Even-Numbered Homework Solutions

Chapter 1

1.1

8. Using the decay-rate parameter you computed in 1.1.7, determine the time since death if:

(a) 88% of the original C-14 is still in the material

The decay-rate parameter from 1.1.7 is λ =ln2

5230. Then use the following equation to determine the time of

death:

0.88 = e−

ln2

5230t⇒ ln(0.88) = − ln2

5230t ⇒ t = 964.54 years

(b) 12% of the original C-14 is still in the material

As above:

0.12 = e−

ln2

5230t⇒ ln(0.12) = − ln2

5230t ⇒ t = 15998.01 years

14. Suppose two students memorize lists according to the model dLdt = 2(1− L).

(a) If one of the students knows one-half of the list at time t = 0 and the other knows none of the list, whichstudent is learning more rapidly at this instant?

First, determinedL

dtfor each student. Call the student who knows one-half of the list Student A and the other

Student B. For Student A,dL

dt= 2(1− 0.5) = 2(0.5) = 1. For Student B,

dL

dt= 2(1− 0) = 2(1) = 2. So,

Student B is learning more rapidly at this instant.

(b) Will the student who starts out knowing none of the list ever catch up to the student who starts out knowingone-half of the list?

Student B will never catch up to Student A with the given model.

16. Using the table given in 1.1.16:

(a) Let s(t) = s0ekt be an exponential function. Show that the graph of ln s(t) as a function of t is a line. What

is its slope and vertical intercept?

Take the natural log of both sides of the equation: ln(s(t)) = ln(s0ekt) = ln(s0) + ln(ekt) = kt+ ln(s0). This

equation is in point-slope form, so the slope is k and the vertical intercept is y = ln(s0). Since s0 = 5669,y = 8.64 is the vertical intercept.

(b) Is spending on education in the U.S. rising exponentially fast? If so, what is the growth-rate coefficient?

In the year 1910, t = 10 and s10 = 10081. Then ln(s(10)) = ln(10081) = k(10) + ln(5669). So:

10k = ln(10081)− ln(5669) ⇒ k = 0.0576.

Yes, the spending on education is rising exponentially fast, and the growth-rate coefficient is k = 0.0575. (Notethat your k may be different, depending on which st you used to determine it.)

1

18. Suppose that the growth-rate parameter k = 0.3 and the carrying capacity N = 2500 in the logistic population modelof Exercise 17. Suppose P (0) = 2500.

(a) If 100 fish are harvested each year, what does the model predict for the long-term behavior of thefish population? In other words, what does a qualitative analysis of the model yield?

If 100 fish are harvested every year, then our model isdP

dt= 0.3P

(

1− P

2500

)

− 100 = −0.00012P 2+0.3P − 100.

SetdP

dt= 0 to get critical values at P ≈ 396 and P ≈ 2104. If 0 < P < 396, then

dP

dt< 0. If 396 < P < 2104

thendP

dt> 0. If P > 2104, then

dP

dt< 0. Since P (0) = 2500, the fish population will decrease until it levels off

around 2104.

(b) If one-third of the fish are harvested each year, what does the model predict for the long-term behavior of thefish population?

If one-third of the fish are harvested every year, then our model isdP

dt= 0.3P

(

1− P

2500

)

− 1

3P . The only

nonnegative critical value is P = 0. If P > 0, thendP

dt< 0. So the population will decrease until the fish are

extinct.

1.2

6. Find the general solution of the differential equation:dy

dt= t4y.

This is a separable differential equation. So:dy

dt= t4y

⇒ 1

ydy = t4dt

⇒∫ 1

ydy =

∫

t4dt

⇒ ln(y) =1

5t5 + C, where C is a constant (and all constants are lumped together on right)

⇒ y = e1

5t5+C

⇒ y = ece1

5t5

⇒ y = Ke1

5t5 , where eC = K, a constant.

30. Solve the initial value problem:dy

dt=

t

y − t2y. y(0) = 4.

Note that this is a separable differential equation. So:dy

dt=

t

y − t2y⇒ dy

dt=

t

y(1− t2)

⇒ y dy =t

(1− t2)dt

⇒∫

y dy =∫ t

(1− t2)dt

⇒ 1

2y2 = −1

2ln(|1− t2|) + C

⇒ y2 = −ln(|1− t2|) + C

⇒ y =√

−ln(|1− t2|) + C.

The initial condition is y(0) = y0 = 4. So, 4 =√

−ln(|1− 0|) + C =√0 + C, so C = 16. Then our final solution

is y =√

−ln(|1− t2|) + 16.

