Math 2280
Summer 2007
Test 2
You will have one hour to complete this test. You may not use a calculator, computer, or any other electronic
device on this test. Please be sure to show as much of your work as possible, as I will determine your score on a
given problem from the work and not necessarily by the answer. Please be neat so that I can grade your answers
easily. All solutions should be real valued solutions.
1. Transform the given system of differential equations into an equivalent system of first order differential
equations.
x′′ + 3x′ + 4x − 2y = 0 y ′′ + 2y ′ − 3x + y = cos t
If we make the substitutions x1 = x, x2 = x′ and y1 = y, y2 = y ′ , then the system becomes
x′1 = x2
x′2 = −3x2 − 4x1 + 2y1
.
y1′ = y2
′
y2 = −2y2 + 3x1 − y1 + cos t
2. Use the Wronskian to determine whether or not the following vector functions are linearly independent:








1
0
0
0






 1 
t 2 
t 3 
t 4 

~x1 (t) = et 
 2  , ~x2 (t) = e  1  , ~x3 (t) = e  9  , ~x4 (t) = e  5  .
4
1
0
0
We have that
W (t) = 0
et
2et
0
0
4et
et
0
0
3et
9et
et
et
2et
5et
4et
= −et (et (e2t − 8e2t )) = −e4t + 8e4t = 7e4t 6= 0,
so the given functions are linearly independent.
3. Find the general solution of the system
′
~x =
3
2
−1
1
~x.
First, we must determine the eigenvalues of the coefficient matrix. These will satisfy
3 − λ −1 = (3 − λ)(1 − λ) + 2 = λ2 − 4λ + 5.
0=
2
1 The roots of this equation are λ = 2 ± i. We also need an eigenvector corresponding to one of these eigenvalues.
Let us choose λ = 2 + i. Then
3−2−i
−1
1−i
−1
1 − 21 − 21 i
=
∼
2
1−2−i
2
−1 − i
0
0
1 1
1+i
1
1
⇒ a=
+ i b ⇒ ~v =
⇒ ~a =
, ~b =
.
2
2
0
2 2
Recalling that
e(p+qi)t (~a + ~bi) = ept [cos qt~a − sin qt~b + i(cos qt~b + sin qt~a)]
1
and using the real and imaginary parts of the solution above as our two linearly independent real valued solutions,
we get
1
1
1
1
− sin t
+ sin t
~x(t) = c1 e2t cos t
+ c2 e2t cos t
.
2
0
0
2
4. Find the general solution to the system below.
x′1 = 3x1 − x2 ,
x′2 = x1 + 5x2
Again, we write the system in matrix form as
′ x1
3 −1
x1
=
.
x′2
1 5
x2
Then the eigenvalues of the coefficient matrix must satisfy
3−λ
−1 0 = = (3 − λ)(5 − λ) + 1 = 15 − 3λ − 5λ + λ2 + 1 = λ2 − 8λ + 16 = (λ − 4)(λ − 4).
1
5−λ Therefore, we have a repeated eigenvalue λ = 4. To find a corresponding eigenvector(s), we solve the homogeneous
system (A − 4I)~v1 = ~0.
−1 −1
1 1
−1
∼
⇒ v1 = −v2 ⇒ ~v1 =
.
1
1
0 0
1
Since we need two linearly independent eigenvectors and there is only one here, we find a generalized eigenvector. Generalized eigenvectors satisfy the equation (A − 4I)~v2 = ~v1 , which becomes the augmented matrix
equation

 

..
..
−1
−1
.
−1
1
1
.
1

∼
 ⇒ v1 = −v2 + 1 ⇒ ~v2 = 0 .
.
.
1
1
1 .. 1
0 0 .. 0
This gives the solution
~x(t) = c1 e
4t
−1
1
+ c2 e
4t
0
1
−1
2
+t
−1
1
.
5. Find a fundamental matrix for the system
~x′ =
2
−4
~x
and then use it to find the solution satisfying the initial value
2
~x(0) =
.
−1
The eigenvalues of the coefficient matrix satisfy
2−λ
−1 = (2 − λ)2 − 4 ⇒ (2 − λ)2 = 4 ⇒ 2 − λ = ±2 ⇒ λ = 2 ± 2 = 4 or 0.
0 = −4
2−λ To find an eigenvector corresponding to λ = 4, we calculate
1
1
−2 −1
1 21
.
⇒ v1 = − v2 ⇒ ~v1 =
∼
−2
0 0
−4 −2
2
Corresponding to λ = 0, we have
1
1
2 −1
1 − 12
⇒ v1 = v2 ⇒ ~v2 =
.
∼
2
−4 2
0 0
2
2
This gives us fundamental matrix
with
Φ(0) =
To find Φ−1 (0), we solve the augmented

