# 2 First-Order Differential Equations

**Topics:**Trigraph, Ordinary differential equation, Gh

**Pages:**93 (24718 words)

**Published:**December 27, 2012

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 and 4 are unstable.

19

Exercises 2.1

23. Solving y ln(y + 2) = 0 we obtain the critical points −1 and 0.

-2

-1

0

From the phase portrait we see that −1 is asymptotically stable and 0 is unstable. 24. Solving yey − 9y = y(ey − 9) = 0 we obtain the critical points 0 and ln 9.

0

ln 9

From the phase portrait we see that 0 is asymptotically stable and ln 9 is unstable. 25. (a) Writing the diﬀerential equation in the form dv k = dt m we see that a critical point is mg/k. mg/k

mg −v k

From the phase portrait we see that mg/k is an asymptotically stable critical point. Thus, limt→∞ v = mg/k. (b) Writing the diﬀerential equation in the form dv k = dt m we see that a critical point is mg k − v2 = k m mg −v k mg +v k

mg/k.

√ mg/k

From the phase portrait we see that limt→∞ v = mg/k. mg/k is an asymptotically stable critical point. Thus,

26. (a) From the phase portrait we see that critical points are α and β. Let X(0) = X0 . α

β

If X0 < α, we see that X → α as t → ∞. If α < X0 < β, we see that X → α as t → ∞. If X0 > β, we see that X(t) increases in an unbounded manner, but more speciﬁc behavior of X(t) as t → ∞ is not known. (b) When α = β the phase portrait is as shown.

α

If X0 < α, then X(t) → α as t → ∞. If X0 > α, then X(t) increases in an unbounded manner. This could happen in a ﬁnite amount of time. That is, the phase portrait does not indicate that X becomes unbounded as t → ∞.

20

Exercises 2.1

(c) When k = 1 and α = β the diﬀerential equation is dX/dt = (α − X)2 . Separating variables and integrating we have dX = dt (α − X)2 1 =t+c α−X 1 α−X = t+c X =α− For X(0) = α/2 we obtain X(t) = α − For X(0) = 2α we obtain X(t) = α − 1 . t − 1/α 1 . t + 2/α 1 . t+c

X

X

α −2/α α/2

t

2α α 1/α

t

For X0 > α, X(t) increases without bound up to t = 1/α. For t > 1/α, X(t) increases but X → α as t → ∞. 27. Critical points are y = 0 and y = c. y x

>

0

<

c

>

c

x t

28. Critical points are y =...

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