Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 17: Differential Equations

95 problems · 0 figures

17.1.2 §17.1 source ↗

Solve y˙=1/(1+t2)⁠.

Answer

y=arctan⁡t+C

17.1.3 §17.1 source ↗

Solve the initial value problem y˙=tn with y(0)=1 and n≥0⁠.

Answer

y=tn+1n+1+1

17.1.4 §17.1 source ↗

Solve y˙=ln⁡t⁠.

Answer

y=tln⁡t−t+C

17.1.5 §17.1 source ↗

Identify the constant solutions (if any) of y˙=tsin⁡y⁠.

Answer

y=nπ⁠, for any integer n⁠.

17.1.6 §17.1 source ↗

Identify the constant solutions (if any) of y˙=tey⁠.

Answer

none

17.1.7 §17.1 source ↗

Solve y˙=t/y⁠.

Answer

y=±t2+C

17.1.8 §17.1 source ↗

Solve y˙=y2−1⁠.

Answer

y=±1⁠, y=(1+Ae2t)/(1−Ae2t)

17.1.9 §17.1 source ↗

Solve y˙=t/(y3−5)⁠. You may leave your solution in implicit form: that is, you may stop once you have done the integration, without solving for y⁠.

Answer

y4/4−5y=t2/2+C

17.1.10 §17.1 source ↗

Find a non-constant solution of the initial value problem y˙=y1/3⁠, y(0)=0⁠, using separation of variables. Note that the constant function y(t)=0 also solves the initial value problem. This shows that an initial value problem can have more than one solution.

Answer

y=(2t/3)3/2

17.1.11 §17.1 source ↗

Solve the equation for Newton's law of cooling leaving M and k unknown.

Answer

y=M+Ae−kt

17.1.12 §17.1 source ↗

After 10 minutes in Jean-Luc's room, his tea has cooled to 40∘ Celsius from 100∘ Celsius. The room temperature is 25∘ Celsius. How much longer will it take to cool to 35∘⁠?

Answer

10ln⁡(15/2)ln⁡5−10≈2.52 minutes

17.1.13 §17.1 source ↗

Solve the logistic equation y˙=ky(M−y)⁠. (This is a somewhat more reasonable population model in most cases than the simpler y˙=ky⁠.) Sketch the graph of the solution to this equation when M=1000⁠, k=0.002⁠, y(0)=1⁠.

Answer

y=M1+Ae−Mkt

17.1.14 §17.1 source ↗

Suppose that y˙=ky⁠, y(0)=2⁠, and y˙(0)=3⁠. What is y⁠?

Answer

y=2e3t/2

17.1.15 §17.1 source ↗

Radioactive substances "decay'', that is, the amount (the mass) of the substance decreases over time. The mass of a radioactive substance obeys the equation y˙=ky where k<0 and y is the mass of the substance at time t⁠. Suppose that initially, the mass of the substance is y(0)=M>0⁠. At what time does half of the mass remain? (This is known as the half life. Note that the half life depends on k but not on M⁠.)

Answer

t=−ln⁡2k

17.1.16 §17.1 source ↗

Bismuth-210 has a half life of five days. If there is initially 600 milligrams, how much is left after 6 days? When will there be only 2 milligrams left?

Answer

600e−6ln⁡2/5≈261 mg; 5ln⁡300ln⁡2≈41 days

17.1.17 §17.1 source ↗

The half life of carbon-14 is 5730 years. If one starts with 100 milligrams of carbon-14, how much is left after 6000 years? How long do we have to wait before there is less than 2 milligrams?

Answer

100e−200ln⁡2/191≈48 mg; 5730ln⁡50ln⁡2≈32339 years

17.1.18 §17.1 source ↗

A certain species of bacteria doubles its population (or its mass) every hour in the lab. The differential equation that models this phenomenon is y˙=ky⁠, where k>0 and y is the population of bacteria at time t⁠. What is y⁠?

Answer

y=y0etln⁡2

17.1.19 §17.1 source ↗

If a certain microbe doubles its population every 4 hours and after 5 hours the total population has mass 500 grams, what was the initial mass?

