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Introduction to differential equations

11 Introduction to differential equations

35 problems · hints, answers and solutions shown beside each one

Q1

A colony of bacteria is treated with a mild antibiotic agent so that the bacteria start to die. It is observed that the population of bacteria as a function of time follows the approximate relationship b(t)=85e−0.5tb(t) = 85 e^{-0.5 t} where tt is time in hours.

Determine the time it takes for half of the bacteria to die; this is called the half-life.

Find how long it takes for 99%99\% of the bacteria to die.

Answer

It takes 2log⁡2≈1.392\log 2 \approx 1.39 hours for half of the bacteria to die, and 2log⁡100≈9.22\log 100 \approx 9.2 hours for 99% of them to die.

Q2

A differential equation is an equation in which some function is related to its own derivative(s).

For each of the following functions, calculate the appropriate derivative, and show that the function satisfies the indicated differential equation.

  1. f(x)=2e−3x,   f′(x)=−3f(x)f(x) = 2 e^{-3x}, \ \ \ f'(x)= -3 f(x)

  2. f(t)=Cekt,   f′(t)=kf(t)f(t) = C e^{kt}, \ \ \ f'(t)= k f(t)

  3. f(t)=1−e−t,   f′(t)=1−f(t)f(t) = 1 - e^{-t}, \ \ \ f'(t)= 1-f(t)

Q3

Consider the function y=f(t)=Cekty = f(t) = Ce^{kt} where CC and kk are constants. For what value(s) of these constants does this function satisfy the equations below?

  1. dydt=−5y{dy \over dt} = - 5 y,

  2. dydt=3y{dy \over dt} = 3 y

Answer

  1. CC any value, k=−5k= -5

  2. CC any value, k=3k = 3

Q4

Check that the function

N(t)=N0ekt=N0e(r−m)tN(t)=N_{0} e^{kt}=N_0e^{(r-m)t}

satisfies the differential equation

dNdt=(r−m)N\diff{N}{t}=(r-m)N

and the initial condition N(0)=N0N(0)=N_0.

Q5

Find a function that satisfies each of the following differential equations.

Note: all your answers should be exponential functions, but they may have different dependent and independent variables. Question 2 may help.

  1. dydt=−y\displaystyle \frac{dy}{dt}= - y,

  2. dcdx=−0.1c\displaystyle \frac{dc}{dx}= -0.1 c and c(0)=20c(0)=20,

  3. dzdt=3z\displaystyle \frac{dz}{dt}= 3 z and z(0)=5z(0)=5.

Answer

  1. y(t)=Ce−t\displaystyle y(t) = C e^{-t};

  2. c(x)=20e−0.1x\displaystyle c(x) = 20 e^{-0.1 x};

  3. z(t)=5e3t\displaystyle z(t) = 5 e^{3t}.

Q6

The per capita birthrate of one species of rodent is 0.050.05 newborns per day. This means that, on average, each member of the population results in 55 newborn rodents every 100100 days. Suppose that over the period of 10001000 days there are no deaths, and that the initial population of rodents is 250250.

  1. Write a differential equation for the population size N(t)N(t) at time tt (in days).

  2. Write down the initial condition that NN satisfies.

  3. Find the solution, i.e. express NN as some function of time tt that satisfies your differential equation and initial condition.

  4. How many rodents are there after 11 year ?

Answer

  1. dNdt=0.05N\displaystyle {dN \over dt} = 0.05 N

  2. N(0)=250N(0)=250

  3. N(t)=250e0.05t\displaystyle N(t) = 250 e^{0.05 t}

  4. 2.1×1010\displaystyle 2.1 \times 10^{10} rodents

Q7

Suppose a population of bacteria starts from a single bacterium, and grows at a rate proportional to the number of bacteria in the population. Suppose further that it takes 20 minutes for the population to double.

Find the appropriate differential equation that describes this growth, the appropriate initial condition, and the exponential function that is the solution to that differential equation. Use units of hours for time tt.

Answer

The population y(t)y(t) after tt hours satisfies

dydt=ky , y(0)=1 , y(13)=2y(0)\diff{y}{t}=ky~,~y(0)=1~,~y\left(\frac13\right)=2y(0)

for some constant kk.

The solution to this initial value problem is

y(t)=e(3log⁡2)t=23t.y(t) = e^{(3\log 2)t}=2^{3t}.
Q8

In Canada, women have only about 2 children during their 4040 years of fertility, and people live to age 8080. In underdeveloped countries, people on average live to age 6060 and women have a child roughly every 44 years between ages 1313 and 4545.

