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Modeling with differential equations

We have seen several ways that differential equations arise in the world around us, from the growth of a population to the temperature of a cup of coffee.

Introduction

We have seen several ways that differential equations arise in the world around us, from the growth of a population to the temperature of a cup of coffee. In this section, we look more closely at how differential equations give us a natural way to describe various phenoma. As we'll see, the key is to understand the different factors that cause a quantity to change.

Exploration

Developing a differential equation

Preview Activity demonstrates the kind of thinking we will be doing in this section. In each of the two examples we considered, there is a quantity, such as the amount of money in the bank account or the amount of salt in the tank, that is changing due to several factors. The governing differential equation states that the total rate of change is the difference between the rate of increase and the rate of decrease.

There are many important lessons to learn from Example. Foremost is how we can develop a differential equation by thinking about the total rate = rate in - rate out model. In addition, we note how we can bring together all of our available understanding (plotting \(\frac{dP}{dt}\) vs. \(P\), creating a slope field, solving the differential equation) to see how the differential equation describes the behavior of a changing quantity.

We can also explore what happens when certain aspects of the problem change. For instance, let's suppose we are at a time when the plastic pollution entering Lake Michigan has stabilized at \(1.5\cdot10^5\) cubic meters, and that new legislation is passed to prevent this type of pollution entering the lake. So, there is no longer any inflow of plastic pollution to the lake. How does the amount of plastic pollution in Lake Michigan now change? For example, how long does it take for the amount of plastic pollution in the lake to halve?

Resetting \(t=0\) at this time, we now have the initial value problem \[\begin{aligned}\end{aligned}\].

It is a straightforward and familiar exercise to find that the solution to this equation is \(P(t) = 1.5\cdot10^5 e^{-t/100}\). The time that it takes for half of the pollution to flow out of the lake is given by \(T\) where \(P(T) = 0.75\cdot10^5\). Thus, we must solve the equation \[\begin{aligned}\end{aligned}\], or \[\begin{aligned}\end{aligned}\].

It follows that \[\begin{aligned}\end{aligned}\]

In the upcoming activities, we explore some other natural settings in which differential equations model changing quantities.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

Differential equations arise in a situation when we understand how various factors cause a quantity to change.
We may use the tools we have developed so farslope fields, Euler's methods, and our method for solving separable equationsto understand a quantity described by a differential equation.

