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Find the general solution?

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1 Answer

Plugging  y = ert 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

yh = c1et+c2tet+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 = yh + y* = c1et+c2tet+c3e-t+(1/2)te-t+3 {edited}