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Chapter 2: First-order equations

43 problems · 4 figures

Exercise 1.1.101 §1.1 source ↗

Solve dydx=ex+x and y(0)=10⁠.

Answer

y=ex+x22+9

Exercise 1.1.102 §1.1 source ↗

Solve x′=1x2⁠, x(1)=1⁠.

Answer

x=(3t−2)1/3

Exercise 1.1.103 §1.1 source ↗

Solve x′=1cos⁡(x)⁠, x(0)=π4⁠.

Answer

x=sin−1⁡(t+1/2)

Exercise 1.1.104 §1.1 source ↗

Sid is in a car traveling at speed 10t+70 miles per hour away from Las Vegas, where t is in hours. At t=0⁠, Sid is 10 miles away from Vegas. How far from Vegas is Sid 2 hours later?

Answer

170

Exercise 1.1.105 §1.1 source ↗

Solve y′=yn⁠, y(0)=1⁠, where n is a positive integer. Hint: You have to consider different cases.

Answer

If n≠1⁠, then y=((1−n)x+1)1/(1−n)⁠. If n=1⁠, then y=ex⁠.

Exercise 1.1.106 §1.1 source ↗

The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. The volume (in centimeters cubed) of a ball of radius r centimeters is (4/3)πr3⁠. The surface area is 4πr2⁠. Set up the differential equation for how the radius r is changing. Then, suppose that at time t=0 minutes, the radius is 10 centimeters. After 5 minutes, the radius is 8 centimeters. At what time t will the snowball be completely melted?

Answer

The equation is r′=−C for some constant C⁠. The snowball will be completely melted in 25 minutes from time t=0⁠.

Exercise 1.1.107 §1.1 source ↗

Find the general solution to y′′′′=0⁠. How many distinct constants do you need?

Answer

y=Ax3+Bx2+Cx+D⁠, so 4 constants.

Exercise 1.2.101 §1.2 source ↗

Sketch the slope field of y′=y3⁠. Can you visually find the solution that satisfies y(0)=0⁠?

Answer (2 parts)

y=0 is a solution such that y(0)=0⁠.

Exercise 1.2.102 §1.2 source ↗

Is it possible to solve y′=xy for y(0)=0⁠? Is the solution unique?

Answer

Yes a solution exists. The equation is y′=f(x,y) where f(x,y)=xy⁠. The function f(x,y) is continuous and ∂f∂y=x⁠, which is also continuous near (0,0)⁠. So a solution exists and is unique. (In fact, y=0 is the solution.)

Exercise 1.2.103 §1.2 source ↗

Is it possible to solve y′=xx2−1 for y(1)=0⁠?

Answer

No, the equation is not defined at (x,y)=(1,0)⁠.

Exercise 1.2.104 §1.2 source ↗

Match equations y′=sin⁡x⁠, y′=cos⁡y⁠, y′=ycos⁡(x) to slope fields. Justify.

a)

b)

c)

Answer

a) y′=cos⁡y⁠, b) y′=ycos⁡(x)⁠, c) y′=sin⁡x⁠. Justification left to reader.

Exercise 1.2.105 §1.2 source ↗

(tricky) Suppose

f(y)={0 if y>0,1 if y≤0.

Does y′=f(y)⁠, y(0)=0 have a continuously differentiable solution? Does Picard apply? Why, or why not?

Answer

Picard does not apply as f is not continuous at y=0⁠. The equation does not have a continuously differentiable solution. Suppose it did. Notice that y′(0)=1⁠. By the first derivative test, y(x)>0 for small positive x⁠. But then for those x⁠, we have y′(x)=f(y(x))=0⁠. It is not possible for y′ to be continuous, y′(0)=1 and y′(x)=0 for arbitrarily small positive x⁠.

Exercise 1.2.106 §1.2 source ↗

Consider an equation of the form y′=f(x) for some continuous function f⁠, and an initial condition y(x0)=y0⁠. Does a solution exist for all x⁠? Why or why not?

Answer

The solution is y(x)=∫x0xf(s)ds+y0⁠, and this does indeed exist for every x⁠.

Exercise 1.3.101 §1.3 source ↗

Solve y′=2xy⁠.

Answer

y=Cex2

Exercise 1.3.102 §1.3 source ↗

Solve x′=3xt2−3t2⁠, x(0)=2⁠.

Answer

x=et3+1

Exercise 1.3.103 §1.3 source ↗

Find an implicit solution for x′=13x2+1⁠, x(0)=1⁠.

Answer

x3+x=t+2

Exercise 1.3.104 §1.3 source ↗

Find an explicit solution to xy′=y2⁠, y(1)=1⁠.

