2 First-Order Differential Equations
2
First-Order Differential Equations
Exercises 2.1
1.
y
2.
y y
x t
x t
3.
y
4.
y y
x
x t
5.
y
6.
y
x
x
7.
y
8.
y
x
x
17
Exercises 2.1
9.
y
10.
y
x
x
11.
y
12.
y
x
x
13.
y
14.
y
x
x
15. Writing the differential equation in the form dy/dx = y(1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown below.
-1 y 5 4 3 2 1 1 2 x
0
1 y (a)
(b)
1
-2
-1
1
2
x
(c)
-2 -1
y
(d)
1 2 x
y 1 -1 -2 -3 2 x
-1
-4 -5
18
Exercises 2.1
16. Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown below.
-1 0 1
(a)
y 5 4 3 2 1 1 2 x
(b)
y
1
-2
-1
1
2
x
(c)
-2 -1
y
(d)
-2 1 2 x -1
y x -1 -2 -3
-1
-4 -5
17. Solving y 2 − 3y = y(y − 3) = 0 we obtain the critical points 0 and 3.
0 3
From the phase portrait we see that 0 is asymptotically stable and 3 is unstable. 18. Solving y 2 − y 3 = y 2 (1 − y) = 0 we obtain the critical points 0 and 1.
0 1
From the phase portrait we see that 1 is asymptotically stable and 0 is semi-stable. 19. Solving (y − 2)2 = 0 we obtain the critical point 2.
2
From the phase portrait we see that 2 is semi-stable. 20. Solving 10 + 3y − y 2 = (5 − y)(2 + y) = 0 we obtain the critical points −2 and 5.
-2 5
From the phase portrait we see that 5 is asymptotically stable and −2 is unstable. 21. Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and 2.
-2 0 2
From the phase portrait we see that 2 is asymptotically stable, 0 is semi-stable, and −2 is unstable. 22. Solving y(2 − y)(4 − y) = 0 we obtain the critical points 0, 2, and 4.
0 2 4
From the phase portrait we see that 2 is asymptotically stable and 0
First-Order Differential Equations
Exercises 2.1
1.
y
2.
y y
x t
x t
3.
y
4.
y y
x
x t
5.
y
6.
y
x
x
7.
y
8.
y
x
x
17
Exercises 2.1
9.
y
10.
y
x
x
11.
y
12.
y
x
x
13.
y
14.
y
x
x
15. Writing the differential equation in the form dy/dx = y(1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown below.
-1 y 5 4 3 2 1 1 2 x
0
1 y (a)
(b)
1
-2
-1
1
2
x
(c)
-2 -1
y
(d)
1 2 x
y 1 -1 -2 -3 2 x
-1
-4 -5
18
Exercises 2.1
16. Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown below.
-1 0 1
(a)
y 5 4 3 2 1 1 2 x
(b)
y
1
-2
-1
1
2
x
(c)
-2 -1
y
(d)
-2 1 2 x -1
y x -1 -2 -3
-1
-4 -5
17. Solving y 2 − 3y = y(y − 3) = 0 we obtain the critical points 0 and 3.
0 3
From the phase portrait we see that 0 is asymptotically stable and 3 is unstable. 18. Solving y 2 − y 3 = y 2 (1 − y) = 0 we obtain the critical points 0 and 1.
0 1
From the phase portrait we see that 1 is asymptotically stable and 0 is semi-stable. 19. Solving (y − 2)2 = 0 we obtain the critical point 2.
2
From the phase portrait we see that 2 is semi-stable. 20. Solving 10 + 3y − y 2 = (5 − y)(2 + y) = 0 we obtain the critical points −2 and 5.
-2 5
From the phase portrait we see that 5 is asymptotically stable and −2 is unstable. 21. Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and 2.
-2 0 2
From the phase portrait we see that 2 is asymptotically stable, 0 is semi-stable, and −2 is unstable. 22. Solving y(2 − y)(4 − y) = 0 we obtain the critical points 0, 2, and 4.
0 2 4
From the phase portrait we see that 2 is asymptotically stable and 0