Math 148 Final Exam Practice Problems
This review should not be used as your sole source for preparation for the exam. You
should also re-work all examples given in lecture, all homework problems, all lab assignment
problems, and all quiz problems.
1. Solve the initial-value problem
dx1
= 4x1 − 7x2 ,
dt
dx2
= 2x2 − 5x2 ,
dt
where x1 (0) = 13 and x2 (0) = 3.
2. Consider the linear system
dx1
= −x1 + 3x2 ,
dt
dx2
= 2x1 − 5x2 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
3. Consider the linear system
dx1
= 2x1 − 3x2 ,
dt
dx2
= 2x1 − x2 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
4. Consider the nonlinear system
dx1
= 3x1 x2 − x1 + x2 ,
dt
dx2
= x22 − x1 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
5. Find all equilibria of the nonlinear system
dx1
= 4x1 (1 − x1 ) − 2x1 x2 ,
dt
dx2
= x2 (2 − x2 ) − x2 .
dt
Determine the stability of each equilibrium.
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6. The nullclines of a general nonlinear system
dx1
= f1 (x1 , x2 )
dt
dx2
= f2 (x1 , x2 )
dt
are illustrated below.
x2
f2  0
f1  0
x1
Assume that both x1 and x2 increase close to the origin and that f1 and f2 change
sign when crossing their respective nullclines. Find the sign structure of the Jacobian
matrix at the equilibrium and determine its stability, if possible.
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Solutions
1. Solve the initial-value problem
dx1
= 4x1 − 7x2 ,
dt
dx2
= 2x2 − 5x2 ,
dt
where x1 (0) = 13 and x2 (0) = 3.
The system can be written in matrix form as
d~x
4 −7
=
~x.
2 −5
dt
To find the eigenvalues of the coefficient matrix, set
4−λ
−7 =
2
−5 − λ (4 − λ)(−5 − λ) + 14 =
λ2 + λ − 6 =
(λ + 3)(λ − 2) =
0
0
0
0.
The eigenvalues are λ1 = −3 and λ2 = 2.
If λ = −3, then
7 −7 v1
7v1 − 7v2
=
= ~0.
2 −2 v2
2v1 − 2v2
It follows that v1 = v2 and so ~v1 = h1, 1i is an eigenvector associated with λ1 = −3.
If λ = 2, then
2 −7 v1
2v1 − 7v2
=
= ~0.
2 −7 v2
2v1 − 7v2
It follows that 2v1 = 7v2 and so ~v2 = h7, 2i is an eigenvector associated with λ2 = 2.
Therefore, the general solution of the initial-value problem is
~x(t) = c1 eλ1 t v~1 + c2 eλ2 t~v2
−3t 1
2t 7
~x(t) = c1 e
+ c2 e
1
2
−3t
2t
c1 e + 7c2 e
~x(t) =
.
c1 e−3t + 2c2 e2t
Applying the initial condition ~x(0) = h13, 3i, we obtain the linear system
c1 + 7c2 = 13
c1 + 2c2 = 3.
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It follows that c1 = −1 and c2 = 2. Therefore, the solution is
−3t
−e + 14e2t
~x(t) =
−e−3t + 4e2t
2. Consider the linear system
dx1
= −x1 + 3x2 ,
dt
dx2
= 2x1 − 5x2 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
The system can be written in matrix form as
d~x
−1 3
=
~x.
2 −5
dt
To find the eigenvalues of the coefficient matrix, set
−1 − λ
3
= 0
2
−5 − λ (−1 − λ)(−5 − λ) − 6 = 0
λ2 + 6λ − 1 = 0.
Using the quadratic formula, the eigenvalues are
1
√ √
√
1
λ=
−6 ± 36 + 4 =
−6 ± 2 10 = −3 ± 10.
2
2
The eigenvalues are real with opposite signs. Therefore, (0, 0) is a saddle point (unstable).
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3. Consider the linear system
dx1
= 2x1 − 3x2 ,
dt
dx2
= 2x1 − x2 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
The system can be written in matrix form as
d~x
2 −3
=
~x.
