Math 211
Homework #10
March 22, 2001
9.2.3.
Find the general solution of the system y = Ay, where
−5 1
A=
−2 −2
Answer: The matrix A has characteristic polynomial p(λ) = λ2 + 7λ + 12 =
)λ + 3)(λ + 4). Hence the eigenvalues are λ1 = −3 and λ2 = −4.
For λ1 = −3 we have
−2 1
A − λ1 I = A + 3I =
.
−2 1
We see that the eigenspace has dimension 1 and is spanned by v1 = (1, 2)T . Thus
we have the solution y1 (t) = eλ1 t v1 = e−3t (1, 2)T .
For λ2 = −4 we have
−1 1
A − λ1 I = A + 4I =
.
−2 2
We see that the eigenspace has dimension 1 and is spanned by v2 = (1, 1)T . Thus
we have the solution y2 (t) = eλ2 t v2 = e−4t (1, 1)T . Thus, the general solution
1
1
+ C2 e−3t
.
y(t) = C1 e−4t
1
2
9.2.9.
Find the solution of the initial value problem for system y = Ay with
−5 1
0
A=
and y(0) =
.
−2 −2
−1
Answer: The system in Exercise 9.2.3 had general solution
1
1
y(t) = C1 e−4t
+ C2 e−3t
.
1
2
Thus, if y(0) = (0, −1)T , then
0
1
1
1
= C1
+ C2
=
−1
1
2
1
The augmented matrix reduces.
1 1
1 2
0
−1
→
1
0
0
1
1
2
1
−1
Therefore, C1 = 1 and C2 = −1, giving particular solution
1
1
y(t) = e−4t
− e−3t
.
1
2
C1
C2
.
9.2.26.
Find a fundamental set of real solutions of the system y = Ay, where
0
4
A=
−2 −4
Answer: The characteristic polynomial is p(λ) = λ2 − 4λ + 8, which has complex
roots λ = −2 ± 2i. For the eigenvalue λ = 2 + 2i, we have the eigenvector w =
(−1 − i, 1)T . The corresponding exponential solution is
−1 − i
z(t) = e(−2+2i)t
1
−1
−1
= e−2t [cos 2t + i sin 2t]
+i
1
0
−1
−1
= e−2t cos 2t ·
− sin 2t ·
1
0
−1
−1
+ ie−2t cos 2t ·
+ sin 2t ·
.
0
1
The real and imaginary parts of z,
y1 (t) = e−2t cos 2t
y2 (t) = e−2t cos 2t
·
·
−1
1
−1
0
− sin 2t ·
+ sin 2t ·
−1
0
−1
1
are a fundamental set of solutions.
9.2.32.
Find the solution of the initial value problem for system y = Ay with
0
4
−1
A=
and y(0) =
−2 −4
2
Answer: A fundamental set of solutions was found in Exercise 9.2.26, so the solution
has the form y(t) = C1 y1 (t) + C2 y2 (t), where
−1
−1
− sin 2t ·
y1 (t) = e−2t cos 2t ·
1
0
−1
−1
y2 (t) = e−2t cos 2t ·
+ sin 2t ·
0
1
At t = 0 we have
−1
2
−1
−1
+ C2
= y(0) = C1
1
0
−1 −1
C1
=
.
1
0
C2
This system can be readily solved, getting C1 = 2 and C2 = −1. Hence the solution
is
y(t) = 2y1 (t) − y2 (t).
9.2.39.
Find the general solution of the system y = Ay, where
3 −1
A=
1 1
Answer: The matrix A has one eigenvalue, λ = 2. However, the nullspace of
1 −1
A − 2I =
1 −1
is generated by the single eigenvector, v1 = (1, 1)T , with corresponding solution
1
y1 (t) = e2t
.
1
To find another solution, we need to find a vector v2 which satisfies (A − 2I )v2 = v1 .
Choose w = (1, 0), which is independent of v1 , and note that
1 −1
1
1
(A − 2I )w =
=
= v1 .
1 −1
0
0
Thus, choose v2 = w = (1, 0)T . Our second solution is
y2 (t) = e2t (v2 + tv1 )
1
1
2t
=e
+t
.
0
1
Thus, the general solution can be written
1
1
1
y(t) = C1 e2t