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn 5% each year.
1. Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to growfirst, you are depositing one millon dollars per year and second, you are earning 5% interest.
2. If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years?
3. The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years?
4. Do you prefer the first or second option? Explain your thinking.
5. At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?
Одкриј го одговорот
1. Let \(A(t)\) be the amount of money (in millions) in the account at time \(t\). We know that the rate of change of \(A\) is affected by the addition of \(1\) million dollars a year, plus the interest earned, which is \(0.05A\). Thus, \[\begin{aligned}\end{aligned}\]
2. If there is no amount of money in the account when you open it, that means \(A(0) = 0\). Solving the initial value problem, we can find how much money will be in the account after 25 years. Note that \(1 + 0.05A = 0.05(20 + A)\), so we have \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Integrating (and noting that \(A(t) > 0\) for all \(t\), \[\begin{aligned}\end{aligned}\], so solving for \(A\) in the standard way, \[\begin{aligned}\end{aligned}\] and thus \(A(t) = Ke^{0.05t} - 20\). Since \(A(0) = 0\), \(K = 20\), and we've shown \[\begin{aligned}\end{aligned}\]. It follows that \(A(25) = 49.80686\) million dollars.
3. For this option, we know that \(\frac{dA}{dt} = 0.05A\), \(A(0) = 10\), and thus by the standard solution to this IVP, \[\begin{aligned}\end{aligned}\] from which we see that \(A(25) = 34.90343\) million dollars.
4. The first option is obviously better because it leads to more money in the long term.
5. The amount of money in the account under the first option overtakes the amount of money in the account under the second option when \(20(e^{0.05t} - 1) = 10e^{0.05t}\). Solving for \(t\), we find that \(10e^{0.05t} = 20\), so \(e^{0.05t} = 2\), and thus \[\begin{aligned}\end{aligned}\].
When a skydiver jumps from a plane, gravity causes her downward velocity to increase at the rate of \(g\approx 9.8\) meters per second squared. At the same time, wind resistance causes her velocity to decrease at a rate proportional to the velocity.
1. Using \(k\) to represent the constant of proportionality, write a differential equation that describes the rate of change of the skydiver's velocity.
2. Find any equilibrium solutions and decide whether they are stable or unstable. Your result should depend on \(k\).
3. Suppose that the initial velocity is zero. Find the velocity \(v(t)\).
4. A typical terminal velocity for a skydiver falling face down is 54 meters per second. What is the value of \(k\) for this skydiver?
5. How long does it take to reach 50% of the terminal velocity?
Одкриј го одговорот
1. Let the skydiver's downward velocity at time \(t\) be given by \(v(t)\). Since gravity causes her downward velocity to increase at the rate of \(g\approx 9.8\), there is a factor of \(+9.8\) meters per second squared in the differential equation for \(\frac{dv}{dt}\), while the factor of wind resistence causing the velocity to decrease contributes the negative factor \(-kv\). Thus, the differential equation is \[\begin{aligned}\end{aligned}\]
2. To find the equilibrium solutions, we set \(\frac{dv}{dt} = 0\) and solve for \(v\). Doing so, \(0 = 9.8 - kv\), so \(v = \frac{9.8}{k}\). Moreover, we note that for \(v \lt \frac{9.8}{k}\), \(\frac{dv}{dt} \gt 0\), while for \(\frac{dv}{dt} \gt \frac{9.8}{k}\), \(\frac{dv}{dt} \lt 0\). Thus, for velocities just above the equilibrium, the velocity will decrease and for velocities just below the equilibrium, the velocity will decrease, making \(v = \frac{9.8}{k}\) a stable equilibrium.
3. We assume that \(v(0) = 0\) and solve the IVP with the differential equation we developed. Taking the differential equation \[\begin{aligned}\end{aligned}\] and separating the variables, we have \[\begin{aligned}\end{aligned}\]. Integrating with respect to \(t\), \[\begin{aligned}\end{aligned}\], and using a standard substitution, we have \[\begin{aligned}\end{aligned}\]. Solving for \(v\), we have \(\ln | 9.8 - kv | = -kt + C_1\) (where \(C_1 = -Ck\), or \[\begin{aligned}\end{aligned}\]. Incorporating the absolute value into the constant \(C_2\) in the usual way, we have for some constant \(A\) that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Setting \(v(0) = 0\), we can also determine \(A\), since then \[\begin{aligned}\end{aligned}\] so \(A = 9.8\). The solution to the IVP is therefore \[\begin{aligned}\end{aligned}\]
4. If the skydiver's terminal velocity is 54 meters per second, this means that \[\begin{aligned}\end{aligned}\] Using the solution we found in (c), we observe that \[\begin{aligned}\end{aligned}\] since \(e^{-kt} \to 0\) as \(t \to \infty\). Thus, \(54 = \frac{9.8}{k}\), so \(k = 9.8/54 \approx 0.181481\).
5. To find how long does it take to reach 50% of the terminal velocity, we set \(v(t) = \frac{9.8 - 9.8e^{-0.181481t}}{0.181481} = 0.5(54)\) and solve for \(t\). We have \[\begin{aligned}\end{aligned}\], so \[\begin{aligned}\end{aligned}\], and thus \(9.8e^{-0.181481t} = 4.9\), which tells us \[\begin{aligned}\end{aligned}\]. Solving for \(t\), \(t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894\) seconds.