Answer

y=11−ln⁡x

Exercise 1.3.105 §1.3 source ↗

Find an implicit solution to y′=sin⁡(x)cos⁡(y)⁠.

Answer

sin⁡(y)=−cos⁡(x)+C

Exercise 1.3.106 §1.3 source ↗

Take Example 1.3.3 with the same numbers: 89 degrees at t=0⁠, 85 degrees at t=1⁠, and ambient temperature of 22 degrees. Suppose these temperatures were measured with a precision of ±0.5 degrees. Given this imprecision, the time it takes the coffee to cool to (exactly) 60 degrees is also only known in a certain range. Find this range. Hint: Think about what kind of error makes the cooling time longer and what shorter.

Example 1.3.3 in the book ↗

Bob made a cup of coffee, and Bob likes to drink coffee only once it reaches 60 degrees Celsius and will not burn him. Initially, right after it was poured in the cup, Bob measured the temperature and the coffee was 89 degrees Celsius. One minute later, Bob measured the coffee again and it had 85 degrees. The temperature of the room (the ambient temperature) is 22 degrees. When should Bob start drinking?

Let T be the temperature of the coffee in degrees Celsius, let A be the ambient (room) temperature, also in degrees Celsius, and let t denote the time in minutes, with t=0 denoting the time when Bob made the coffee. Newton’s law of cooling states that the rate at which the temperature of the coffee is changing is proportional to the difference between the ambient temperature and the temperature of the coffee. That is,

dTdt=k(A−T),

for some positive constant k⁠. For our setup A=22⁠, T(0)=89⁠, T(1)=85⁠. Let C and D denote arbitrary constants, and notice that T−A>0 since the coffee is clearly warmer than the room. We separate variables and integrate.

1T−AdTdt=−k,ln⁡(T−A)=−kt+C,T−A=De−kt,T=A+De−kt.

That is, T=22+De−kt⁠. We plug in the first condition: 89=T(0)=22+D⁠, and hence D=67⁠. So T=22+67e−kt⁠. The second condition says 85=T(1)=22+67e−k⁠. Solving for k⁠, we get k=−ln⁡85−2267≈0.0616⁠. We solve for the time t that gives a temperature of 60 degrees. Namely, we solve

60=22+67e−0.0616t

to get t= − ln⁡60−22670.0616≈ 9.21 minutes. So Bob can begin to drink the coffee at just over 9 minutes from the time Bob made it. That is probably about the amount of time it took us to calculate how long it would take. See Figure 1.7.

Two plots of the coffee temperature. On the left, the downwards solution is shown for time 0 to 12.5 as the temperature goes from 89 to below 60. There are horizontal lines at 89, 85, and 60. A vertical line at t equals 1 intersects the solution at temperature 85. A vertical line at t equals 9.21 intersects the solution at temperature 60. The right-hand plot is over longer period of time to t equals 80. The solution exponentially decays towards a horizontal line at temperature 22.

Graphs of the coffee temperature function T(t)⁠. On the left, horizontal lines are drawn at temperatures 60, 85, and 89. Vertical lines are drawn at t=1 and t=9.21⁠. Notice that the temperature of the coffee hits 85 at t=1⁠, and 60 at t≈9.21⁠. On the right, the graph is over a longer period of time, with a horizontal line at the ambient temperature 22.

Figure 1.7 in the book ↗

Two plots of the coffee temperature.  On the left, the downwards
solution is shown for time 0 to 12.5 as the temperature goes
from 89 to below 60.  There are horizontal lines at 89, 85,
and 60.  A vertical line at t equals 1 intersects the solution
at temperature 85.  A vertical line at t equals 9.21 intersects the
solution at temperature 60.  The right-hand plot is over longer
period of time to t equals 80.  The solution exponentially decays
towards a horizontal line at temperature 22.

Graphs of the coffee temperature function T(t)⁠. On the left, horizontal lines are drawn at temperatures 60, 85, and 89. Vertical lines are drawn at t=1 and t=9.21⁠. Notice that the temperature of the coffee hits 85 at t=1⁠, and 60 at t≈9.21⁠. On the right, the graph is over a longer period of time, with a horizontal line at the ambient temperature 22.

Answer

The range is approximately 7.45 to 12.15 minutes.

Exercise 1.3.107 §1.3 source ↗

A population x of rabbits on an island is modeled by x′=x−(1/1000)x2⁠, where the independent variable t is time in months. At time t=0⁠, there are 40 rabbits on the island.

a) Find the solution to the equation with the initial condition.

b) How many rabbits are on the island in 1 month, 5 months, 10 months, 15 months (round to the nearest integer)?

Answer

a) x=1000etet+24⁠. b) 102 rabbits after one month, 861 after 5 months, 999 after 10 months, 1000 after 15 months.

