Math 2280-001
Quiz 6 SOLUTIONS
February 27, 2015
1) Find all solutions x t to
x## t C 4 x t = cos 3 t
(5 points)
Hint: You will need to find a particular solution as well as all solutions to the homogeneous problem, as
steps in completing this exercise.
For the homogeneous problem the characteristic polynomial is p r = r2 C 4 = r C 2 i r K 2 i , so
the homogeneous solutions are
xH t = c1 cos 2 t C c2 sin 2 t .
For a particular solution we try xP t = A cos 3 t . (We don't need any sin 3 t terms because we only
have even derivatives on the left side of the operator, i.e. L preserves the 1-dimensional subspace that is the
span of cos 3 t .)
4 xP t = A cos 3 t
C 0 xP # t =K3 A sin 3 t
C 1 xP ## t =K9 A cos 3 t
__________________________
L xP = cos 3 t 4 A K 9 A
=K5 A cos 3 t .
1
Thus we want K5 A = 1, i.e. A =K . The complete solution is
5
1
x t = xP C xH =K cos 3 t C c1 cos 2 t C c2 sin 2 t .
5
2) What form would the undetermined coefficients "guess" for the particular solution take, in the
following two problems? (Do not attempt to find the undetermined coefficients.)
a) x## t C 4 x t = cos 2 t
because cos 2 t solves the homogeneous problem and because p r = r C 2 i
case II of undetermined coefficients and try
xP t = d1 t cos 2 t C d2 t sin 2 t .
1
rK2 i
1
(3 points)
we apply
b) x## t C 0.2 x# t C 4 x t = cos 2 t .
(2 points)
Because the roots of the characteristic polynomial do not include r =G 2 i this is case I of undetermined
coefficients:
xP t = d1 cos 2 t C d2 sin 2 t .