Answer

500e−5ln⁡2/4≈210 g

17.2.1 §17.2 source ↗

Find the general solution of each equation.

y˙+5y=0

Answer

y=Ae−5t

17.2.2 §17.2 source ↗

Find the general solution of each equation.

y˙−2y=0

Answer

y=Ae2t

17.2.3 §17.2 source ↗

Find the general solution of each equation.

y˙+y1+t2=0

Answer

y=Ae−arctan⁡t

17.2.4 §17.2 source ↗

Find the general solution of each equation.

y˙+t2y=0

Answer

y=Ae−t3/3

17.2.5 §17.2 source ↗

Solve the initial value problem.

y˙+y=0⁠, y(0)=4

Answer

y=4e−t

17.2.6 §17.2 source ↗

Solve the initial value problem.

y˙−3y=0⁠, y(1)=−2

Answer

y=−2e3t−3

17.2.7 §17.2 source ↗

Solve the initial value problem.

y˙+ysin⁡t=0⁠, y(π)=1

Answer

y=e1+cos⁡t

17.2.8 §17.2 source ↗

Solve the initial value problem.

y˙+yet=0⁠, y(0)=e

Answer

y=e2e−et

17.2.9 §17.2 source ↗

Solve the initial value problem.

y˙+y1+t4=0⁠, y(0)=0

Answer

y=0

17.2.10 §17.2 source ↗

Solve the initial value problem.

y˙+ycos⁡(et)=0⁠, y(0)=0

Answer

y=0

17.2.11 §17.2 source ↗

Solve the initial value problem.

ty˙−2y=0⁠, y(1)=4

Answer

y=4t2

17.2.12 §17.2 source ↗

Solve the initial value problem.

t2y˙+y=0⁠, y(1)=−2⁠, t>0

Answer

y=−2e(1/t)−1

17.2.13 §17.2 source ↗

Solve the initial value problem.

t3y˙=2y⁠, y(1)=1⁠, t>0

Answer

y=e1−t−2

17.2.14 §17.2 source ↗

Solve the initial value problem.

t3y˙=2y⁠, y(1)=0⁠, t>0

Answer

y=0

17.2.15 §17.2 source ↗

A function y(t) is a solution of y˙+ky=0⁠. Suppose that y(0)=100 and y(2)=4⁠. Find k and find y(t)⁠.

Answer

k=ln⁡5⁠, y=100e−tln⁡5

17.2.16 §17.2 source ↗

A function y(t) is a solution of y˙+tky=0⁠. Suppose that y(0)=1 and y(1)=e−13⁠. Find k and find y(t)⁠.

Answer

k=−12/13⁠, y=exp⁡(−13t1/13)

17.2.17 §17.2 source ↗

A bacterial culture grows at a rate proportional to its population. If the population is one million at t=0 and 1.5 million at t=1 hour, find the population as a function of time.

Answer

y=106etln⁡(3/2)

17.2.18 §17.2 source ↗

A radioactive element decays with a half-life of 6 years. If a block of the element has mass 10 kilograms at t=0⁠, find the amount of the element at time t⁠.

Answer

y=10e−tln⁡(2)/6

17.3.1 §17.3 source ↗

Find the general solution of the equation.

y˙+4y=8

Answer

y=Ae−4t+2

17.3.2 §17.3 source ↗

Find the general solution of the equation.

y˙−2y=6

Answer

y=Ae2t−3

17.3.3 §17.3 source ↗

Find the general solution of the equation.

y˙+ty=5t

Answer

y=Ae−(1/2)t2+5

17.3.4 §17.3 source ↗

Find the general solution of the equation.

y˙+ety=−2et

Answer

y=Ae−et−2

17.3.5 §17.3 source ↗

Find the general solution of the equation.

y˙−y=t2

Answer

y=Aet−t2−2t−2

17.3.6 §17.3 source ↗

Find the general solution of the equation.

