

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

So, the homogeneous solution is

y_h = c₁e^t+c₂te^t+c₃e^-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* = c₁e^t+c₂te^t+c₃e^-t+(1/2)te^-t+3 {edited}