Exercise 1.4.101 §1.4 source ↗

Solve y′+3x2y=x2⁠.

Answer

y=Ce−x3+1/3

Exercise 1.4.102 §1.4 source ↗

Solve y′+2sin⁡(2x)y=2sin⁡(2x)⁠, y(π/2)=3⁠.

Answer

y=2ecos⁡(2x)+1+1

Exercise 1.4.103 §1.4 source ↗

Suppose a water tank is being pumped out at 3Lmin. The water tank starts at 10L of clean water. Water with a toxic substance is flowing into the tank at 2Lmin, with concentration 20tgL at time t⁠. When the tank is half empty, how many grams of toxic substance are in the tank (assuming perfect mixing)?

Answer

250 grams

Exercise 1.4.104 §1.4 source ↗

There are bacteria on a plate and a toxic substance is being added that slows down the rate of growth of the bacteria. That is, suppose that dPdt=(2−0.1t)P⁠. If P(0)=1000⁠, find the population at t=5⁠.

Answer

P(5)= 1000e2×5−0.05×52= 1000e8.75≈ 6.31×106

Exercise 1.4.105 §1.4 source ↗

A cylindrical water tank has water flowing in at I cubic meters per second. Let A be the area of the cross section of the tank in square meters. Suppose water is flowing out from the bottom of the tank at a rate proportional to the height of the water level. Set up the differential equation for h⁠, the height of the water, introducing and naming constants that you need. You should also give the units for your constants.

Answer

Ah′=I−kh⁠, where k is a constant with units m2s⁠.

Exercise 1.5.101 §1.5 source ↗

Solve xy′+y+y2=0⁠, y(1)=2⁠.

Answer

y=23x−2

Exercise 1.5.102 §1.5 source ↗

Solve xy′+y+x=0⁠, y(1)=1⁠.

Answer

y=3−x22x

Exercise 1.5.103 §1.5 source ↗

Solve y2y′=y3−3x⁠, y(0)=2⁠.

Answer

y=(7e3x+3x+1)1/3

Exercise 1.5.104 §1.5 source ↗

Solve 2yy′=ey2−x2+2x⁠.

Answer

y=±x2−ln⁡(C−x)

Exercise 1.6.101 §1.6 source ↗

Let x′=(x−1)(x−2)x2⁠.

a) Sketch the phase diagram and find critical points.

b) Classify the critical points.

c) If x(0)=0.5⁠, then find limt→∞⁡x(t)⁠.

Answer

a) 0, 1, 2 are critical points. b) x=0 is unstable (semistable), x=1 is stable, and x=2 is unstable. c) 1

Exercise 1.6.102 §1.6 source ↗

Let x′=e−x⁠.

a) Find and classify all critical points.

b) Find limt→∞⁡x(t) given any initial condition.

Answer

a) There are no critical points. b) ∞

Exercise 1.6.103 §1.6 source ↗

Assume that a population of fish in a lake satisfies dxdt=kx(M−x)⁠. Now suppose that fish are continually added at A fish per unit of time.

a) Find the differential equation for x⁠.

b) What is the new limiting population?

Answer

a) dxdt=kx(M−x)+A b) kM+(kM)2+4Ak2k

Exercise 1.6.104 §1.6 source ↗

Suppose dxdt=(x−α)(x−β) for two numbers α<β⁠.

a) Find the critical points, and classify them.

For b), c), d), find limt→∞⁡x(t) based on the phase diagram.

a) x(0)<α⁠,

b) α<x(0)<β⁠,

c) β<x(0)⁠.

Answer

a) α is a stable critical point, β is an unstable one. b) α⁠, c) α⁠, d) ∞ or DNE.

Exercise 1.7.101 §1.7 source ↗

Let x′=sin⁡(xt)⁠, and x(0)=1⁠. Approximate x(1) using Euler’s method with step sizes 1, 0.5, 0.25. Use a calculator and compute up to 4 decimal digits.

Answer

Approximately: 1.0000, 1.2397, 1.3829

Exercise 1.7.102 §1.7 source ↗

Let x′=2t⁠, and x(0)=0⁠.

a) Approximate x(4) using Euler’s method with step sizes 4, 2, and 1.

b) Solve exactly, and compute the errors.

c) Compute the factor by which the errors changed.

Answer

a) 0, 8, 12 b) x(4)=16⁠, so errors are: 16, 8, 4. c) Factors are 0.5 and 0.5.

Exercise 1.7.103 §1.7 source ↗

Let x′=xext+1⁠, and x(0)=0⁠.

a) Approximate x(4) using Euler’s method with step sizes 4, 2, and 1.

b) Guess an exact solution based on part a) and compute the errors.