+ C2 e2t
+t
1
0
1
1
1
= e2t (C1 + C2 t)
+ C2
0
1
9.2.45.
Find the solution of the initial value problem for system y = Ay where
3 −1
2
A=
and y(0) =
.
1 1
−1
Answer: From Exercise 9.2.39,
1
1
2t
y(t) = e (C1 + C2 t)
.
+ C2
0
1
If y(0) = (2, −1)T , then
2
−1
The augmented matrix reduces,
1 1
1 0
1
1
= C1
+ C2
.
1
0
2
−1
→
1
0
0
1
−1
3
,
and C1 = −1 and C2 = 3. Thus, the particular solution is
1
1
2t
+3
y(t) = e (−1 + 3t)
1
0
2 + 3t
2t
=e
.
−1 + 3t
9.3.30.
Classify the equilibrium point of each of the system y = Ay where
−7 10
A=
−5 8
based on the position of (T , D) in the trace-determinant plane. Verify your result by
creating a phase portrait with your numerical solver.
Answer: For
A=
−7
−5
10
8
we have D = det(A) = −6 < 0. Hence the origin is a saddle point. The following
phase plane diagram is typical of a saddle.
y
5
0
−5
−5
9.3.31.
0
x
5
Classify the equilibrium point of each of the system y = Ay where
4
3
A=
−15 −8
based on the position of (T , D) in the trace-determinant plane. Verify your result by
creating a phase portrait with your numerical solver.
Answer: If
A=
4
−15
3
−8
,
then the trace is T = −4 and the determinant is D = 13 > 0. Further, T 2 − 4D =
(−4)2 − 4(13) = −36 < 0, so the equilibrium point at the origin is a spiral sink.
Further,
4
−15
3
−8
1
4
=
,
0
−15
so the motion is counterclockwise. A hand sketch follows.
The phase portrait, drawn in a numerical solver, follows.
10
y
5
0
−5
−10
−10
9.3.32.
−5
0
x
5
10
Classify the equilibrium point of each of the system y = Ay where
3
2
A=
−4 −1
based on the position of (T , D) in the trace-determinant plane. Verify your result by
creating a phase portrait with your numerical solver.
Answer: For
A=
3
−4
2
−1
we have T = tr(A) = 2, and D = det(A) = 5. The discriminant is T 2 −4D = −16.
Hence the equilibrium point is a spiral source. The phase portrait, drawn in a numerical
solver, follows.
y
5
0
−5
−5
0
x
5
Degenerate nodes. Degenerate nodal sinks are equilibrium points characterized by
the fact that all solutions tend to the equilibrium point as t → ∞, all tangent to the
same line. The solutions of degenerate nodal sources tend to the equilibrium point as
t → −∞, all tangent to the same line. Exercises 9.3.36–9.3.39 discuss degenerate
nodes.
9.3.36. The system
y =
1
−1
4
−3
y
has a repeated eigenvalue, λ = −1, but only one eigenvector, v1 = (2, −1)T .
(a)
Explain why this system lies on the boundary separating nodal sinks from spiral sinks
in the trace-determinant plane.
Answer: For
A=
1
−1
4
−3
we have T = tr(A) = −2 and D = det(A) = 1. Since the discriminant T 2 −4D = 0
the point (T , D) lies on the parabola that divides nodal sinks from spiral sinks in the
trace-determinant plane.
(b) The general solution (see Example 2.45 in section 9.2) can be written
0
2
.
+ C2
y(t) = e−t (C1 + C2 t)
1/2
−1