2y˙+y=t

Answer

y=Ae−t/2+t−2

17.3.7 §17.3 source ↗

Find the general solution of the equation.

ty˙−2y=1/t⁠, t>0

Answer

y=At2−13t

17.3.8 §17.3 source ↗

Find the general solution of the equation.

ty˙+y=t⁠, t>0

Answer

y=ct+23t

17.3.9 §17.3 source ↗

Find the general solution of the equation.

y˙cos⁡t+ysin⁡t=1⁠, −π/2<t<π/2

Answer

y=Acos⁡t+sin⁡t

17.3.10 §17.3 source ↗

Find the general solution of the equation.

y˙+ysec⁡t=tan⁡t⁠, −π/2<t<π/2

Answer

y= Asec⁡t+tan⁡t+ 1− tsec⁡t+tan⁡t

17.4.1 §17.4 source ↗

Compute the Euler approximations for the initial value problem for 0≤t≤1 and Δt=0.2⁠.

y˙=t/y⁠, y(0)=1

Answer

y(1)≈1.355

17.4.2 §17.4 source ↗

Compute the Euler approximations for the initial value problem for 0≤t≤1 and Δt=0.2⁠.

y˙=t+y3⁠, y(0)=1

Answer

y(1)≈40.31

17.4.3 §17.4 source ↗

Compute the Euler approximations for the initial value problem for 0≤t≤1 and Δt=0.2⁠.

y˙=cos⁡(t+y)⁠, y(0)=1

Answer

y(1)≈1.05

17.4.4 §17.4 source ↗

Compute the Euler approximations for the initial value problem for 0≤t≤1 and Δt=0.2⁠.

y˙=tln⁡y⁠, y(0)=2

Answer

y(1)≈2.30

17.5.4 §17.5 source ↗

Solve the initial value problem y¨−ω2y=0⁠, y(0)=1⁠, y˙(0)=1⁠, assuming ω≠0⁠.

Answer

ω+12ωeωt+ω−12ωe−ωt

17.5.5 §17.5 source ↗

Solve the initial value problem 2y¨+18y=0⁠, y(0)=2⁠, y˙(0)=15⁠.

Answer

2cos⁡(3t)+5sin⁡(3t)

17.5.6 §17.5 source ↗

Solve the initial value problem y¨+6y˙+5y=0⁠, y(0)=1⁠, y˙(0)=0⁠.

Answer

−(1/4)e−5t+(5/4)e−t

17.5.7 §17.5 source ↗

Solve the initial value problem y¨−y˙−12y=0⁠, y(0)=0⁠, y˙(0)=14⁠.

Answer

−2e−3t+2e4t

17.5.8 §17.5 source ↗

Solve the initial value problem y¨+12y˙+36y=0⁠, y(0)=5⁠, y˙(0)=−10⁠.

Answer

5e−6t+20te−6t

17.5.9 §17.5 source ↗

Solve the initial value problem y¨−8y˙+16y=0⁠, y(0)=−3⁠, y˙(0)=4⁠.

Answer

(16t−3)e4t

17.5.10 §17.5 source ↗

Solve the initial value problem y¨+5y=0⁠, y(0)=−2⁠, y˙(0)=5⁠.

Answer

−2cos⁡(5t)+5sin⁡(5t)

17.5.11 §17.5 source ↗

Solve the initial value problem y¨+y=0⁠, y(π/4)=0⁠, y˙(π/4)=2⁠.

Answer

−2cos⁡t+2sin⁡t

17.5.12 §17.5 source ↗

Solve the initial value problem y¨+12y˙+37y=0⁠, y(0)=4⁠, y˙(0)=0⁠.

Answer

e−6t(4cos⁡t+24sin⁡t)

17.5.13 §17.5 source ↗

Solve the initial value problem y¨+6y˙+18y=0⁠, y(0)=0⁠, y˙(0)=6⁠.

Answer

2e−3tsin⁡(3t)

17.5.14 §17.5 source ↗

Solve the initial value problem y¨+4y=0⁠, y(0)=3⁠, y˙(0)=2⁠. Put your answer in the form developed at the end of example 17.5.3.

Answer

2cos⁡(2t−π/6)

17.5.15 §17.5 source ↗

Solve the initial value problem y¨+100y=0⁠, y(0)=5⁠, y˙(0)=50⁠. Put your answer in the form developed at the end of example 17.5.3.

Answer

52cos⁡(10t−π/4)

17.5.16 §17.5 source ↗

Solve the initial value problem y¨+4y˙+13y=0⁠, y(0)=1⁠, y˙(0)=1⁠. Put your answer in the form developed at the end of example 17.5.3.