Compare the per capita birth and mortality rates and the predicted population growth or decay in each of these scenarios, using arguments analogous to those of Section 11.4.2.

Find the growth rate kk in percent per year and the doubling time for the growing population.

Q9

A population of animals has a per-capita birth rate of b=0.08b=0.08 per year and a per-capita death rate of m=0.01m=0.01 per year. The population density, P(t)P(t) is found to satisfy the differential equation

dP(t)dt=bP(t)−mP(t)\frac{dP(t)}{dt}= b P(t) - m P(t)

  1. If the population is initially P(0)=1000P(0)=1000, find how big the population is in 55 years.

  2. When does the population double?

Answer

  1. P(5)=1000e0.35≈1419P(5)=1000e^{0.35} \approx 1419

  2. t=log⁡20.07≈9.9t=\frac{\log 2}{0.07} \approx 9.9 years

Q10

  1. The population y(t)y(t) of a certain microorganism grows continuously and follows an exponential behaviour over time. Its doubling time is found to be 0.270.27 hours. What differential equation would you use to describe its growth?

    Note: you must find the value of the rate constant, kk, using the doubling time.

  2. With exposure to ultra-violet radiation, the population ceases to grow, and the microorganisms continuously die off. It is found that the half-life is then 0.10.1 hours. What differential equation would now describe the population?

Answer

  1. dydt=(log⁡20.27)y\diff{y}{t}=\left(\frac{\log 2}{0.27}\right)y

  2. dydt=(−log⁡20.1)y\diff{y}{t}=\left(-\frac{\log 2}{0.1}\right)y

Q11

A bacterial population grows at a rate proportional to the population size at time t.t. Let y(t)y(t) be the population size at time t.t. By experiment it is determined that the population at t=10t=10 min is 15,00015,000 and at t=30t=30 min it is 20,000.20,000.

  1. What was the initial population?

  2. What is the population at time t=60t=60min?

Answer

  1. 75003≈129907500\sqrt3 \approx 12990

  2. 160,00031.5≈30792\frac{160,000}{3^1.5} \approx 30792 bacteria

Q12

Two populations are studied. Population 1 is found to obey the differential equation

dy1dt=0.2y1\diff{y_1}{t} = 0.2 y_{1}

and population 2 obeys

dy2dt=−0.3y2\diff{y_2}{t} = -0.3 y_{2}

where tt is time in years.

  1. Which population is growing and which is declining?

  2. Find the doubling time (respectively half-life)
    associated with the given population.

  3. If the initial levels of the two populations were y1(0)=100y_{1}(0)=100 and y2(0)=10,000y_{2}(0)=10,000, how big would each population be at time tt?

  4. At what time would the two populations be exactly equal?

Answer

  1. y1y_{1} growing, y2y_{2} declining

  2. y1y_1 has doubling time log⁡20.2≈3.5\frac{\log 2}{0.2} \approx 3.5 years; y2y_2 has half-life log⁡20.3≈2.3\frac{\log 2}{0.3}\approx 2.3 years

  3. y1(t)=100e0.2t\displaystyle y_{1}(t) = 100 e^{0.2 t}, y2(t)=10000e−0.3t\displaystyle y_{2}(t) = 10000 e^{-0.3 t}

  4. 2log⁡100≈9.22\log 100\approx 9.2 years

Q13

The human population on Earth
doubles roughly every 5050 years. In October 20002000 there were 6.16.1 billion humans on earth.

  1. Determine what the human population would be 500500 years later under the uncontrolled growth scenario.

  2. How many people would have to inhabit each square kilometer of the planet for this population to fit on earth? (Take the circumference of the earth to be 40,00040,000 km for the purpose of computing its surface area and assume that the oceans have dried up.)

Answer

  1. 62464000000006246400000000 people (6.2464×10126.2464\times 10^{12})

  2. 3904π≈122653904\pi \approx 12265 people per square km

Q14

Two lakes have populations of fish, but the conditions are quite different in these lakes. In the first lake, the fish population is growing and satisfies the differential equation

dydt=0.2y\diff{y}{t}=0.2 y

where tt is time in years. At time t=0t=0 there were 500500 fish in this lake. In the second lake, the population is dying due to pollution. Its population satisfies the differential equation

dydt=−0.1y,\diff{y}{t}=-0.1 y,

and initially there were 40004000 fish in this lake.

At what time are the fish populations in the two lakes identical?