2

32. Solve the given initial-value problem:dy

dt= ty2 + 2y2, y(0) = 1.

Simplify tody

dt= y2(t+ 2). Then note that this is a separable differential equation. So:

dy

dt= y2(t+ 2) ⇒ 1

y2dy = t+ 2 dt

⇒ −1

y=

1

2t2 + 2t+ C

⇒ −2

y= t2 + 4t+ C

⇒ y =−2

t2 + 4t+ C.

The initial condition is y(0) = 1. So, 1 =−2

02 + 4(0) + C3, so C = −2. Then our final solution is y =

−2

t2 + 4t− 2.

34. Solve the given initial-value problem:dy

dt=

1− y2

y, y(0) = −2.

Note that this is a separable differential equation. So:dy

dt=

1− y2

y⇒ y

1− y2dy = dt

⇒ −1

2ln(1− y2) = t+ C

⇒ ln(1− y2) = −2t+ C

⇒ 1− y2 = Ce−2t

⇒ y2 = 1− Ce−2t

The initial condition is y(0) = −2. So, (−2)2 = 4 = 1− Ce−2(0) = 1− C, so C = −3. Then our final solutionis y = −

√1 + 3e−2t.

40. (a) Suppose N = 200, k1 = 0.1, k2 = 0.1, E = 400 and C(0) = 150. What will the person’s cholesterol level beafter 2 days on this diet?

We havedC

dt= k1(C0 − C) + k2E ⇒ 1

k1(C0 − C) + k2EdC = dt

⇒ −1

k1ln |k1(C0 − C) + k2E| = t+B, where B is a constant of integration

⇒ ln |k1(C0 − C) + k2E| = −k1(t+B)

⇒ k1(C0 − C) + k2E = Be−k1t

⇒ −k1C = Be−k1t − k2E − k1C0

⇒ C(t) =−1

k1(Be−k1t − k2E − k1C0).

Now we can plug in the parameters and initial condition to find B:

C(0) = 150 =−1

0.1(B − 0.1(400)− 0.1(200)) ⇒ B = 45.

So, C(t) = −10(45e−0.1t − 60), which means that C(2) = 231.57.

(b) With the initial conditions as above, what will the person’s cholesterol level be after 5 days on this diet?C(5) = −10(45e−0.5 − 60) = 327.06.

(c) What will the person’s cholesterol level be after a long time on this diet?

Simplify C(t) to C(t) =−450

e0.1t+ 600. As t → ∞,

−450

e0.1t→ 0, so after a long time on this diet the person’s

cholesterol level will be 600.

3

(d) Let E = 100. The initial cholesterol level at the starting time of the new diet is the result of part (c). What willthe person’s cholesterol level be after 1 day on the new diet, after 5 days on the new diet, and after a very longtime on the new diet?

Find our new B:

C(0) = 600 =−1

0.1(B − 0.1(100)− 0.1(200)) ⇒ B = −30.

Then C(1) = −10(−30e−0.1t−30) = 300(e−0.1t+1) = 300(e−0.1+1) = 571.45 and C(5) = 300(e−0.5+1) = 481.96.As t → ∞, C(t) → 300.

(e) Change k2 to k2 = 0.075. With the cholesterol level from part (c), what will the person’s cholesterol level be after1 day, after 5 days, and after a very long time?

Find our new B:

C(0) = 600 =−1

0.1(B − 0.075(400)− 0.1(200)) ⇒ B = −10.

Then C(1) = 100e−0.1t + 500 = 100e−0.1 + 500 = 590.48 and C(5) = 100e−0.5 + 500 = 560.65. As t → ∞,C(t) → 500.

42. Let Q(t) represent the number of teaspoons of hot sauce at time t in minutes. Then Q(0) = 12. Note thatdQ

dt

represents the amount of hot sauce per cup at time t, sodQ

dt=

−Q

32, since

1

32of the total amount of hot sauce

will be in each cup, and there are 32 cups in 2 gallons. This is a separable differential equation:

dQ

dt=

−Q

32⇒ 1

QdQ =

−1

32dt

⇒ ln(y) =−t

32+ C

⇒ y = Ke−t

32 .

Now we utilize our initial condition: 12 = Ke0 = K. So, our equation is y = 12e−t

32 . We are trying to dilute thechili to the intended number of teaspoons, which is 4. So,

4 = 12e−t

32 ⇒ ln

(

1

3

)

=−t

32⇒ t = 35.155 minutes.

