Applied differential equations

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HW#3 - Solutions

Section 1.4

6. dw

dt = (3 −w)(w + 1), w(0) = 0, 0 ≤ t ≤ 5, ∆t = 0.5

k tk wk f(tk, wk)

0 0 0 (3 − 0)(0 + 1) = 3 1 0.5 0 + 3 × 0.5 = 1.5 (3 − 1.5)(1.5 + 1) = 3.75 2 1.0 1.5 + 3.75 × 0.5 = 3.375 (3 − 3.375)(3.375 + 1) = −1.641 3 1.5 3.375 − 1.641 × 0.5 = 2.555 (3 − 2.555)(2.555 + 1) = 1.583 4 2.0 2.555 + 1.583 × 0.5 = 3.347 (3 − 3.347)(3.347 + 1) = −1.506 5 2.5 3.347 − 1.506 × 0.5 = 2.594 (3 − 2.594)(2.594 + 1) = 1.461 6 3.0 2.594 + 1.461 × 0.5 = 3.324 (3 − 3.324)(3.324 + 1) = −1.400 7 3.5 3.324 − 1.400 × 0.5 = 2.624 (3 − 2.624)(2.624 + 1) = 1.364 8 4.0 2.624 + 1.364 × 0.5 = 3.306 (3 − 3.306)(3.306 + 1) = −1.315 9 4.5 3.306 − 1.315 × 0.5 = 2.648 (3 − 2.648)(2.648 + 1) = 1.285 10 5.0 2.648 + 1.285 × 0.5 = 3.290 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗

t 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

w(t)

0.5

1.5

2.5

3.5

Therefore, w(5) ≈ 3.290 .

8. dy

dt = e2/y, y(1) = 2, 1 ≤ t ≤ 3, ∆t = 0.5

k tk yk f(tk,yk)

0 1 2 e2/2 = e1 ≈ 2.718 1 1.5 2 + 2.718 × 0.5 = 3.359 e2/3.359 ≈ 1.814 2 2.0 3.359 + 1.814 × 0.5 = 4.266 e2/4.266 ≈ 1.5984 3 2.5 4.266 + 1.598 × 0.5 = 5.065 e2/5.065 ≈ 1.484 4 3.0 5.065 + 1.484 × 0.5 = 5.807 ∗∗∗∗∗∗∗

t 0.5 1 1.5 2 2.5 3

y(t)

Therefore, y(3) ≈ 5.807 .

Section 1.5

2. See the following figure.

t −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−1

y = 2

y = 4

y = 0

y(0) = 1

Notice that the given IC y(0) = 1 is between two equilibrium solutions y2(t) = 2 and y3(t) = 0. Since f satisfies the hypothesis of the Uniqueness Theorem in the entire ty-plane, the solution to the equation with this initial condition must stay strictly between these two equilibrium solutions. That is, if y(t) is a solution to dy/dt = f(t,y), then we can conclude that

0 < y(t) < 2 for all t .

4. See the following figure.

t −4 −3 −2 −1 1 2 3 4

−2

−1

y1(t) = −1

y2(t) = 1 + t 2

y(0) = 0

Since the given IC y(0) = 0 is between y1(t) = −1 and y2(t) = 1 + t2 and f satisfies the hypothesis of the Uniqueness Theorem in the entire ty-plane, the solution to the equation with this initial condition must stay strictly between these two solutions. Specifically, if y(t) is a solution to dy/dt = f(t,y), then y(t) is bounded below by y1(t) = −1 and bounded above by y2(t) = 1 + t

2. Hence we can conclude that

−1 < y(t) < 1 + t2 for all t .

For Exercises 6 and 8, first let us determine the equilibrium solutions for the given autonomous differential equation. To this end, setting the right-hand side of the equation equal to 0 and solving for y, we obtain

y(y − 1)(y − 3) = 0 y = 0, 1, 3.

Let us graph these equilibrium solutions on the same set of axes.

t −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−2

−1

y = 3

y = 1

y = 0

y(0) = 4

y(0) = −1

6. Since the given IC y(0) = 4 is above the equilibrium solution y = 3, the Existence and Uniqueness Theorem tells us that if y(t) is a solution to the given autonomous ODE dy/dt = y(y − 1)(y − 3) with y(0) = 4, then

y(t) > 3 for all t .

8. Since the given IC y(0) = −1 is below the equilibrium solution y = 0, the Existence and Uniqueness Theorem tells us that if y(t) is a solution to the given autonomous ODE dy/dt = y(y − 1)(y − 3) with y(0) = −1, then

y(t) < 0 for all t .