Answer

a) 0, 0, 0 b) x=0 is a solution so errors are: 0, 0, 0.

Exercise 1.7.104 §1.7 source ↗

Consider dydx=x+y⁠, y(0)=1⁠.

a) Approximate y(1) using the improved Euler’s method (see above) with step sizes h=1/4 and h=1/8⁠.

b) Approximate y(1) using Euler’s method with h=1/4 and h=1/8⁠.

c) Solve exactly, find the exact value of y(1)⁠.

d) Compute the errors, and the factors by which the errors changed.

Answer

a) Improved Euler: y(1)≈3.3897 for h=1/4⁠, y(1)≈3.4237 for h=1/8⁠, b) Standard Euler: y(1)≈2.8828 for h=1/4⁠, y(1)≈3.1316 for h=1/8⁠, c) y=2ex−x−1⁠, so y(1) is approximately 3.4366⁠. d) Approximate errors for improved Euler: 0.046852 for h=1/4 and 0.012881 for h=1/8⁠. For standard Euler: 0.55375 for h=1/4 and 0.30499 for h=1/8⁠. Factor is approximately 0.27 for improved Euler and 0.55 for standard Euler.

Exercise 1.8.101 §1.8 source ↗

Solve the following exact equations, implicit general solutions will suffice:

a) cos⁡(x)+yexy+xexyy′=0

b) (2x+y)dx+(x−4y)dy=0

c) ex+eydydx=0

d) (3x2+3y)dx+(3y2+3x)dy=0

Answer

a) exy+sin⁡(x)=C b) x2+xy−2y2=C c) ex+ey=C d) x3+3xy+y3=C

Exercise 1.8.102 §1.8 source ↗

Find the integrating factor for the following equations making them into exact equations:

a) 1ydx+3ydy=0

b) dx−e−x−ydy=0

c) (cos⁡(x)y2+1y)dx+ xy2dy= 0

d) (2y+y2x)dx+(2y+x)dy=0

Answer

a) Integrating factor is y⁠, equation becomes dx+3y2dy=0⁠. b) Integrating factor is ex⁠, equation becomes exdx−e−ydy=0⁠. c) Integrating factor is y2⁠, equation becomes (cos⁡(x)+y)dx+xdy=0⁠. d) Integrating factor is x⁠, equation becomes (2xy+y2)dx+(x2+2xy)dy=0⁠.

Exercise 1.8.103 §1.8 source ↗

a) Show that every separable equation y′=f(x)g(y) can be written as an exact equation, and verify that it is indeed exact.

b) Rewrite y′=xy as an exact equation, solve it, and verify that the solution is the same as it was in Example 1.3.1.

Example 1.3.1 in the book ↗

Take the equation

y′=xy.

Note that y=0 is a solution. We will remember that fact and assume y≠0 from now on, so that we can divide by y⁠. Write the equation as dydx=xy or dyy=xdx⁠. Then

∫dyy=∫xdx+C.

We compute the antiderivatives to get

ln⁡|y|=x22+C,

or

|y|=ex22+C=ex22eC=Dex22,

where D>0 is some constant. Because y=0 is also a solution and because of the absolute value, we can write:

y=Dex22,

for any number D (including zero or negative).

We check:

y′=Dxex22=x(Dex22)=xy.

Yay!

Answer

a) The equation is −f(x)dx+1g(y)dy=0⁠, and this is exact because M=−f(x)⁠, N=1g(y)⁠, so My=0=Nx⁠. b) −xdx+1ydy=0⁠, leads to potential function F(x,y)=−x22+ln⁡|y|⁠, solving F(x,y)=C leads to the same solution as the example.

Exercise 1.9.101 §1.9 source ↗

Solve

a) −5ux+ut=0⁠,   u(x,0)=11+x2⁠,

b) 2ux+ut=0⁠,   u(x,0)=cos⁡(x)⁠.

Answer

a) u=11+(x+5t)2 b) u=cos⁡(x−2t)

Exercise 1.9.102 §1.9 source ↗

Solve ux+ut+tu=0⁠, u(x,0)=cos⁡(x)⁠.

Answer

u=cos⁡(x−t)e−t2/2

Exercise 1.9.103 §1.9 source ↗

Solve ux+ut=5⁠, u(x,0)=x⁠.

Answer

u=x+4t

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Exercises and answers are adapted from Notes on Diffy Qs: Differential Equations for Engineers by Jiří Lebl, which is free to read and download, and are used under the CC BY-NC-SA 4.0 licence. They are taken from the book’s LaTeX source and set for the web, so they may differ slightly from the printed page. This adaptation is released under the same licence and is neither affiliated with nor endorsed by Lebl or Jiří Lebl.