Section 7.1 An Introduction to Differential Equations
¶Motivating Questions
What is a differential equation and what kinds of information can it tell us?
How do differential equations arise in the world around us?
What do we mean by a solution to a differential equation?
In previous chapters, we have seen that a function's derivative tells us the rate at which the function is changing. The Fundamental Theorem of Calculus helped us determine the total change of a function over an interval from the function's rate of change. For instance, an object's velocity tells us the rate of change of that object's position. By integrating the velocity over a time interval, we can determine how much the position changes over that time interval. If we know where the object is at the beginning of that interval, we have enough information to predict where it will be at the end of the interval.
In this chapter, we introduce the concept of differential equations. A differential equation is an equation that provides a description of a function's derivative, which means that it tells us the function's rate of change. Using this information, we would like to learn as much as possible about the function itself. Ideally we would like to have an algebraic description of the function. As we'll see, this may be too much to ask in some situations, but we will still be able to make accurate approximations.
Preview Activity 7.1.1.
The position of a moving object is given by the function \(s(t)\text{,}\) where \(s\) is measured in feet and \(t\) in seconds. We determine that the velocity is \(v(t) = 4t + 1\) feet per second.
How much does the position change over the time interval \([0,4]\text{?}\)
Does this give you enough information to determine \(s(4)\text{,}\) the position at time \(t=4\text{?}\) If so, what is \(s(4)\text{?}\) If not, what additional information would you need to know to determine \(s(4)\text{?}\)
Suppose you are told that the object's initial position \(s(0) = 7\text{.}\) Determine \(s(2)\text{,}\) the object's position 2 seconds later.
If you are told instead that the object's initial position is \(s(0) = 3\text{,}\) what is \(s(2)\text{?}\)
If we only know the velocity \(v(t)=4t+1\text{,}\) is it possible that the object's position at all times is \(s(t) = 2t^2 + t - 4\text{?}\) Explain how you know.
Are there other possibilities for \(s(t)\text{?}\) If so, what are they?
If, in addition to knowing the velocity function is \(v(t) = 4t+1\text{,}\) we know the initial position \(s(0)\text{,}\) how many possibilities are there for \(s(t)\text{?}\)
Subsection 7.1.1 What is a differential equation?
A differential equation is an equation that describes the derivative, or derivatives, of a function that is unknown to us. For instance, the equation
describes the derivative of a function \(y(x)\) that is unknown to us.
As many important examples of differential equations involve quantities that change in time, the independent variable in our discussion will frequently be time \(t\text{.}\) In the preview activity, we considered the differential equation
Knowing the velocity and the starting position of a moving object, we were able to find its position at any later time.
Because differential equations describe the derivative of a function, they give us information about how that function changes. Our goal will be to use this information to predict the value of the function in the future; in this way, differential equations provide us with something like a crystal ball.
Differential equations arise frequently in our every day world. For instance, you may hear a bank advertising:
Your money will grow at a 3% annual interest rate with us.
This innocuous statement is really a differential equation. Let's translate: \(A(t)\) will be amount of money you have in your account at time \(t\text{.}\) The rate at which your money grows is the derivative \(dA/dt\text{,}\) and we are told that this rate is \(0.03 A\text{.}\) This leads to the differential equation
This differential equation has a slightly different feel than the previous equation \(\frac{ds}{dt} = 4t+1\text{.}\) In the earlier example, the rate of change depends only on the independent variable \(t\text{,}\) and we may find \(s(t)\) by integrating the velocity \(4t+1\text{.}\) In the banking example, however, the rate of change depends on the dependent variable \(A\text{,}\) so we'll need some new techniques in order to find \(A(t)\text{.}\)
Activity 7.1.2.
Express the following statements as differential equations. In each case, you will need to introduce notation to describe the important quantities in the statement so be sure to clearly state what your notation means.
The population of a town grows continuously at an annual rate of 1.25%.
A radioactive sample loses 5.6% of its mass every day.
You have a bank account that continuously earns 4% interest every year. At the same time, you withdraw money continually from the account at the rate of $1000 per year.
