### Solutions - Columbia Math

```Ordinary Differential Equations, Spring 2015
Solutions to Homework 10
Maximal grade for HW10: 100 points
Section 9.2: 24. Solve the system of equations dx/dt = y, dy/dt =
−x + x3 /6 and sketch the phase portrait.
Solution: We have dy/dx = (−x + x3 /6)/y, so
Z
Z
y2
ydy = (−x + x3 /6)dx,
= −x2 /2 + x4 /24 + C,
2
so
p
−x2 + x4 /12 + 2C.
√
√
The system has singular points (0, 0), ( 6, 0), (− 6, 0). The linearization
has the form
0
1
A(x, y) =
,
−1 + x2 /3 0
y=±
so at (0, 0) we get
A(0, 0) =
0 1
, λ2 + 1 = 0,
−1 0
√
so (0, 0) is a center; at (± 6, 0) one has
√
0 1
A(± 6, 0) =
, λ2 − 2 = 0,
2 0
√
√
so both points ( 6, 0), (− 6, 0) are saddles. The phase portrait is shown in
Figure 1.
Section 9.3: 6. Find all singular points and sketch the phase portrait
for the system
dx/dt = x − x2 − xy, dy/dt = 3y − xy − 2y 2 .
Solution: We have
dx/dt = x(1 − x − y), dy/dt = y(3 − x − 2y).
1
There are four different cases for singular points: (a) x = 0, y = 0; (b)
x = 0, 3 − x − 2y = 0, so y = 3/2; (c) y = 0, 1 − x − y = 0, so x = 1; (d)
1 − x − y = 0, 3 − x − 2y = 0, so y = 2, x = −1. The singular points are
(0, 0), (0, 3/2), (1, 0), (−1, 2). The linearization has the form
1 − 2x − y
−x
A(x, y) =
,
−y
3 − x − 4y
so at (0, 0) we get
1 0
A(0, 0) =
, (λ − 1)(λ − 3) = 0,
0 3
so the eigenvalues are λ1 = 1, λ2 = 3 and (0, 0) is an unstable node; at
(0, 3/2) we get
−1/2 0
A(0, 3/2) =
, (λ + 1/2)(λ + 3) = 0,
−3/2 −3
so the eigenvalues are λ1 = −1/2, λ2 = −3 and (0, 3/2) is a stable node; at
(1, 0) we get
−1 −1
A(1, 0) =
, (λ + 1)(λ − 2) = 0,
0
2
so the eigenvalues are λ1 = −1, λ2 = 2 and (1, 0) is a saddle point; at (−1, 2)
we get
1
1
A(−1, 2) =
, (λ − 1)(λ + 4) + 2 = λ2 + 3λ − 2 = 0,
−2 −4
√
so the eigenvalues are λ1,2 = −3±2 17 and (−1, 2) is a saddle point. The phase
portrait is shown in Figure 2.
7. Find all singular points and sketch the phase portrait for the system
dx/dt = 1 − y, dy/dt = x2 − y 2 .
Solution: We have y = 1, x2 = y 2 = 1, so x = ±1. The linearization
has the form
0 −1
A(x, y) =
,
2x −2y
2
so at (−1, 1) we get
A(−1, 1) =
0 −1
, λ(λ + 2) − 2 = λ2 + 2λ − 2 = 0,
−2 −2
√
√
the eigenvalues are λ1,2 = −2±2 12 = −1 ± 3, so (−1, 1) is a saddle point; at
(1, 1) we get
0 −1
A(1, 1) =
, λ(λ + 2) + 2 = λ2 + 2λ + 2 = 0,
2 −2
√
the eigenvalues are λ1,2 = −2±2 −4 = −1±i, so (−1, 1) is a stable spiral point.
The phase portrait is shown in Figure 3.
19. Solve the nonlinear system and its linearization at (0, 0) and sketch
the corresponding phase portraits:
dx/dt = y, dy/dt = x + 2x3 .
Solution: The linearized system has the form x0 = y, y 0 = x, so dy/dx =
x/y,
Z
Z
ydy =
xdx, y 2 /2 = x2 /2 + C.
Similarly, for the nonlinear system we have
Z
Z
ydy = (x + 2x3 )dx, y 2 /2 = x2 /2 + x4 /2 + C.
The matrix of the linearlized system equals
0 1
,
1 0
it has eigenvalues λ1 = 1, λ2 = −1, and eigenvectors v1 = (1, 1) and v2 =
(−1, 1) (so this is a saddle point) The phase portraits for both systems are
shown in Figure 4.
20. Solve the nonlinear system and its linearization at (0, 0) and sketch
the corresponding phase portraits:
dx/dt = x, dy/dt = −2y + x3 .
3
Solution: The linearized system has the form x0 = x, y 0 = −2y, so we
get x = Aet , y = Be−2t , and (0, 0) is a saddle point. In the nonlinear system,
we can still write x = Aet , so y 0 + 2y = A3 e3t . We can look for a special
solution of this nonhomogeneous linear equation in the form y = Be3t , then
y 0 = 3Be3t , and
y 0 + 2y = 3Be3t + 2Be3t = A3 e3t ⇒ B = A3 /5.
3
Therefore the general solution is y(t) = A5 e3t + Ce−2t . For C = 0, we get
y = x3 /5. The phase portraits for both systems are shown in Figure 5.
4
4
2
0
-2
-4
-4
-2
0
2
Figure 1: Graph for problem 24
5
4
4
2
0
-2
-4
-4
-2
0
2
Figure 2: Graph for problem 6
6
4
4
2
0
-2
-4
-4
-2
0
2
Figure 3: Graph for problem 7
7
4
4
4
2
2
0
0
-2
-2
-4
-4
-4
-2
0
2
4
-4
-2
0
2
4
0
2
4
Figure 4: Graph for problem 19
4
4
2
2
0
0
-2
-2
-4
-4
-4
-2
0
2
4
-4
-2
Figure 5: Graph for problem 20
8
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