Answer

2e−2tcos⁡(3t−π/4)

17.5.17 §17.5 source ↗

Solve the initial value problem y¨−8y˙+25y=0⁠, y(0)=3⁠, y˙(0)=0⁠. Put your answer in the form developed at the end of example 17.5.3.

Answer

5e4tcos⁡(3t+ arcsin⁡(4/5))

17.5.18 §17.5 source ↗

A mass-spring system my¨+by˙+ky has k=29⁠, b=4⁠, and m=1⁠. At time t=0 the position is y(0)=2 and the velocity is y˙(0)=1⁠. Find y(t)⁠.

Answer

(2cos⁡(5t)+ sin⁡(5t))e−2t

17.5.19 §17.5 source ↗

A mass-spring system my¨+by˙+ky has k=24⁠, b=12⁠, and m=3⁠. At time t=0 the position is y(0)=0 and the velocity is y˙(0)=−1⁠. Find y(t)⁠.

Answer

−(1/2)e−2tsin⁡(2t)

17.6.1 §17.6 source ↗

Find the general solution to the differential equation.

y¨−10y˙+25y=cos⁡t

Answer

Ae5t+ Bte5t+ (6/169)cos⁡t− (5/338)sin⁡t

17.6.2 §17.6 source ↗

Find the general solution to the differential equation.

y¨+22y˙+2y=10

Answer

Ae−2t+Bte−2t+5

17.6.3 §17.6 source ↗

Find the general solution to the differential equation.

y¨+16y=8t2+3t−4

Answer

Acos⁡(4t)+ Bsin⁡(4t)+ (1/2)t2+ (3/16)t− 5/16

17.6.4 §17.6 source ↗

Find the general solution to the differential equation.

y¨+2y=cos⁡(5t)+sin⁡(5t)

Answer

Acos⁡(2t)+ Bsin⁡(2t)− (cos⁡(5t)+sin⁡(5t))/23

17.6.5 §17.6 source ↗

Find the general solution to the differential equation.

y¨−2y˙+2y=e2t

Answer

et(Acos⁡t+Bsin⁡t)+e2t/2

17.6.6 §17.6 source ↗

Find the general solution to the differential equation.

y¨−6y+13=1+2t+e−t

Answer

Ae6t+Be−6t+2−t/3−e−t/5

17.6.7 §17.6 source ↗

Find the general solution to the differential equation.

y¨+y˙−6y=e−3t

Answer

Ae−3t+Be2t−(1/5)te−3t

17.6.8 §17.6 source ↗

Find the general solution to the differential equation.

y¨−4y˙+3y=e3t

Answer

Aet+Be3t+(1/2)te3t

17.6.9 §17.6 source ↗

Find the general solution to the differential equation.

y¨+16y=cos⁡(4t)

Answer

Acos⁡(4t)+ Bsin⁡(4t)+ (1/8)tsin⁡(4t)

17.6.10 §17.6 source ↗

Find the general solution to the differential equation.

y¨+9y=3sin⁡(3t)

Answer

Acos⁡(3t)+ Bsin⁡(3t)− (1/2)tcos⁡(3t)

17.6.11 §17.6 source ↗

Find the general solution to the differential equation.

y¨+12y˙+36y=6e−6t

Answer

Ae−6t+Bte−6t+3t2e−6t

17.6.12 §17.6 source ↗

Find the general solution to the differential equation.

y¨−8y˙+16y=−2e4t

Answer

Ae4t+Bte4t−t2e4t

17.6.13 §17.6 source ↗

Find the general solution to the differential equation.

y¨+6y˙+5y=4

Answer

Ae−t+Be−5t+(4/5)

17.6.14 §17.6 source ↗

Find the general solution to the differential equation.

y¨−y˙−12y=t

Answer

Ae4t+ Be−3t+ (1/144)− (t/12)

17.6.15 §17.6 source ↗

Find the general solution to the differential equation.

y¨+5y=8sin⁡(2t)

Answer

Acos⁡(5t)+ Bsin⁡(5t)+ 8sin⁡(2t)