Thus, it takes approximately 35.155 minutes and 36 cups (since one cup per minute is taken from the pot) toproperly dilute the chili.

4

1.3

8.dy

dt= 2y − t

10.dy

dt= (t+ 1)y

5

12. Suppose the constant function y(t) = 2 for all t is a solution of the differential equationdy

dt= f(t, y).

(a) What does this tell you about the function f(t, y)?At y = 2, f(t, y) = 0 for all t.

(b) What does this tell you about the slope field?y = 2 has slope 0.

(c) What does this tell you about solutions with initial conditions y(0) 6= 2?Solutions with initial conditions y(0) 6= 2 will not cross the line y = 2.

16. Determine the equation that corresponds to each slope field.(a) iii (b) viii (c) v (d) vi

1.4

6.dw

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

k tk wk mk

0 0 0 7871 0.5 1.5 54152 1.0 3.38 75073 1.5 2.55 75604 2.0 3.35 63445 2.5 2.59 63446 3.0 3.32 63447 3.5 2.62 63448 4.0 3.31 63449 4.5 2.65 634410 5.0 3.29 6344

6

8.dy

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

k tk wk mk

0 1 2 7871 1.5 3.35 54152 2.0 4.27 75073 2.5 5.07 75604 3.0 5.81 6344

1.5

10. Consider the differential equationdy

dt= 2

√

|y|.

(a) Show that y(t) = 0 for all t is an equilibrium solution.dy

dt= 2

√

|y| ⇒ 2√

|y| = 0 ⇒√

|y| = 0 ⇒ y(t) = 0

(b) Find all solutions.Case 1: y > 0

y > 0 ⇒ |y| = y ⇒ dy

dt= 2

√y

⇒∫ 1

2√ydy =

∫

dt

⇒ 1

2(2)

√y = t+ C, where C is a constant

⇒ √y = t+ C

⇒ y = (t+ C)2

Case 2: y < 0

y > 0 ⇒ |y| = −y ⇒ dy

dt= 2

√−y

⇒∫ 1

2√−y

dy =∫

dt

⇒ −√−y = t+D, where D is a constant⇒ √−y = −t−D

⇒ −y = (−t−D)2 = (t+D)2

⇒ y = −(t+D)2

7

12.

(a) Show that y1(t) =1

t− 1and y2(t) =

1

t− 2are solutions of

dy

dt= −y2.

We have:d

dt

(

1

t− 1

)

=−1

(t− 1)2= −

(

1

t− 1

)2

= −y21 . The same steps show that y2 is also a solution of the

differential equation.

(b) What can you say about the solutions ofdy

dt= −y2 for which the initial condition y(0) satisfies the inequality

−1 < y(0) < −1

2?

Note that y1(0) =1

0− 1= −1 and y2(0) =

1

0− 2= −1

2. So, −1 < y(0) < −1

2implies that

y1(0) < y(0) < y2(0), so1

t− 1< y(t) <

1

t− 2.

14.dy

dt=

1

(y + 1)(t− 2), y(0) = 0

(a) Find a formula for the solution.

dy

dt=

1

(y + 1)(t− 2)

⇒∫

(y + 1)dy =∫ 1

t− 2dt

⇒ 1

2y2 + y = ln|t− 2|+ C, where C is a constant

Since y(0) = 0, 0 = ln|0− 2|+ C = ln(2) + C ⇒ C = −ln(2). So:

1

2y2 + y = ln|t− 2| − ln(2) = ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

Now we put all terms on the left side and multiply by two to get rid of the fraction coefficient:

1

2y2 + y − ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

= 0

⇒ 2y2 + 2y − 2ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

= 0

Use the quadratic formula to find a formula for y:

y =

−2±√

22 − 4(1)(−2)ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

2(1)=

−2±√

4 + 8ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

2=

−2± 2

√

1 + 2ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

2

= −1±√

1 + 2ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

(b) State the domain of definition of the solution.

Since

√

1 + 2ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

cannot be negative, we have:

1 + 2ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

≥ 0

⇒ ln

∣

∣

∣

∣

t− 2

2

∣

∣

∣

∣

≥ −1

2

⇒ t− 2

2≥ e

−1

2

8

⇒ t− 2 ≥ 2e−1

2

⇒ t ≥ 2e−1

2 + 2

Thus, the domain is all real numbers such that t ≥ 2e−1

2 + 2.

1.6

2. Sketch the phase lines for the given differential equation. Identify the equilibrium points as sinks, sources, or nodes.