A cup of hot chocolate is sitting in a 70\(^\circ\) room. The temperature of the hot chocolate cools continuously by 10% of the difference between the hot chocolate's temperature and the room temperature every minute.
A can of cold soda is sitting in a 70\(^\circ\) room. The temperature of the soda warms continuously at the rate of 10% of the difference between the soda's temperature and the room's temperature every minute.
Small hints for each of the prompts above.
Let \(P\) be the population \(t\) the time in years; \(\frac{dP}{dt} = 0.0125P\text{.}\)
Let \(m\) be the mass \(t\) the time in days; \(\frac{dm}{dt} = -0.056m\text{.}\)
Let \(B\) be the balance \(t\) be time in years; \(\frac{dB}{dt} = 0.04B - 1000\text{.}\)
Let \(t\) be time in minutes \(H\) the temperature of the hot chocolate; \(\frac{dH}{dt} = -0.1(H - 70)\text{.}\)
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Let \(t\) be time in minutes and \(H\) the temperature of the soda;
\begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}
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Let \(P\) be the population of the town and let \(t\) be the time in years. The population of the town grows continuously at an annual rate of \(1.25\%\) means that
\begin{equation*} \frac{dP}{dt} = 0.0125P\text{.} \end{equation*} -
Let \(m\) be the mass of a radioactive sample and let \(t\) be the time in days. The sample loses 5.6% f its mass every day means that
\begin{equation*} \frac{dm}{dt} = -0.056m\text{.} \end{equation*} -
Let \(B\) be the balance (in dollars) in the account and let \(t\) be the time in years. The money in the account continuously earns 4% interest every year and $1000 is withdrawn every year means that
\begin{equation*} \frac{dB}{dt} = 0.04B - 1000\text{.} \end{equation*} -
The temperature of a room is \(70^\circ\text{.}\) Let \(t\) be time in minutes. The temperature \(H\) of a cup of hot chocolate cools continuosly by 10% of the difference between the hot chocolate's temperature and the room temperature every minute means that
\begin{equation*} \frac{dH}{dt} = -0.1(H - 70)\text{.} \end{equation*} -
The temperature of a room is \(70^\circ\text{.}\) Let \(t\) be time in minutes. The temperature \(H\) of a can of soda warms continuosly by 10% of the difference between the soda's temperature and the room temperature every minute means that
\begin{equation*} \frac{dH}{dt} = 0.1(70 - H) = -0.1(H - 70)\text{.} \end{equation*}
Subsection 7.1.2 Differential equations in the world around us
Differential equations give a natural way to describe phenomena we see in the real world. For instance, physical principles are frequently expressed as a description of how a quantity changes. A good example is Newton's Second Law, which says:
The product of an object's mass and acceleration equals the force applied to it.
For instance, when gravity acts on an object near the earth's surface, it exerts a force equal to \(mg\text{,}\) the mass of the object times the gravitational constant \(g\text{.}\) We therefore have
where \(v\) is the velocity of the object, and \(g = 9.8\) meters per second squared. Notice that this physical principle does not tell us what the object's velocity is, but rather how the object's velocity changes.
Activity 7.1.3.
Shown below are two graphs depicting the velocity of falling objects. On the left is the velocity of a skydiver, while on the right is the velocity of a meteorite entering the Earth's atmosphere.
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Begin with the skydiver's velocity and use the given graph to measure the rate of change \(dv/dt\) when the velocity is \(v=0.5, 1.0, 1.5, 2.0\text{,}\) and \(2.5\text{.}\) Plot your values on the graph below. You will want to think carefully about this: you are plotting the derivative \(dv/dt\) as a function of velocity.
Now do the same thing with the meteorite's velocity: use the given graph to measure the rate of change \(dv/dt\) when the velocity is \(v=3.5,4.0,4.5\text{,}\) and \(5.0\text{.}\) Plot your values on the graph above.
You should find that all your points lie on a line. Write the equation of this line being careful to use proper notation for the quantities on the horizontal and vertical axes.