2 −1
dt
To find the eigenvalues of the coefficient matrix, set
2−λ
−3
= 0
2
−1 − λ (2 − λ)(−1 − λ) + 6 = 0
λ2 − λ + 4 = 0.
Using the quadratic formula, the eigenvalues are
λ=
√ √
1
1
1 ± 3i .
1± 1−4 =
2
2
The eigenvalues are complex conjugates with positive real part. Therefore, (0, 0) is an
unstable spiral.
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4. Consider the nonlinear system
dx1
= 3x1 x2 − x1 + x2 ,
dt
dx2
= x22 − x1 .
dt
Determine the stability of (0, 0) and classify the equilibrium.
The Jacobian matrix for the system is
3x2 − 1 3x1 + 1
J(x1 , x2 ) =
.
−1
2x2
At the equilibrium (0, 0), we have
−1 1
J(0, 0) =
.
−1 0
To find the eigenvalues, set
−1 − λ 1
−1
−λ
= 0
λ2 + λ + 1 = 0.
Using the quadratic formula, the eigenvalues are
√ √
1
1
−1 ± 1 − 4 =
−1 ± 3i .
λ=
2
2
The eigenvalues are complex with negative real part. Therefore, (0, 0) is a stable spiral.
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5. Find all equilibria of the nonlinear system
dx1
= 4x1 (1 − x1 ) − 2x1 x2 ,
dt
dx2
= x2 (2 − x2 ) − x2 .
dt
Determine the stability of each equilibrium.
To find the equilibria of the system, let
4x1 (1 − x1 ) − 2x1 x2 = 0
x2 (2 − x2 ) − x2 = 0.
Simplifying each equation, we obtain
x1 [4(1 − x1 ) − 2x2 ] = 0
x2 (1 − x2 ) = 0.
Using the first equation, we find that x1 = 0 or 4x1 + 2x2 = 4. Similarly, the second
equation yields x2 = 0 or x2 = 1. Therefore, the equilibria are (0, 0), (0, 1), (1, 0), and
(1/2, 1).
The Jacobian matrix for the system is
4 − 8x1 − 2x2 −2x1
J(x1 , x2 ) =
.
0
1 − 2x2
At the equilibrium (0, 0),
4 0
J(0, 0) =
.
0 1
The eigenvalues are λ1 = 4 and λ2 = 1. Therefore, (0, 0) is an unstable node.
At the equilibrium (0, 1),
2 0
J(0, 1) =
.
0 −1
The eigenvalues are λ1 = 2 and λ2 = −1. Therefore, (0, 1) is a saddle point (unstable).
At the equilibrium (1, 0),
−4 −2
J(1, 0) =
.
0
1
The eigenvalues are λ1 = −4 and λ2 = 1. Therefore, (1, 0) is a saddle point (unstable).
At the equilibrium (1/2, 1),
J
1
−2 −1
,1 =
.
0 −1
2
The eigenvalues are λ1 = −2 and λ2 = −1. Therefore, (1/2, 1) is a stable node.
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6. The nullclines of a general nonlinear system
dx1
= f1 (x1 , x2 )
dt
dx2
= f2 (x1 , x2 )
dt
are illustrated below.
x2
f2  0
f1  0
x1
Assume that both x1 and x2 increase close to the origin and that f1 and f2 change
sign when crossing their respective nullclines. Find the sign structure of the Jacobian
matrix at the equilibrium and determine its stability, if possible.
Since x1 and x2 are increasing close to the origin, f1 > 0 and f2 > 0 near the origin. Using the assumption that f1 and f2 change sign when crossing their respective
nullclines, we obtain the following:
x2
f2  0
f1  0
f2  0
f1  0
f2  0
f1  0
f2  0
f1  0
f2  0
f1  0
x1
Therefore, the sign structure of the Jacobian matrix at the equilibrium (x∗1 , x∗2 ) is given
by
+ −
∗
∗
J(x1 , x2 ) =
.
− −
The sign of trJ(x∗1 , x∗2 ) cannot be determined, and det J(x∗1 , x∗2 ) < 0. By the RouthHurwitz Criterion, the equilibrium (x∗1 , x∗2 ) is unstable.
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