Answer

103log⁡8≈6.93\frac{10}{3}\log 8\approx 6.93 years

Q15

When chemists say that a chemical reaction follows “first order kinetics”, they mean that the concentration of the reactant at time tt, i.e. c(t)c(t), satisfies an equation of the form dcdt=−rc{dc \over dt}= -rc where rr is a rate constant, here assumed to be positive. Suppose the reaction mixture initially has concentration 11M (“11 molar”) and that after 11 hour there is half this amount.

  1. Find the “half life” of the reactant.

  2. Find the value of the rate constant rr.

  3. Determine how much is left after 22 hours.

  4. When is only 10%10\% of the initial amount be left?

Answer

  1. 1 hour

  2. r=log⁡(2)r= \log(2)

  3. 0.25 M

  4. t=log⁡2(10)≈3.322t=\log_2(10)\approx 3.322 hours

Q16

In a chemical reaction, a substance SS is broken down. The concentration of the substance is observed to change at a rate proportional to the current concentration. It was observed that 11 Mole/litre of SS decreased to 0.50.5 Moles/litre in 1010 minutes.

  1. How long does it take until only 0.250.25 Moles per litre remain?

  2. How long does it take until only 1%1\% of the original concentration remains?

Answer

  1. 2020 min

  2. 20log⁡2(10)≈66.4420\log_2(10) \approx 66.44 min

Q17

If  10%10\% of a radioactive substance remains after one year, find its half-life.

Answer

τ=log⁡(2)log⁡(10)\displaystyle \tau = \frac{\log(2)}{\log(10)}

Q18

Carbon 1414, or 14C^{14}C, has a half-life of 57305730 years. This means that after 57305730 years, a sample of Carbon 1414, which is a radioactive isotope of carbon, has lost one half of its original radioactivity.

  1. Estimate how long it takes for the sample to fall to roughly 0.0010.001 of its original level of radioactivity.

  2. Each gram of 14C^{14}C has an activity given here in units of 12 decays per minute. After some time, the amount of radioactivity decreases. For example, a sample 57305730 years old has only one half the original activity level, i.e. 66 decays per minute. If a 1 gm sample of material is found to have 4545 decays per hour, approximately how old is it?

    Note: 14C^{14}C is used in radiocarbon dating, a process by which the age of materials containing carbon can be estimated. W. Libby received the Nobel prize in chemistry in 19601960 for developing this technique.

Answer

  1. 5730log⁡(10000log⁡(2)≈57100\frac{5730\log(10000}{\log(2)}\approx 57100 years

  2. 2292022920 years

Q19

Strontium-90 is a radioactive isotope with a half-life of 2929 years. If you begin with a sample of 800800 units, how long does it take for the amount of radioactivity of the strontium sample to be reduced to:

  1. 400400 units?

  2. 200200 units?

  3. 11 unit?

Answer

  1. 29 years

  2. 58 years

  3. 29log⁡(800)log⁡(2)≈279.7\frac{29\log(800)}{\log(2)}\approx279.7 years

Q20

Cobalt 60 is a radioactive substance with half life 5.35.3 years. It is used in medical applications (radiology). How long does it take for 80%80\% of a sample of this substance to decay?

Answer

5.3log⁡5log⁡2≈12.3\frac{5.3\log 5}{\log 2}\approx12.3 years

Q21

A barrel initially contains 2 kg2~kg of salt dissolved in 20 L20~L of water. If water flows in at the rate of 0.4 L0.4~L per minute and the well-mixed salt water solution flows out at the same rate, how much salt is present after 88 minutes?

Hint

If there are SS kg of salt in the entire barrel, then 0.4 litres of barrel water contains S⋅0.420S\cdot\frac{0.4}{20} kg of salt.

Answer

2e−8/50≈1.7043  kg2e^{-8/50}\approx 1.7043~\text{ kg}

Q22

Assume the atmospheric pressure yy at a height xx meters above the sea level satisfies the relation

dydx=ky\diff{y}{x}=ky

for some constant kk. If one day at a certain location the atmospheric pressures are 760760 and 675675 torr (unit for pressure) at sea level and at 10001000 meters above sea level, respectively, find the value of the atmospheric pressure at 600600 meters above sea level.

Answer

y=760(675760)3/5≈707.8 torry=760\left(\frac{675}{760}\right)^{3/5} \approx 707.8 {\rm ~torr}

Q23

Water draining from a container. In Example 11.2.3 in the text, we verified that the function h(t)=(h0−kt2)2h(t)=\left(\sqrt{h_0}-k\frac{t}{2}\right)^2 is a solution to the differential equation

dhdt=−kh.\diff{h}{t}=-k\sqrt{h}.