Predict the behavior of the solution in the phase plane as t → ∞ and as t → −∞.
Answer: The general solution can be written
0
2
.
+ C2
y(t) = e−t (C1 + C2 t)
1/2
−1
We will write
y(t) = te−t
C1
t
2
−1
+
C2
t
0
1/2
+ C1
2
−1
.
Since the factor te−t → 0 as t → ∞ we see that y(t) → 0 as t → ∞. If we now
look at the other factor, we see that it approaches C1 (2, −1)T as t → ∞. Hence
we see that y(t) → 0 and becomes tangent to the half-line through C1 (2, −1)T as
t → ∞.
Next as t → −∞, the factor te−t → −∞, and the other factor again approaches
C1 (2, −1)T . Hence we see that as t → −∞, y(t) → ∞, and in the process becomes
parallel to the half-line generated by −C1 (2, −1)T in the process.
Use your numerical solver to sketch the half-line solutions. Then sketch exactly one
solution in each region separated by the half-line solutions. Explain how the behavior
you see agrees with your findings in part (b).
Answer: The following figure shows the half-line solutions and one other in each
sector. The solutions clearly exhibit the behavior predicted in part (a).
5
y
(c)
0
−5
−5
0
x
5
Let A be a real 2 × 2 matrix with one eigenvalue λ and suppose that the eigenspace
of λ has dimension 1. The matrix in Exercise 9.3.36 is an example. According to
Theorem 2.42 in Section 9.2, a fundamental set of solutions for the system y = Ay
is given by
y1 (t) = eλt v1
and
λt
y2 (t) = e [v2 + tv1 ],
where v1 is a nonzero eigenvector, and v2 satisfies (A − λI )v2 = v1 . The general
solution can be written as
y(t) = eλt [(C1 + C2 t)v1 + C2 v2 ],
9.3.37. Assume that the eigenvalue λ is negative.
(a)
Describe the exponential solutions.
Answer: There is one exponential solution, eλt v1 . Because λ < 0, this solution
decays to the equilibrium point at the origin along the half line generated by Cv1 .
(b)
Describe the behavior of the general solution as t → ∞.
Answer: The general solution is
y(t) = eλt [(C1 + C2 t)v1 + C2 v2 ].
Because λ < 0, the terms eλt (C1 + C2 t)v1 and C2 eλt v2 both decay to zero. However,
the first term is larger for large values of t. Thus, as t → ∞, y(t) ≈ eλt (C1 +C2 t)v1 ≈
teλt C2 v1 , which implies that the solution approaches zero tangent to the half-line
generated by C2 v1 .
(c)
Describe the behavior of the general solution as t → −∞.
Answer: Because λ < 0, the terms eλt (C1 + C2 t)v1 and C2 eλt v2 get infinitely large
in magnitude as t → −∞. However, the first term is larger in magnitude for negative
values of t that are large in magnitude. Thus, as t → −∞, y(t) ≈ eλt (C1 + C2 t)v1 ≈
teλt C2 v1 ,. Since t < 0 as t → −∞, this implies that the solution eventually parallels
the half-line generated by −C2 v1 .
(d)
Is the equilibrium point at the origin a degenerate nodal sink or source?
Answer: Degenerate nodal sink.
9.3.38.
Redo Exercise 9.3.37 under the assumption that the eigenvalue λ is positive.