During the first few years of life, the rate at which a baby gains weight is proportional to the reciprocal of its weight.
1. Express this fact as a differential equation.
2. Suppose that a baby weighs 8 pounds at birth and 9 pounds one month later. How much will he weigh at one year?
3. Do you think this is a realistic model for a long time?
Одкриј го одговорот
1. Since the baby's weight, \(w(t)\) (in pounds at month \(t\)), grows at a rate proportional to its reciprocal, we have \[\begin{aligned}\end{aligned}\].
2. Given that \(W(0) = 8\) and \(W(1) = 9\), we can both solve the IVP and determine \(k\). To start, we separate variables and solve the differential equation by writing \[\begin{aligned}\end{aligned}\] and integrating to get \[\begin{aligned}\end{aligned}\]. It follows that \[\begin{aligned}\end{aligned}\] and therefore \(w(t) = \sqrt{2kt + C_1}\) (where \(C_1 = 2C\) and we take the positive square root because we know \(w(t) \gt 0\)). Applying the initial condition \(w(0) = 8\), we find that \(8 = \sqrt{C_1}\), so \(C_1 = 64\) and \[\begin{aligned}\end{aligned}\]. To find \(k\), we use the additional information that \(w(1) = 9\), so \[\begin{aligned}\end{aligned}\] which implies that \(81 = 2k + 64\) and thus \(k = 17/2\). Thus, \[\begin{aligned}\end{aligned}\]. At one year, according to the model the baby weighs \(w(12) = \sqrt{268} \approx 16.37\) pounds.
3. While the function \(w(t) = \sqrt{7t+64}\) grows without bound as \(t\) increases, the model is unrealistic more because the function initially doesn't grow fast enough. For example, the model predicts that at age \(10\), \(w(120) \approx 30.06\) pounds, and at age \(20\), \(w(120) = 41.76\) pounds. A realistic model would need to grow much more quickly early in life and then level off around age \(20\).
Congratulations, you just won the lottery! In one option presented to you, you will be paid one million dollars a year for the next 25 years. You can deposit this money in an account that will earn 5% each year.
1. Set up a differential equation that describes the rate of change in the amount of money in the account. Two factors cause the amount to growfirst, you are depositing one millon dollars per year and second, you are earning 5% interest.
2. If there is no amount of money in the account when you open it, how much money will you have in the account after 25 years?
3. The second option presented to you is to take a lump sum of 10 million dollars, which you will deposit into a similar account. How much money will you have in that account after 25 years?
4. Do you prefer the first or second option? Explain your thinking.
5. At what time does the amount of money in the account under the first option overtake the amount of money in the account under the second option?
Одкриј го одговорот
1. Let \(A(t)\) be the amount of money (in millions) in the account at time \(t\). We know that the rate of change of \(A\) is affected by the addition of \(1\) million dollars a year, plus the interest earned, which is \(0.05A\). Thus, \[\begin{aligned}\end{aligned}\]
2. If there is no amount of money in the account when you open it, that means \(A(0) = 0\). Solving the initial value problem, we can find how much money will be in the account after 25 years. Note that \(1 + 0.05A = 0.05(20 + A)\), so we have \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Integrating (and noting that \(A(t) > 0\) for all \(t\), \[\begin{aligned}\end{aligned}\], so solving for \(A\) in the standard way, \[\begin{aligned}\end{aligned}\] and thus \(A(t) = Ke^{0.05t} - 20\). Since \(A(0) = 0\), \(K = 20\), and we've shown \[\begin{aligned}\end{aligned}\]. It follows that \(A(25) = 49.80686\) million dollars.
3. For this option, we know that \(\frac{dA}{dt} = 0.05A\), \(A(0) = 10\), and thus by the standard solution to this IVP, \[\begin{aligned}\end{aligned}\] from which we see that \(A(25) = 34.90343\) million dollars.
4. The first option is obviously better because it leads to more money in the long term.
5. The amount of money in the account under the first option overtakes the amount of money in the account under the second option when \(20(e^{0.05t} - 1) = 10e^{0.05t}\). Solving for \(t\), we find that \(10e^{0.05t} = 20\), so \(e^{0.05t} = 2\), and thus \[\begin{aligned}\end{aligned}\].
When a skydiver jumps from a plane, gravity causes her downward velocity to increase at the rate of \(g\approx 9.8\) meters per second squared. At the same time, wind resistance causes her velocity to decrease at a rate proportional to the velocity.
1. Using \(k\) to represent the constant of proportionality, write a differential equation that describes the rate of change of the skydiver's velocity.
2. Find any equilibrium solutions and decide whether they are stable or unstable. Your result should depend on \(k\).
3. Suppose that the initial velocity is zero. Find the velocity \(v(t)\).
4. A typical terminal velocity for a skydiver falling face down is 54 meters per second. What is the value of \(k\) for this skydiver?