Based on the meaning of the problem, for how long does this solution remain valid?

Answer

Until t=2h0k\displaystyle t=\frac{2\sqrt{h_0}}{k}.

Q24

Verifying a solution. Verify that the function y(t)=1−(1−y0)e−ty(t) = 1-(1-y_{0}) e^{-t} satisfies the initial value problem (differential equation and initial condition)

dydt=1−y,y(0)=y0\diff{y}{t}=1-y,\qquad y(0)=y_0

(equation 11.4.3 in the text).

Q25

Linear differential equation. Consider the differential equation

dydt=a−by\displaystyle \frac{dy}{dt}= a - by

where aa, bb are constants.

  • Show that the function

    y(t)=ab−Ce−bt\displaystyle y(t) = \frac{a}{b} - C e^{-bt}

    satisfies the above differential equation for any constant CC.

  • Show that by setting

    C=ab−y0\displaystyle C= \frac{a}{b} - y_{0}

    we also satisfy the initial condition

    y(0)=y0.y(0)=y_{0}.

    Remark: you have shown that the function

    y(t)=(y0−ab)e−bt+ab\displaystyle y(t) = \left(y_{0} - {a \over b}\right)e^{-bt} + {a \over b}

    is a solution to the initial value problem (i.e differential equation plus initial condition)

    dydt=a−by,   y(0)=y0.{dy \over dt} = a - by, \ \ \ y(0)=y_{0}.
Q26

Verifying a solution. Show that the function

y(t)=11−ty(t)=\frac{1}{1-t}

is a solution to the differential equation and initial condition

dydt=y2,y(0)=1.\frac{dy}{dt}=y^2,\quad y(0)=1.

Comment on what happens to this solution as tt approaches 11.

Q27

Verifying solutions. For each of the following, show the given function yy is a solution to the given differential equation.

  • t⋅dydt=3y\displaystyle t \cdot \frac{dy}{dt} = 3y, y=2t3y=2t^{3}.

  • d2ydt2+y=0\displaystyle \frac{d^{2}y}{dt^{2}}+y=0, y=−2sin⁡t+3cos⁡ty=-2\sin t + 3\cos t.

  • d2ydt2−2dydt+y=6et\displaystyle \frac{d^{2}y}{dt^{2}}-2\frac{dy}{dt}+y=6e^{t}, y=3t2ety=3t^{2} e^{t}.

Q28

Verifying a solution. Show the function determined by the equation 2x2+xy−y2=C2x^{2}+xy-y^{2}=C, where CC is a constant and 2y≠x2y \ne x, is a solution to the differential equation (x−2y)dydx=−4x−y\displaystyle (x-2y) \frac{dy}{dx} = -4x-y.

Q29

Determining the constant. Find the constants CC, C1C_1, and/or C2C_2 that satisfy the given initial conditions.

  1. 2x2−3y2=C2x^{2}-3y^{2}=C, y∣x=0=2y |_{x=0} = 2.

  2. y=C1e5t+C2te5ty=C_{1} e^{5t} + C_{2} t e^{5t}, y∣t=0=1y |_{t=0} =1 and dydt∣t=0=0\diff{y}{t} |_{t=0} = 0.

  3. y=C1cos⁡(t−C2)y=C_{1} \cos (t - C_{2}), y∣t=π2=0y |_{t=\frac{\pi}{2}} =0 and dydt∣t=π2=1\diff{y}{t} |_{t=\frac{\pi}{2}} = 1.

Answer

  1. C=−12C=-12

  2. C1=1C_{1} = 1, C2=−5C_{2} = -5

  3. For any integer nn, we can have C2=πnC_2=\pi n, and $C_1=\begin{cases} 1 & \text{ if }n\text{ odd}\ -1 & \text{ if }n\text{ even}\ \end{cases}.Soforexample,oneoptionis. So for example, one option is C_{1}=-1,, C_{2} = 0;anotheroptionis; another option is C_1=1andandC_2=\pi$.

Q30

Checking a solution. Check that the differential equation (11.4.4) has the right sign, so that a hot object cools off in a colder environment.

That is:

dTdt=k(E−T(t)),where k>0\diff{T}{t}=k(E-T(t)),\quad\text{where } k>0

describes the change in temperature TT of an object over time, where EE is the (constant) temperature of the environment, and kk is a constant.

Q31

Infant weight gain. During the first year of its life, the weight of a baby is given by

y(t)=3t+64y(t) = \sqrt{3t + 64}

where tt is measured in some convenient unit.