The relationship you just found is a differential equation. Write a complete sentence that explains its meaning.
By looking at the differential equation, determine the values of the velocity for which the velocity increases.
By looking at the differential equation, determine the values of the velocity for which the velocity decreases.
By looking at the differential equation, determine the values of the velocity for which the velocity remains constant.
Hints for each of the prompts above.
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For the skydiver:
\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*} -
For the meteorite:
\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp \approx -0.3 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp \approx -0.6\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp \approx -0.9 \amp \left. \frac{dv}{dt}\right|_{(v = 5)} \amp \approx -1.2 \end{align*}A graph of the points from parts (a) and (b) is shown in the following diagram:
\(\frac{dv}{dt} = -0.6v + 1.8\text{.}\)
The rate of change of velocity with respect to time is a linear function of velocity.
\(0 \lt v \lt 3\text{.}\)
\(3 \lt v \lt 5\text{.}\)
\(v = 3\text{.}\)
Each graph is a graph of velocity \(v\) as a function of time \(t\text{.}\) To estimate \(\frac{dv}{dt}\) at a particular time, we must estimate the slope of the tangent line for the graph of \(v\) at that point. For parts (a) and (b), we draw tangent lines and measure the slopes of the tangent lines. These values are approximations.
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For the skydiver:
\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 0.5)} \amp \approx 1.5 \amp \left. \frac{dv}{dt}\right|_{(v = 1)} \amp \approx 1.2 \amp \left. \frac{dv}{dt}\right|_{(v = 1.5 )} \amp \approx 0.9\\ \left. \frac{dv}{dt}\right|_{(v = 2)} \amp \approx 0.6 \amp \left. \frac{dv}{dt}\right|_{(v = 2.5)} \amp \approx 0.3 \end{align*} -
For the meteorite:
\begin{align*} \left. \frac{dv}{dt}\right|_{(v = 3.5)} \amp \approx -0.3 \amp \left. \frac{dv}{dt}\right|_{(v = 4)} \amp \approx -0.6\\ \left. \frac{dv}{dt}\right|_{(v = 4.5)} \amp \approx -0.9 \amp \left. \frac{dv}{dt}\right|_{(v = 5)} \amp \approx -1.2 \end{align*}A graph of the points from parts (a) and (b) is shown in the following diagram:
Using the values in parts (a) and (b), we obtain \(\frac{dv}{dt} = -0.6v + 1.8\text{.}\)
This differential equation means that the rate of change of velocity with respect to time is a linear function of velocity.
The velocity increases when \(\frac{dv}{dt} \gt 0\text{.}\) So the velocity increases when \(0 \lt v \lt 3\text{.}\)
The velocity decreases when \(\frac{dv}{dt} \lt 0\text{.}\) So the velocity decreases when \(3 \lt v \lt 5\text{.}\)
The velocity will remain constant when \(\frac{dv}{dt} = 0\text{,}\) which is when \(v = 3\text{.}\)
The point of this activity is to demonstrate how differential equations model processes in the real world. In this example, two factors influence the velocities: gravity and wind resistance. The differential equation describes how these factors influence the rate of change of the velocities.
Subsection 7.1.3 Solving a differential equation
A differential equation describes the derivative, or derivatives, of a function that is unknown to us. By a solution to a differential equation, we mean simply a function that satisies this description.
For instance, the first differential equation we looked at is
which describes an unknown function \(s(t)\text{.}\) We may check that \(s(t) = 2t^2+t\) is a solution because it satisfies this description. Notice that \(s(t) = 2t^2+t+4\) is also a solution.
If we have a candidate for a solution, it is straightforward to check whether it is a solution or not. Before we demonstrate, however, let's consider the same issue in a simpler context. Suppose we are given the equation \(2x^2 - 2x = 2x+6\) and asked whether \(x=3\) is a solution. To answer this question, we could rewrite the variable \(x\) in the equation with the symbol \(\Box\text{:}\)
To determine whether \(x=3\) is a solution, we can investigate the value of each side of the equation separately when the value \(3\) is placed in \(\Box\) and see if indeed the two resulting values are equal. Doing so, we observe that
and
Therefore, \(x=3\) is indeed a solution.