 
..
1
1
.
1
0

∼ 1 1
..
−2 2 . 0 1
0 4
e4t
−2e4t
Φ(t) =
1
2
1 1
−2 2
matrix system
 
..
. 1 0   1 0
∼
..
0 1
. 2 1
,
.
..
.
..
.

− 41 
−1
⇒ Φ (0) =
1
2
1
2
1
4
1
2
1
2
− 14
2
−1
1
4
.
Finally, we have that
~x(t) = Φ(t)Φ
−1
1 4t 1
1
− 41
2
e +
e4t
1
2
= 2 4t 2
(0)~x0 =
1
1
−1
−e
+1
−2e4t 2
2
4
5 4t 3 e4t + 1 + 41 e4t − 41
4e + 4
=
.
=
−2e4t + 2 − 21 e4t − 21
− 25 e4t + 32
− 41 e4t + 14
1 4t
1
2e + 2
6. Use the method of undetermined coefficients to find a particular solution to the nonhomogeneous problem
x′ = x + 2y + 3,
y ′ = 2x + y − 2.
As a matrix equation, the system is
x′
y′
=
1
2
2
1
x
y
+
3
−2
.
The eigenvalues of the coefficient matrix satisfy
1−λ
2 0=
= (1 − λ)2 − 4 ⇒ (1 − λ)2 = 4 ⇒ 1 − λ = ±2 ⇒ λ = 1 ± 2 = 3 or − 1.
2
1−λ Thus, we take as our guess
~xp (t) =
a1
a2
+
=
b1
b2
b1
b2
t.
Differentiating with respect to t, we have
~x′p (t)
.
Also, we have that
A~xp + f~(t) =
1 2
2 1
a1
a2
+
1 2
2 1
b1
b2
t+
3
−2
=
a1 + 2a2 + 3
2a1 + a2 − 2
+
b1 + 2b2
2b1 + b2
t.
Equating the ”coefficients” of the various powers of t gives us
a1 + a2 + 3 = b 1
b + 2b2 = 0
and 1
.
2b1 + b2 = 0
2a1 + a2 − 2 = b2
Since the coefficient matrix is nonsingular, the second set of equations implies that b1 = b2 = 0. In this case, the
first set of equations corresponds to the augmented matrix equation

 
 

..
..
..
7
3 .
 1 2 . −3  ∼  1 2 . −3  ∼  1 0 .
.
.
.
2 1 .. 2
0 −3 .. 8
0 1 .. − 83
3
This gives us that
~xp (t) =
7
3
− 38
.
7. Without completely solving the system below, determine the type (node, center, etc.) of the fixed point at the
origin, and say whether it is asymptotically stable, stable, or unstable.
dx
= y,
dt
dy
= −5x − 4y
dt
As a matrix equation, the system is
x′
y′
The eigenvalues of the coefficient matrix satisfy
−λ
1
0 = −5 −4 − λ
=
0
1
−5 −4
x
y
.
= −λ(−4 − λ) + 5 = λ2 + 4λ + 5
The quadratic formula tells us that the eigenvalues are
p
√
−4 ± 16 − 4(1)(5)
−4 ± −4
λ=
=
= −2 ± i.
2
2
Since the coefficient matrix has complex conjugate eigenvalues with negative real part, the fixed point at the
origin is a stable spiral, which is asymptotically stable.
8. Given the nonlinear system
dy
dx
= x20 + y,
= − sin 5x − y 8 − 4y,
dt
dt
find the linearization about the fixed point at the origin and classify the fixed point of the linearization as in
problem 7.
The linearization of the system will have coefficient matrix J(0, 0), where
20x19
1
J(x, y) =
.
−5 cos 5x −8y 7 − 4
This gives that
J(0, 0) =
0
1
−5 −4
.
Since this is that same matrix as in problem 7, we have that the linearization has a stable spiral fixed point at
the origin, and this is of course an asymptotically stable fixed point.
4