1. (20 pts) Indicate the form in which you'd be looking for a solution to

y'''' + 2y''' + 2y'' + y' = 3et + 2te-t + e-tsin(t).

Do not solve the equation.

Solution

First write the characteristic equation:

r4 + 2r3 + 2r2 + r = r (r3 + 2r2 + 2r + 1) = 0.

So we get one root: r1 = 0. The equation r3 + 2r2 + 2r + 1 = 0 has an obvious root r2 = -1. It must be divisible by (r + 1). The long division gives

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

The latter (quadratic) polynomial has two roots: r3, 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:

  1. 3et is of the Aet 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: yp1 = Aet.

  2. 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: yp2 = e-tt(B + Ct).

  3. e-tsin(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: yp3 = e-t(Dsin(t) + Ecos(t)).

Answer

y(t) = C1 + C2e-t + e-t/2(C3sin(sqrt(3)·t/2) + C4cos(sqrt(3)·t/2)) + Aet + e-tt(B + Ct) + e-t(Dsin(t) + Ecos(t)).