Note thatdy

dt= y2 − 4y − 12 = (y + 2)(y − 6). Setting this equal to zero gives y = −2 and y = 6 as equilibrium

points. If y < −2, you will find thatdy

dtis positive. For −2 < y < 6,

dy

dtis negative. For y > 6,

dy

dtis positive.

Thus, y = 6 is a source and y = −2 is a sink. Drawing the phase line is straightforward.

14.

14.jpg

9

34.

34.jpg

1.8

12. Solve the given initial value problem:dy

dt− 2y = 7e2t, y(0) = 3

Following the example on page 120, we guess that the general solution of the homogeneous equation is y(t) = ke2t.We guess yp(t) = αe2t, with α as the undetermined coefficient. But, like the example, we find that upon substituting yp(t)

intody

dt− 2y, we get zero. Instead, try yp(t) = αte2t. The derivative of yp(t) is α(1 + 2t)e2t. Then we have:

dy

dt− 2yp = α(1 + 2t)e2t − 2αte2t = αe2t ⇒ α = 7.

So, y(t) = ke2t + 7te2t. Use the initial condition to find k:

y(0) = 3 ⇒ 3 = ke0 ⇒ k = 3.

The solution is thus y(t) = 3e2t + 7te2t.

18. Consider the nonhomogeneous linear equationdy

dt= −y + 2.

(a) Compute an equilibrium solution for this equation.dy

dt= 0 ⇒ 0 = −y + 2 ⇒ y = 2

Thus y = 2 is an equilibrium solution.

(b) Verify that y(t) = 2− e−t is a solution for this equation.

dy

dt=

d

dt(2− e−t) = 0− (−1)e−t = e−t

Now note that e−t = 2− (2− e−t) = 2− y. Thus, y(t) is a solution.

22. Find the general solution and the solution that satisfies the initial condition y(0) = 0.

10

dy

dt+ y = t3 + sin(3t)

Guess that y1(t) = αsin(3t) + βcos(3t) and y2(t) = at3 + bt2 + ct+ d. Solve for α and β:

dy

dt+ y = sin(3t)

⇒ 3αcos(3t)− 3βsin(3t) + αsin(3t) + βcos(3t) = sin(3t)

⇒ (3α+ β)cos(3t) + (−3β + α− 1)sin(3t) = 0

Equating coefficients gives:

3α+ β = 0 and −3β + α− 1 = 0 ⇒ α =1

10and β =

−3

10.

Now we need to solve for a, b, c, and d:

dy

dt+ y = t3

⇒ (3at2 + 2bt+ c) + (at3 + bt2 + ct+ d)− t3 = 0

⇒ (a+ 1)t3 + (3a+ b)t2 + (2b+ c)t+ (c+ d) = 0

Equating coefficients gives:a = 1, 3a+ b = 0, 2b+ c = 0, and c+ d = 0 ⇒ b = −3, c = 6, and d = −6.

Now, we know that yh(t) is given by yh(t) = ke−t. So,

y(t) = ke−t +1

10sin(3t)− 3

10cos(3t) + t3 − 3t2 + 6t− 6.

Our initial condition y(0) = 0 yields k =63

10.

1.9

8. Solve the given initial-value problem.

dy

dt=

1

t+ 1y + 4t2 + 4t, y(1) = 10

Note that µ(t) = e

∫

−

(

1

t+ 1

)

dt =1

1 + tby natural log rules. Then:

1

t+ 1

dy

dt− y

(t+ 1)2=

4t2 + 4t

1 + t

⇒ d

dt

(

y

t+ 1

)

= 4t

⇒∫

(

y

t+ 1

)

d

dt=

∫

4t dt

⇒ y

t+ 1= 2t2 + C, where C is a constant

⇒ y = (2t2 + C)(t+ 1)

Using the initial value y(1) = 10, we get:10 = (2 + C)(2) = 2C + 4 ⇒ C = 3.

16. Determine the general solution to the equation and express it with as few integrals as possible.

11

dy

dt= y + 4cos(t2)

The integrating factor is given by µ(t) = e∫

−1dt = e−t. Then we have:

(e−t)dy

dt− (e−t)y = e−t4cos(t2)

⇒ e−tyd

dt= e−t4cos(t2)

⇒∫

e−tyd

dt= 4

∫

cos(t2)e−tdt

⇒ e−ty = 4∫

cos(t2)e−tdt

⇒ y = 4et∫

cos(t2)e−tdt

12