18.� (a) Since we are assuming that the (time) rate of growth of the volume of a raindrop is proportional to its surface area, we have

dt = Cs = C(4πr2) = 4πCr2, (1)

where C is a proportionality constant. Now, solving the volume equation ( i.e., v = 4

3 πr3 )

for r yields

r =

( 3v

4π

)1/3 . (2)

(Check this.) Substituting (2) into (1) and simplifying the right-hand side, we obtain

dt = 4πC

[( 3v

4π

)1/3]2

= 4πC

( 3v

4π

)2/3 = 4πC

( 3

4π

)2/3 ︸︷︷︸ Just a constant!

v2/3

= kv2/3 ,

where k ≡ 4πC (

3 4π

)2/3 is a (modified) proportionality constant, as desired.

(b) Let f(v) = kv2/3 be the right-hand side of the resulting differential equation in part (a). It is easy to see that f is continuous for all v. However, notice that

∂f

∂v =

3 kv−1/3 =

3v1/3 ,

which fails to be continuous at v = 0. This shows that the differential equation dv/dt = kv2/3 does not satisfy the hypothesis of the Uniqueness Theorem.

Section 1.6

1. It is easy to see that the equilibrium points are y = 0 and y = 2. Let f(y) = 3y(y − 2) be the right-hand side of the given differential equation. There are three regions to consider: (i) y > 2, (ii) 0 < y < 2, and (iii) y < 0.

• Region 1 (y > 2), pick y = 3. Then f(3) = 3(3)(3 − 2) = 9 > 0. So on this interval, y is increasing and thus the arrow points up.

• Region 2 (0 < y < 2), pick y = 1. Then f(1) = 3(1)(1 − 2)) = −3 < 0. So on this interval, y is decreasing and thus the arrow points down.

• Region 3 (y < 0), pick y = −1. Then f(−1) = 3(−1)(−1 − 2) = 9 > 0. So on this interval, y is increasing and thus the arrow points up.

The following is a sketch of the phase line.

2 ⇒ source

0 ⇒ sink

2. First, locate the equilibrium points. Setting the right-hand side of the given differential equation equal to 0 and solving for y yields

y2 − 4y − 12 = 0 (y − 6)(y + 2) = 0

y = 6,−2.

Let f(y) = y2 −4y−12 be the right-hand side of the given differential equation. There are three regions to consider: (i) y > 6, (ii) −2 < y < 6, and (iii) y < −2.

• Region 1 (y > 6), pick y = 7. Then f(7) = 72 − 4(7) − 12 = 9 > 0. So on this interval, y is increasing and thus the arrow points up.

• Region 2 (−2 < y < 6), pick y = 0. Then f(0) = 02 − 4(0) − 12 = −12 < 0. So on this interval, y is decreasing and thus the arrow points down.

• Region 3 (y < −2), pick y = −3. Then f(−3) = (−3)2 − 4(−3) − 12 = 9 > 0. So on this interval, y is increasing and thus the arrow points up.

The following is a sketch of the phase line.

6 ⇒ source

−2 ⇒ sink

5. Find the equilibrium points. To this end, setting the right-hand side of the given differential equation equal to 0 yields

(1 −w) sin(w) = 0. So either (i) 1−w = 0 or (ii) sin(w) = 0. Solving (i) for w yields w = 1 and solving (ii) for w yields w = 0,±π,±2π,. . . (i.e., the integral multiplies of π). Let us plot these equilibrium points on the w-line:

2π ≈ 6.28

π ≈ 3.14

1 0

−π ≈−3.14

−2π ≈−6.28

Let f(w) = (1−w) sin(w) be the right-hand side of the given differential equation. Technically, there are infinite many regions we need to consider (since there are infinitely many equilibrium points for the given differential equation) but obviously it’s impossible to sketch an infinitely long phase line so we are only going to consider the following regions: (i) w > 2π ≈ 6.28 (but < 3π ≈ 9.42), (ii) 3.14 ≈ π < w < 2π ≈ 6.28, (iii) 1 < w < π ≈ 3.14, (iv) 0 < w < 1, (v) −3.14 ≈−π < w < 0, (vi) −6.28 ≈−2π < w < −π ≈−3.14, and (vii) w < −2π ≈−6.28 (but > −3π ≈−9.42)

• Region 1 (w > 2π ≈ 6.28 (but < 3π ≈ 9.42)), pick w = 7. Then f(7) = (1 − 7) sin(7) ≈ −3.94 < 0. So on this interval, w is decreasing and thus the arrow points down.

