Math 2280
Quiz 8
Consider the initial value problem
′
~x =
−3 −2
9
3
~x, ~x(0) =
1
−1
.
The eigenpairs of the coefficient matrix are
−1 + i
−1 − i
3i,
, −3i,
.
3
3
Form a fundamental matrix and use it to solve the IVP.
In order to find a fundamental matrix, we must find two linearly independent solutions. Since we have complex
conjugate eigenvalues, we choose one and use the real and imaginary parts of eλt~v as our two solutions. We have
that
eλt~v = e(p+qi)t (~a + ~bi) = ept (cos qt~a − sin qt~b) + iept (cos qt~b + sin qt~a).
In this case, p = 0, q = 3, ~a = [−1 3]T , and ~b = [1 0]T . Therefore,
− cos 3t − sin 3t cos 3t − sin 3t
Φ(t) =
3 cos 3t
3 sin 3t
and
Φ(0) =
−1 1
3 0
.
To find the inverse of this matrix, we augment it with the identity and row reduce:

 
 
..
..
..
 −1 1 . 1 0  ∼  1 −1 . −1 0  ∼  1 0 . 0
.
.
.
0 1 .. 1
3 0 .. 0 1
0 3 .. 3 1
Therefore, the solution to the initial value problem is
− cos 3t − sin 3t cos 3t − sin 3t
0
~x(t) = Φ(t)Φ−1 (0)~x0 =
3 cos 3t
3 sin 3t
1
1 1
− cos 3t − sin 3t cos 3t − sin 3t
−3
cos 3t + 13 sin 3t + 32 cos 3t −
= 3
=
2
− cos 3t + 2 sin 3t
3 cos 3t
3 sin 3t
3
1
3
1
3
1
3
1
3

.
1
−1
2
cos 3t − 13 sin 3t
3 sin 3t
=
.
− cos 3t + 2 sin 3t
2. If A is the coefficient matrix from the previous problem, find a particular solution to
~x′ = A~x + f~(t),
where f (t) = [0 1]T .
Since we are given the eigenvalues and eigenvectors of A, we know that f~(t) is not part of the complimentary
solution. We take as our guess
A
~xp (t) =
.
B
Then
0
0
=
~x′p (t) =
and
A~xp + f~(t) =
−3 −2
9
3
A
B
+
0
1
1
−3A − 2B
9A + 3B + 1
⇒
−3A − 2B = 0
.
9A + 3B = −1
We form the augmented matrix for this system of equations and row reduce:

 
 
..
..
2
−3
−2
.0
1
.
0
3

∼
∼ 1 0
..
..
9
3 .−1
0 1
0 −3 . −1
Therefore,
~xp (t) =
2
2
9
− 31
.

..
2
.
9 .
..
. − 31