  1. Show that yy satisfies the differential equation

    dydt=ky\frac{dy}{dt} = \frac{k}{y}

    where kk is some positive constant.

  2. What is the value for kk?

  3. Suppose we adopt this differential equation as a model for human growth.
    State concisely (that is, in one sentence) one feature about this differential equation which makes it a reasonable model.
    State one feature which makes it unreasonable.

Answer

(b) k=3/2k=3/2.

Q32

Lake Fishing. Fish Unlimited is a company that manages the fish population in a private lake. They restock the lake (that is, they add fish to the lake) at constant rate.  NN fishers are allowed
to fish in the lake per day. The population of fish in the lake, F(t)F(t) is found to satisfy the differential equation

dFdt=I−αNF\begin{equation}\frac{dF}{dt}=I -\alpha NF \end{equation}

where FF is measured in individual fish, and tt is measured in days.

  1. At what rate are fish added per day according to Eqn. (11.0.1)? Give both value and units.

  2. What is the average number of fish caught by one fisher? Give both the value and units.

  3. What is being assumed about the fish birth and mortality rates in Eqn. (11.0.1)?

  4. If the fish input and number of fishers are constant, what is the steady state level of the fish population in the lake?

  5. At time t=0t=0 the company stops restocking the lake with fish. Give the revised form of the differential equation (11.0.1) that takes this into account, assuming the same level of fishing as before. How long would it take for the fish to fall to 25%25\% of their initial level?

  6. When the fish population drops to the level FlowF_{low}, fishing is stopped and the lake is restocked with fish at the same constant rate (Eqn (11.0.1), with α=0\alpha=0.) Write down the revised version of Eqn. (11.0.1) that takes this into account. How long would it take for the fish population to double?

Answer

  1. Input rate is II fish per day

  2. αF\alpha F fish caught per fisher per day.

  3. Birth and mortality are neglected, or assumed to exactly cancel out.

  4. Steady state level F=I/αNF=I/\alpha N

  5. New differential equation is dFdt=−αNF\diff{F}{t}=-\alpha NF; it would take 2log⁡(2)/αN2\log(2)/\alpha N days for the population to fall to 25% of its initial level.

  6. New differential equation is dFdt=I\diff{F}{t}=I; it would take t=Flow/It=F_{low}/I days to double.

Q33

Tissue culture. Cells in a tissue culture produce a cytokine (a chemical that controls the growth of other cells) at a constant rate of 1010 nano-Moles per hour (nM/h). The chemical has a half-life of 20 hours.

Give a differential equation (DE) that describes this chemical production and decay. Solve this DE assuming that at t=0t=0 there is no cytokine. [1nM=10−910^{-9}M].

Q34

Determining constants. Find those constants a,ba,b so that y=exy=e^{x} and y=e−xy=e^{-x} are both solutions of the differential equation

y′′+ay′+by=0.y^{\prime\prime}+ay^{\prime}+by=0.
Answer

a=0a=0, b=−1b=-1

Q35

Biochemical reaction. A biochemical reaction in which a substance SS is both produced and consumed is investigated. The concentration c(t)c(t) of SS changes during the reaction, and is seen to follow the differential equation

dcdt=Kmaxck+c−rc\frac{dc}{dt} = K_{\text{max}} \frac{c}{k+c} - r c

where Kmax,k,rK_{\text{max}}, k, r are positive constants with certain convenient units. The first term is a concentration-dependent production term and the second term represents consumption of the substance.

  1. What is the maximal rate at which the substance is produced? At what concentration is the production rate 50%50\% of this maximal value?

  2. If the production is turned off, the substance decays. How long would it take for the concentration to drop by 50%50\%?

  3. At what concentration does the production rate just balance the consumption rate?

Answer

  1. KmaxK_{\text{max}}, c=kc=k

  2. ln⁡(2)/r\ln(2)/r

  3. c=0c=0, c=Kmaxr−k\displaystyle c = \frac{K_{\text{max}}}{r} -k

Source

Questions 1 - 22 are adapted from Keshet, Chapter 11. The selected answers provided here are used with permission. Many answers are given in exact terms, e.g. “103log⁡8\frac{10}{3}\log 8," as well as with decimal approximations. You should be able to solve all questions exactly without use of a calculator, unless otherwise instructed; the decimal approximations are included because they are easier to understand in the context of a model.

Question 35 is from Keshet, Chapter 13. The remaining questions are from Keshet, Chapter 12.

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From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.