Answer: In general everything moves in the opposite direction in comparison to the
situation in Exercise 9.3.37.
As t → ∞ the exponential solution tends to ∞ along the half-line generated by C1 v1 .
As t → ∞ the general solution tends to ∞ and becomes parallel to the half-line
generated by C2 v1 .
As t → −∞ the general solution tends to 0 tangent to the half-line generated by
−C2 v1 .
The origin is a degenerate nodal source.
9.3.39. Where do linear degenerate sources and sinks fit on the trace-determinant plane?
Answer: Because the linear degenerate nodal sources and sinks have only one eigenvalue, and because the eigenvalues are given by
√
T ± T 2 − 4D
,
λ1 , λ2 =
2
we must have T 2 − 4D = 0. Therefore, the degenerate nodal sources and sinks lie
on the parabola T 2 − 4D = 0 in the trace-determinant plane. This positioning on
the boundary between the nodal sink and sources and the spiral sinks and sources
is significant. The solutions attempt to spiral, but they cannot. The presence of the
half-line solutions prevents them from spiralling (solutions cannot cross).
9.4.8.
Find the general solution of the system
y =
Answer: For
−3
3
2
A=
we have
A − λI =
−1
3
0
0
2
0
−3
3
2
−1
3
0
0
2
0
y
−3 − λ
0
3
2−λ
2
0
−1
3
−λ
We can compute the characteristic polynomial p(λ) = det(A − λI ) by expanding
along the second column to get
p(λ) = (2 − λ) det
−3 − λ −1
2
−λ
= −(λ − 2)(λ2 + 3λ + 2)
= −(λ − 2)(λ + 1)(λ + 2).
Hence the eigenvalues are λ1 = −2, λ2 = −1, and λ3 = 2.
For λ1 = −2 we have
A − λ1 I = A + 2I =
−1
3
2
0
4
0
−1
3
2
The nullspace is generated by the vector v1 = (−1, 0, 1)T .
For λ2 = −1 we have
A − λ2 I = A + I =
−2
3
2
0
3
0
−1
3
1
The nullspace is generated by the vector v2 = (−1, 0, 2)T .
For λ3 = 2 we have
A − λ3 I = A − 2I =
−5
3
2
0
0
0
−1
3
−2
The nullspace is generated by the vector v3 = (0, 1, 0)T .
Thus we have three exponential solutions:
−1
y1 (t) = e v1 = e
0
1
−1
λ2 t
−t
y2 (t) = e v2 = e
0
2
0
y3 (t) = eλ3 t v3 = e2t 1
0
λ1 t
−2t
Since the three eigenvalues are distinct, these solutions are linearly independent, and
form a fundamental system of solutions. The general solution is
y(t) = C1 y1 (t) + C2 y2 (t) + C3 y3 (t).
9.4.14.
Find the solution of the initial value problem for the system in Exercise 9.4.8 with
initial value y(0) = (1, −1, 2)T .
Answer: The solution has the form
y(t) = C1 y1 (t) + C2 y2 (t) + C3 y3 (t).
where y1 , y2 , and y3 are the fundamental set of solutions found in Exercise 9.4.8.
Hence we must have
1
−1 = y(0)
2
= C1 y1 (0) + C2 y2 (0) + C3 y3 (0)
= C1 v1 + C2 v2 + C3 v3
C1
= [v1 , v2 , v3 ] C2 ,
C3
where v1 , v2 , and v3 are the eigenvectors of A found in Exercise 9.4.8. To solve the
system we form the augmented matrix
−1 −1 0 1
[v1 , v2 , v3 , y(0)] =
0
0 1 −1
1
2 0 2
This is reduced to the row echelon form
1 1 0
0 1 0
0 0 1
−1
3
−1
.
Backsolving, we find that C3 = −1, C2 = 3, and C1 = −4. Hence the solution is
