(20 pts) Indicate the form in which you'd be looking for a solution to

y'''' + 2y''' + 2y'' + y' = 3e^{t} + 2te^{-t} + e^{-t}sin(t).
Do not solve the equation.

### Solution

First write the characteristic equation:

r^{4} + 2r^{3} + 2r^{2} + r = r (r^{3} + 2r^{2} + 2r + 1) = 0.
So we get one root: r_{1} = 0. The equation r^{3} + 2r^{2} + 2r + 1 = 0 has an obvious root r_{2} = -1. It must be divisible by (r + 1). The long division gives

r^{3} + 2r^{2} + 2r + 1 = (r + 1)(r^{2} + r + 1).
The latter (quadratic) polynomial has two roots: r_{3, 4} = (-1 ± sqrt(1 - 4))/2 = (-1 ± i·sqrt(3))/2.

Now we have to consider the right hand side one piece at a time:

**3e**^{t} is of the Ae^{t} kind with an eye on r = 1 as a root of the characteristic equation. There is none. Therefore, we should try a partial solution in the same form: y_{p1} = Ae^{t}.

**2te**^{-t} is of the e^{-t}(B + Ct) kind with an eye on r = -1 as a root of the characteristic equation. There's one such root. We therefore should upgrade our guess by a factor of t: y_{p2} = e^{-t}t(B + Ct).

**e**^{-t}sin(t) is of e^{-t}(Dsin(t) + Ecos(t)) kind with an eye on r = -1 ± i as the roots of the characteristic equation. There are no such roots. We therefore should be looking for a particular solution in the same form: y_{p3} = e^{-t}(Dsin(t) + Ecos(t)).

### Answer

y(t) = C_{1} + C_{2}e^{-t} + e^{-t/2}(C_{3}sin(sqrt(3)·t/2) + C_{4}cos(sqrt(3)·t/2)) + Ae^{t} + e^{-t}t(B + Ct) + e^{-t}(Dsin(t) + Ecos(t)).