We will do the same thing with differential equations. Consider the differential equation
Let's ask whether \(v(t) = 3 - 2e^{-0.5t}\) is a solution 1 . Using this formula for \(v\text{,}\) observe first that
and
Since \(\frac{dv}{dt}\) and \(1.5 - 0.5v\) agree for all values of \(t\) when \(v = 3-2e^{-0.5t}\text{,}\) we have indeed found a solution to the differential equation.
Activity 7.1.4.
Consider the differential equation
Which of the following functions are solutions of this differential equation?
\(v(t) = 1.5t - 0.25t^2\text{.}\)
\(v(t) = 3 + 2e^{-0.5t}\text{.}\)
\(v(t) = 3\text{.}\)
\(v(t) = 3 + Ce^{-0.5t}\) where \(C\) is any constant.
hints for each of the prompts above.
\(v(t) = 1.5t - 0.25t^2\) is not a solution to the given DE.
\(v(t) = 3 + 2e^{-0.5t}\) is a solution to the given DE.
\(v(t) = 3\) is a solution to the given DE.
\(v(t) = 3 + Ce^{-0.5t}\) is a solution to the given DE for any choice of \(C\text{.}\)
Since \(v(t) = 1.5t - 0.25t^2\text{,}\) \(\frac{dv}{dt} = 1.5 - 0.5t\text{.}\) We observe that for this choice of \(v\text{,}\) \(1.5 - 0.5v = 1.5 - 0.5(1.5t - 0.25t^2) = 1.5 - 0.75t + 0.125t^2 \ne 1.5 - 0.5t = \frac{dv}{dt}\text{,}\) and thus \(v(t) = 1.5t - 0.25t^2\) is not a solution to the given differential equation.
Since \(v(t) = 3 + 2e^{-0.5t}\text{,}\) \(\frac{dv}{dt} = 2e^{-0.5t}(-0.5) = -e^{-0.5t}\text{.}\) Hence, for this choice of \(v\text{,}\) \(1.5 - 0.5v = 1.5 - 0.5(3+2e^{-0.5t}) = 1.5 - 1.5 - e^{-0.5t} = -e^{-0.5t} = \frac{dv}{dt}\text{.}\) Therefore \(v(t) = 3 + 2e^{-0.5t}\) is a solution to the given differential equation.
Since \(v(t) = 3\text{,}\) \(\frac{dv}{dt} = 0\text{.}\) Hence, for this choice of \(v\text{,}\) \(1.5 - 0.5v = 1.5 - 0.5(3) = 1.5 - 1.5 = 0 = \frac{dv}{dt}\text{.}\) Therefore \(v(t) = 3\) is a solution to the given differential equation.
Since \(v(t) = 3 + Ce^{-0.5t}\) (where \(C\) is a constant), \(\frac{dv}{dt} = Ce^{-0.5t}(-0.5) = -0.5C e^{-0.5t}\text{.}\) Hence, for this choice of \(v\text{,}\) \(1.5 - 0.5v = 1.5 - 0.5(3+Ce^{-0.5t}) = 1.5 - 1.5 - 0.5Ce^{-0.5t} = -0.5Ce^{-0.5t} = \frac{dv}{dt}\text{.}\) Therefore \(v(t) = 3 + Ce^{-0.5t}\) is a solution to the given differential equation for any choice of \(C\text{.}\)
This activity shows us something interesting. Notice that the differential equation has infinitely many solutions, which are parametrized by the constant \(C\) in \(v(t) = 3+Ce^{-0.5t}\text{.}\) In Figure 7.1.3, we see the graphs of these solutions for a few values of \(C\text{,}\) as labeled.