• Region 2 (3.14 ≈ π < w < 2π ≈ 6.28), pick w = 4. Then f(4) = (1 − 4) sin(4) ≈ 2.27 > 0. So on this interval, w is increasing and thus the arrow points up.

• Region 3 (1 < w < π ≈ 3.14), pick w = 2. Then f(2) = (1 − 2) sin(2) ≈ −0.91 < 0. So on this interval, w is decreasing and thus the arrow points down.

• Region 4 (0 < w < 1), pick w = 0.5. Then f(0.5) = (1 − 0.5) sin(0.5) ≈ 0.24 > 0. So on this interval, w is increasing and thus the arrow points up.

• Region 5 (−3.14 ≈ −π < w < 0, pick w = −1. Then f(−1) = (1 − (−1)) sin(−1) ≈ −1.68 < 0. So on this interval, w is decreasing and thus the arrow points down.

• Region 6 (−6.28 ≈−2π < w < −π ≈−3.14), pick w = −4. Then f(−4) = (1−(−4)) sin(−4) ≈ 3.78 > 0. So on this interval, w is increasing and thus the arrow points up.

• Region 7 (w < −2π ≈−6.28 (but > −3π ≈−9.42), pick w = −7. Then f(−7) = (1−(−7)) sin(−7) ≈−5.26 < 0. So on this interval, w is decreasing and thus the arrow points down.

See the following phase line.

2π ⇒ sink

π ⇒ source

1 ⇒ sink 0 ⇒ source

−π ⇒ sink

−2π ⇒ source

9. Find the equilibrium points. To this end, setting the right-hand side of the given differential equation equal to 0 and solving for y yields

1 + cos(y) = 0

cos(y) = −1 y = . . . ,−3π,−π,π, 3π,. . .

(See the following graph of the right-hand side of the given differential equation f(y) = 1 + cos(y).)

y −7π/2 −3π −5π/2 −2π −3π/2 −π −π/2 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2

f(y)

−3

−2

−1

3 f(y) = 1 + cos(y)

According to the graph of f(y) = 1 + cos(y), we can see that f(y) ≥ 0 for all y and therefore, the arrows point up on each interval.

See the following phase line.

5π ⇒ node

3π ⇒ node

π ⇒ node

−π ⇒ node

−3π ⇒ node

14. See the following figure.

t −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7

−4

−3

−2

−1

(i)

(iv)

(iii)

(ii)

(0, 1) (1, 0)

(0, 6)

(0, 4)

30. First, observe that there are two equilibrium points, namely y = a and y = b (i.e., where the graph of f(y) crosses the horizontal axis) and a < b. Also observe that: (i) f(y) < 0 for y < a so the arrow points down on the phase line below y = a, (ii) f(y) > 0 for a < y < b so the arrow points up between y = a and y = b, and (iii) f(y) < 0 again for y > b so the arrow points down above y = b on the phase line. Putting all these together, we can sketch the phase line for the given autonomous differential equation as follows.

32. The following is the sketch of the phase line. (Similar argument as in Exercise 30 above.)

34. This is the reverse of Exercises 30 and 32. First, let us label the equilibrium points of the given phase line as follow (assuming that y = 0 is in the middle of the segment shown as reference):

y = 0

(This means that a,b < 0 and c,d > 0.) Based on the given phase line, we see that the solution y(t) is increasing for c < y < b and a < y < b so the graph of f(y) must stay above the horizontal axis. On the other hand, the solution y(t) is decreasing for y > d, c < y < d, and y < a so the graph of f(y) must stay below the horizontal axis. The following is a rough sketch of the graph of f(y).

f(y)

a b c d

39.� (a) Note that since we are not told whether the population increases or decreases for P between 0 and 10, there are two possible phase lines we could sketch.

Case 1. The population increases for 0 < P < 10.

P = 50

P = 10

P = 0

Case 2. The population decreases for 0 < P < 10.

P = 50

P = 10

P = 0

(b) Case 1. The population increases for 0 < P < 10.

f(P)

0 10 50

Case 2. The population decreases for 0 < P < 10.

f(P)

0 10 50

f(P) = −P(P − 10)2(P − 50) (This is just one possible answer.)

Case 2. The population decreases for 0 < P < 10.

f(P) = −P(P − 10)(P − 50) (This is just one possible answer.)