y(t) = −4y1 (t) + 3y2 (t) − y3 (t).
9.4.25.
Find the general solution of
y =
−7
2
3
−13
3
8
0
0
−2
−7
2
3
−13
3
8
0
0
−2
y
Answer: In system y = Ay, where
A=
we have
A − λI =
,
−7 − λ −13
0
2
3−λ
0
3
8
−2 − λ
.
We can compute the characteristic polynomial by expanding along the third column.
We get
p(λ) = det(A − λI )
= (−2 − λ) det
−7 − λ
2
−13
3−λ
= −(λ + 2)(λ2 + 4λ + 5).
Hence we have one real eigenvalue λ1 = −2, and the quadratic λ2 + 4λ + 5 has
complex roots λ2 = −2 + i, and λ2 = −2 − i.
For the eigenvalue λ1 = −2, we look for a vector in the nullspace (eigenspace) of
A − λ1 I = A + 2I =
−5
2
3
−13
5
8
0
0
0
.
The eigenspace is generated by v1 = (0, 0, 1)T . Thus, one solution is
y1 (t) = e
−2t
0
0 .
1
For the eigenvalue λ2 = −2 + i, we look for a vector in the nullspace (eigenspace) of
A − λ1 I = A + (2 − i)I =
−5 − i
2
3
−13
5−i
8
0
0
−i
.
The eigenspace is generated by (−5 + i, 2, 3 − i)T . Thus, we have the complex
conjugate solutions
z(t) = e
(−2+i)t
−5 + i
2
3−i
and
z(t) = e
(−2−i)t
−5 − i
2
3+i
.
Using Euler’s formula, we find the real and imaginary parts of the solution z(t).
z(t) = e
=e
−2t it
e
−2t
(cos t + i sin t)
= e−2t
−5 + i
2
3−i
−5 cos t − sin t
2 cos t
3 cos t + sin t
1
+i 0
−1
cos t − 5 sin t
+ ie−2t
2 sin t
− cos t + 3 sin t
−5
2
3
Thus we have the solutions
−5 cos t − sin t
y2 (t) = Re(z(t)) = e
2 cos t
3 cos t + sin t
cos t − 5 sin t
−2t
y3 (t) = Im(z(t)) = e
2 sin t
− cos t + 3 sin t
−2t
and
The general solution is
y(t) = C1 y1 (t) + C2 y2 (t) + C3 y3 (t).
9.4.38.
Use a computer to help you find a fundamental set of solutions to the system y = Ay,
where
−7
2
−4
A = 42 −11 18
38 −10 18
Answer: Using MATLAB we proceed as follows.
A = [-7 2 -4;42 -11 18;38 -10 18];
ee = eig(A)
ee =
1
+
2i
1
2i
-2
We see that A has two complex conjugate eigenvalues 1 ± 2i and one real eigenvalue
−2. Let’s look at the real eigenvalue first.
v3 = null(A - ee(3)*eye(3),’r’)
v3 =
-1/1501199875790165
2
1
v3(1) = 0
v3 =
0
2
1
Thus we have the real solution
0
y3 (t) = e
2 .
1
Next we look at the complex eigenvalue 1 + 2i
−2t
w = null(A - ee(1)*eye(3),’r’)
w =
-1/4
+
1/4i
3/4
+
3/4i
1
w=4*w
w =
-1
+
1i
3
+
3i
4
Thus we have the eigenvector w = (−1 + i, 3 + 3i, 4)T . We also have the complex
valued solution.
−1 + i
(1+2i)t
z(t) = e
3 + 3i .
4
Expanding using Euler’s formula, we get
−1
1
t
z(t) = e [cos 2t + i sin 2t]
3 +i 3
4
0
−1
1
= et cos 2t · 3 − sin 2t · 3
4
0
1
−1
t
+ ie cos 2t · 3 + sin 2t · 3
0
4
Since the real and imaginary parts of z(t) are solutions we get two real solutions
−1
1
t
y1 (t) = Re(z(t)) = e cos 2t · 3 − sin 2t · 3
and
4
0
1
−1
t
y2 (t) = Im(z(t)) = e cos 2t · 3 + sin 2t · 3
.
0
4
The functions y1 , y2 , and y3 form a fundamental set of solutions.
9.4.43.
Use a computer to help you find a fundamental set of solutions to the system y = Ay,
where