Notice that the value of \(C\) is connected to the initial value of the velocity \(v(0)\text{,}\) since \(v(0) = 3+C\text{.}\) In other words, while the differential equation describes how the velocity changes as a function of the velocity itself, this is not enough information to determine the velocity uniquely: we also need to know the initial velocity. For this reason, differential equations will typically have infinitely many solutions, one corresponding to each initial value. We have seen this phenomenon before: given the velocity of a moving object \(v(t)\text{,}\) we cannot uniquely determine the object's position function unless we also know its initial position.
If we are given a differential equation and an initial value for the unknown function, we say that we have an initial value problem. For instance,
is an initial value problem. In this problem, we know the value of \(v\) at one time and we know how \(v\) is changing. Consequently, there should be exactly one function \(v\) that satisfies the initial value problem.
This demonstrates the following important general property of initial value problems.
Initial value problems that are “well behaved” have exactly one solution, which exists in some interval around the initial point.
We won't worry about what “well behaved” means—it is a technical condition that will be satisfied by all the differential equations we consider.
To close this section, we note that differential equations may be classified based on certain characteristics they may possess. You may see many different types of differential equations in a later course in differential equations. For now, we would like to introduce a few terms that are used to describe differential equations.
A first-order differential equation is one in which only the first derivative of the function occurs. For this reason,
is a first-order equation while
is a second-order equation.
A differential equation is autonomous if the independent variable does not appear in the description of the derivative. For instance,
is autonomous because the description of the derivative \(dv/dt\) does not depend on time. The equation
however, is not autonomous.
Subsection 7.1.4 Summary
A differential equation is simply an equation that describes the derivative(s) of an unknown function.
Physical principles, as well as some everyday situations, often describe how a quantity changes, which lead to differential equations.
A solution to a differential equation is a function whose derivatives satisfy the equation's description. Differential equations typically have infinitely many solutions, parametrized by the initial values.
Exercises 7.1.5 Exercises
¶1. Matching solutions with equations.
2. Finding constant to complete solution.
3. Choosing solution of \(dy/dt=k(1-Ay)\).
4.
Suppose that \(T(t)\) represents the temperature of a cup of coffee set out in a room, where \(T\) is expressed in degrees Fahrenheit and \(t\) in minutes. A physical principle known as Newton's Law of Cooling tells us that
Supposes that \(T(0)=105\text{.}\) What does the differential equation give us for the value of \(\frac{dT}{dt}\vert_{T=105}\text{?}\) Explain in a complete sentence the meaning of these two facts.
Is \(T\) increasing or decreasing at \(t=0\text{?}\)
What is the approximate temperature at \(t=1\text{?}\)
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On the graph below, make a plot of \(dT/dt\) as a function of \(T\text{.}\)
For which values of \(T\) does \(T\) increase? For which values of \(T\) does \(T\) decrease?
What do you think is the temperature of the room? Explain your thinking.
Verify that \(T(t) = 75 + 30e^{-t/15}\) is the solution to the differential equation with initial value \(T(0) = 105\text{.}\) What happens to this solution after a long time?
\(\frac{dT}{dt}\vert_{T=105} = -2\text{;}\) when \(T = 105\text{,}\) the coffee's temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
\(T\) decreasing at \(t=0\text{.}\)
\(T(1) \approx 103\) degrees F.
For \(T \lt 75\text{,}\) \(T\) increases. For \(T \gt 75\text{,}\) \(T\) decreases.
Room temperature is \(75\) degrees F.
Substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\) in the differential equation \(\frac{dT}{dt}= -\frac1{15}T+5\) and verify the equality holds; \(T(0) = 75 + 30e^0 = 75 + 30 = 105\text{;}\) \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\text{.}\)
Since \(\frac{dT}{dt}= -\frac1{15}T+5\) and \(T(0)=105\text{,}\) \(\frac{dT}{dt}\vert_{T=105} = -\frac{1}{15}(105) + 5 = -7 + 5 = -2\text{.}\) The fact that \(T(0) = 105\) tells us the coffee's temperature at time \(t = 0\) is \(T = 105\) degrees F, and \(\frac{dT}{dt}\vert_{T=105} = -2\) tells us that at the instant \(T = 105\) (i.e. when \(t = 0\)), the coffee's temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
At \(t = 0\text{,}\) \(T = 105\text{,}\) and since \(\frac{dT}{dt}\vert_{T=105}\) is negative, \(T\) decreasing at \(t=0\text{.}\)
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Using the linear approximation \(T(1) \approx T(0) + \frac{dT}{dt}\vert_{t=0} (1-0)\) along with the given information, it follows that
\begin{equation*} T(1) \approx 105 + (-2)(1) = 103 \end{equation*}degrees F.