17.6.16 §17.6 source ↗

Find the general solution to the differential equation.

y¨−4y=4e2t

Answer

Ae2t+Be−2t+te2t

17.6.17 §17.6 source ↗

Solve the initial value problem.

y¨−y=3t+5⁠, y(0)=0⁠, y˙(0)=0

Answer

4et+e−t−3t−5

17.6.18 §17.6 source ↗

Solve the initial value problem.

y¨+9y=4t⁠, y(0)=0⁠, y˙(0)=0

Answer

−(4/27)sin⁡(3t)+(4/9)t

17.6.19 §17.6 source ↗

Solve the initial value problem.

y¨+12y˙+37y=10e−4t⁠, y(0)=4⁠, y˙(0)=0

Answer

e−6t(2cos⁡t+20sin⁡t)+ 2e−4t

17.6.20 §17.6 source ↗

Solve the initial value problem.

y¨+6y˙+18y=cos⁡t−sin⁡t⁠, y(0)=0⁠, y˙(0)=2

Answer

(− 23325cos⁡(3t)+ 592975sin⁡(3t))e−3t+ 23325cos⁡t− 11325sin⁡t

17.6.21 §17.6 source ↗

Solve the initial value problem.

Find the solution for the mass-spring equation y¨+4y˙+29y=689cos⁡(2t)⁠.

Answer

e−2t(Asin⁡(5t)+ Bcos⁡(5t))+ 8sin⁡(2t)+ 25cos⁡(2t)

17.6.22 §17.6 source ↗

Solve the initial value problem.

Find the solution for the mass-spring equation 3y¨+12y˙+24y=2sin⁡t⁠.

Answer

e−2t(Asin⁡(2t)+ Bcos⁡(2t))+ (14/195)sin⁡t− (8/195)cos⁡t

17.7.1 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨+y=tan⁡t

Answer

Asin⁡(t)+ Bcos⁡(t)− cos⁡tln⁡|sec⁡t+ tan⁡t|

17.7.2 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨+y=e2t

Answer

Asin⁡(t)+Bcos⁡(t)+15e2t

17.7.3 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨+4y=sec⁡t

Answer

Asin⁡(2t)+ Bcos⁡(2t)+ cos⁡t− sin⁡tcos⁡tln⁡|sec⁡t+ tan⁡t|

17.7.4 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨+4y=tan⁡t>

Answer

Asin⁡(2t)+ Bcos⁡(2t)+ 12sin⁡(2t)sin2⁡(t)+ 12sin⁡(2t)ln⁡|cos⁡t|− t2cos⁡(2t)+ 14sin⁡(2t)cos⁡(2t)

17.7.5 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨+y˙−6y=t2e2t

Answer

Ae2t+ Be−3t+ t315e2t− (t25−2t25+2125)e2t5

17.7.6 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨−2y˙+2y=ettan⁡(t)

Answer

Aetsin⁡t+ Betcos⁡t− etcos⁡tln⁡|sec⁡t+ tan⁡t|

17.7.7 §17.7 source ↗

Find the general solution to the differential equation using variation of parameters.

y¨− 2y˙+ 2y= sin⁡(t)cos⁡(t) (This is rather messy when done by variation of parameters; compare to undetermined coefficients.)

Answer

Aetsin⁡t+ Betcos⁡t− 110cos⁡t(cos3⁡t+ 3sin3⁡t− 2cos⁡t− sin⁡t)+ 110sin⁡t(sin3⁡t− 3cos3⁡t− 2sin⁡t+ cos⁡t)= Aetsin⁡t+ Betcos⁡t+ 110cos⁡(2t)− 120sin⁡(2t)

End of the book.

Exercises and answers are adapted from Calculus: Early Transcendentals by David Guichard and friends, 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 online edition 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 David Guichard or Whitman College. The book was first written by David Guichard. Its chapters 1–9 modify and expand notes by Neal Koblitz of the University of Washington; it includes exercises and examples from H. Jerome Keisler’s Elementary Calculus: An Approach Using Infinitesimals, from which its chapter on differential equations is largely derived, and some exercises from the OpenStax Calculus books; and Albert Schueller, Barry Balof and Mike Wills contributed additional material.