1
4
1 −5
 −6 −10 −2 10 
A=

3
4
−1 −5
−3 −4 −1 3
Answer: Using MATLAB we proceed as follows.
A=[1 4 1 -5;-6 -10 -2 10;3 4 -1 -5;-3 -4 -1 3];
format rat
ee = eig(A)
-1
-2
-2
-2
Thus A has eigenvalues −1 and −2, with −2 repeated three times. Let’s look at the
eigenspace for −1.
null(A-ee(1)*eye(4),’r’)
ans =
-1
2
-1
1
Thus the eigenspace has dimension 1 and is spanned by the vector (−1, 2, −1, 1)T .
Hence we get the solution


−1
 2 
y1 (t) = e−t 
.
−1
1
Next we look at the eigenspace for −2.
null(A-ee(2)*eye(4),’r’)
ans =
-4/3
-1/3
1
0
0
1
0
0
V=3*ans
V =
-4
-1
3
0
0
3
0
0
5/3
0
0
1
5
0
0
3
Thus the eigenspace has dimension 3 and it is spanned by the indicated column
vectors. We get three linearly independent solutions.




 
−4
−1
5
3
0
0






−2t
−2t
y2 (t) = e−2t 
 , y3 (t) = e 
 , and y4 (t) = e   .
0
3
0
0
0
3
The functions y1 , y2 , y3 , and y4 form a fundamental set of solutions.
9.4.46.
Use a computer to help you find the solution to the system y = Ay, with the given
matrix A in Exercise 9.4.38 and the initial value y(0) = (−2, 2, 5)T .
Answer: In Exercise 9.4.38 we found the fundamental set of solutions
−1
1
t
y1 (t) = Re(z(t)) = e cos 2t · 3 − sin 2t · 3
,
4
0
1
−1
t
y2 (t) = Im(z(t)) = e cos 2t · 3 + sin 2t · 3
, and
0
4
0
−2t
y3 (t) = e
2 .
1
Our solution has the form y(t) = C1 y1 (t) + C2 y2 (t) + C3 y3 (t). At t = 0 we have
−2
−1
1
0
= y(0) = C1 3 + C2 3 + C3 2
2
5
4
0
1
−1 1 0
C1
=
3 3 2
C2 .
4 0 1
C3
We can use MATLAB to solve this system of equations
V=[-1 1 0;3 3 2;4 0 1];
y0 = [-2 2 5]’;
C = V\y0
C =
1
-1
1
Thus our solution is
y(t) = y1 (t) − y2 (t) + y3 (t).
9.4.51.
Use a computer to help you find the solution to the system y = Ay, with the given
matrix A in Exercise 9.4.43 and the initial value y(0) = (−1, 5, 2, 4)T .
Answer: In Exercise 9.4.43 we found the fundamental set of solutions




−1
−4
 2 
−2t  3 
y1 (t) = e−t 
 , y2 (t) = e 
,
−1
0
1
0


 
−1
5
0
0




−2t
y3 (t) = e−2t 
 , and y4 (t) = e   .
3
0
0
3
Our solution has the form y(t) = C1 y1 (t) + C2 y2 (t) + C3 y3 (t) + C4 y4 (t). At t = 0
we have






 


−1
−4
−1
5
−1
 2 
 3 
 0 
0
 5 
 + C2 
 + C3 
 + C4  

 = y(0) = C1 
−1
0
3
0
2
1
0
0
3
4

 
−1 −4 −1 5
C1
3
0 0   C2 
 2
=
 .
2
0
3 0
C3
1
0
0 3
C4
We can use MATLAB to solve this system as follows.
V=[-1 -4 -1 5;2 3 0 0; -1 0 3 0; 1 0 0 3];
y0=[-1 5 2 4]’;
C = V\y0
C =
1
1
1
1
Thus, C1 = C2 = C3 = C4 = 1, leading to
y(t) = y1 (t) + y2 (t) + y3 (t) + y4 (t)


−e−t
−t
−2t
 2e + 3e

=  −t
.
−e + 3e−2t
−t
−2t
e + 3e