5. How long does it take to reach 50% of the terminal velocity?
Одкриј го одговорот
1. Let the skydiver's downward velocity at time \(t\) be given by \(v(t)\). Since gravity causes her downward velocity to increase at the rate of \(g\approx 9.8\), there is a factor of \(+9.8\) meters per second squared in the differential equation for \(\frac{dv}{dt}\), while the factor of wind resistence causing the velocity to decrease contributes the negative factor \(-kv\). Thus, the differential equation is \[\begin{aligned}\end{aligned}\]
2. To find the equilibrium solutions, we set \(\frac{dv}{dt} = 0\) and solve for \(v\). Doing so, \(0 = 9.8 - kv\), so \(v = \frac{9.8}{k}\). Moreover, we note that for \(v \lt \frac{9.8}{k}\), \(\frac{dv}{dt} \gt 0\), while for \(\frac{dv}{dt} \gt \frac{9.8}{k}\), \(\frac{dv}{dt} \lt 0\). Thus, for velocities just above the equilibrium, the velocity will decrease and for velocities just below the equilibrium, the velocity will decrease, making \(v = \frac{9.8}{k}\) a stable equilibrium.
3. We assume that \(v(0) = 0\) and solve the IVP with the differential equation we developed. Taking the differential equation \[\begin{aligned}\end{aligned}\] and separating the variables, we have \[\begin{aligned}\end{aligned}\]. Integrating with respect to \(t\), \[\begin{aligned}\end{aligned}\], and using a standard substitution, we have \[\begin{aligned}\end{aligned}\]. Solving for \(v\), we have \(\ln | 9.8 - kv | = -kt + C_1\) (where \(C_1 = -Ck\), or \[\begin{aligned}\end{aligned}\]. Incorporating the absolute value into the constant \(C_2\) in the usual way, we have for some constant \(A\) that \[\begin{aligned}\end{aligned}\] and thus \[\begin{aligned}\end{aligned}\]. Setting \(v(0) = 0\), we can also determine \(A\), since then \[\begin{aligned}\end{aligned}\] so \(A = 9.8\). The solution to the IVP is therefore \[\begin{aligned}\end{aligned}\]
4. If the skydiver's terminal velocity is 54 meters per second, this means that \[\begin{aligned}\end{aligned}\] Using the solution we found in (c), we observe that \[\begin{aligned}\end{aligned}\] since \(e^{-kt} \to 0\) as \(t \to \infty\). Thus, \(54 = \frac{9.8}{k}\), so \(k = 9.8/54 \approx 0.181481\).
5. To find how long does it take to reach 50% of the terminal velocity, we set \(v(t) = \frac{9.8 - 9.8e^{-0.181481t}}{0.181481} = 0.5(54)\) and solve for \(t\). We have \[\begin{aligned}\end{aligned}\], so \[\begin{aligned}\end{aligned}\], and thus \(9.8e^{-0.181481t} = 4.9\), which tells us \[\begin{aligned}\end{aligned}\]. Solving for \(t\), \(t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894\) seconds.
During the first few years of life, the rate at which a baby gains weight is proportional to the reciprocal of its weight.
1. Express this fact as a differential equation.
2. Suppose that a baby weighs 8 pounds at birth and 9 pounds one month later. How much will he weigh at one year?
3. Do you think this is a realistic model for a long time?
Одкриј го одговорот
1. Since the baby's weight, \(w(t)\) (in pounds at month \(t\)), grows at a rate proportional to its reciprocal, we have \[\begin{aligned}\end{aligned}\].
2. Given that \(W(0) = 8\) and \(W(1) = 9\), we can both solve the IVP and determine \(k\). To start, we separate variables and solve the differential equation by writing \[\begin{aligned}\end{aligned}\] and integrating to get \[\begin{aligned}\end{aligned}\]. It follows that \[\begin{aligned}\end{aligned}\] and therefore \(w(t) = \sqrt{2kt + C_1}\) (where \(C_1 = 2C\) and we take the positive square root because we know \(w(t) \gt 0\)). Applying the initial condition \(w(0) = 8\), we find that \(8 = \sqrt{C_1}\), so \(C_1 = 64\) and \[\begin{aligned}\end{aligned}\]. To find \(k\), we use the additional information that \(w(1) = 9\), so \[\begin{aligned}\end{aligned}\] which implies that \(81 = 2k + 64\) and thus \(k = 17/2\). Thus, \[\begin{aligned}\end{aligned}\]. At one year, according to the model the baby weighs \(w(12) = \sqrt{268} \approx 16.37\) pounds.
3. While the function \(w(t) = \sqrt{7t+64}\) grows without bound as \(t\) increases, the model is unrealistic more because the function initially doesn't grow fast enough. For example, the model predicts that at age \(10\), \(w(120) \approx 30.06\) pounds, and at age \(20\), \(w(120) = 41.76\) pounds. A realistic model would need to grow much more quickly early in life and then level off around age \(20\).

Symbols used here

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\approx

approximately equal
Equal to the precision shown, not exactly.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

\frac{\partial u}{\partial t},\ \nabla^2 u

partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Modeling with differential equations

How can we use differential equations to describe phenomena in the world around us?
How can we use differential equations to better understand these phenomena?

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

Обиди се со себе.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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