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We note that \(\frac{dT}{dt}\) is a linear function of \(T\) with slope \(-\frac{1}{15}\) and vertical intercept \((0,5)\text{,}\) so its graph is as shown below. Note particularly that \(\frac{dT}{dt}\vert_{T = 75} = 0\text{.}\)
For \(T \lt 75\text{,}\) \(\frac{dT}{dt} \gt 0\text{,}\) so for these \(T\)-values, \(T\) increases. For \(T \gt 75\text{,}\) \(\frac{dT}{dt} \lt 0\text{,}\) so for these \(T\)-values, \(T\) decreases.
In the previous question, we observed that \(75\) degrees F is the critical temperature that changes the behavior of \(T\text{.}\) There are several reasons to think that \(75\) degrees F is room temperature: if the coffee's temperature is above room temperature (\(75\)), its temperature will decrease to room temperature, while if the coffee is below room temperature, its temperature will increase to room temperature. Moreover, if the coffee's temperature is \(75\) (which would match the room temperature), then its rate of change is \(0\) and the coffee's temperature doesn't change.
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Verify that \(T(t) = 75 + 30e^{-t/15}\) is the solution to the differential equation with initial value \(T(0) = 105\text{.}\) What happens to this solution after a long time? Assuming that \(T(t) = 75 + 30e^{-t/15}\text{,}\) we observe that
\begin{equation*} \frac{dT}{dt} = 30e^{-t/15} \left(-\frac{1}{15} \right) = -2e^{-t/15}\text{.} \end{equation*}In addition,
\begin{equation*} -\frac{1}{15}T + 5 = -\frac{1}{15}(75 + 30e^{-t/15}) + 5 = -5 - 2e^{-t/15} + 5 = -2e^{-t/15}\text{.} \end{equation*}The two preceding equations show that
\begin{equation*} \frac{dT}{dt}= -\frac1{15}T+5 \end{equation*}when we substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\text{,}\) and thus this function is a solution to the differential equation. Moreover, \(T(0) = 75 + 30e^0 = 75 + 30 = 105\text{,}\) so the function satisfies the initial condition, too. Finally, because \(e^{-t/15} \to 0\) as \(t \to \infty\text{,}\) it follows that \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\text{,}\) which shows that the solution function's value tends to room temperature as time increases without bound, as we expect.
5.
Suppose that the population of a particular species is described by the function \(P(t)\text{,}\) where \(P\) is expressed in millions. Suppose further that the population's rate of change is governed by the differential equation
where \(f(P)\) is the function graphed below.
For which values of the population \(P\) does the population increase?
For which values of the population \(P\) does the population decrease?
If \(P(0) = 3\text{,}\) how will the population change in time?
If the initial population satisfies \(0\lt P(0)\lt 1\text{,}\) what will happen to the population after a very long time?
If the initial population satisfies \(1\lt P(0)\lt 3\text{,}\) what will happen to the population after a very long time?
If the initial population satisfies \(3\lt P(0)\text{,}\) what will happen to the population after a very long time?
This model for a population's growth is sometimes called “growth with a threshold.” Explain why this is an appropriate name.
\(1 \lt P \lt 3\text{.}\)
\(P \lt 1\) and \(3 \lt P \lt 4\text{.}\)
\(P\) will not change at all.
The population will decrease toward \(P = 0\) with \(P\) always being positive.
The population will increase toward \(P = 3\) with \(P\) always being between \(1\) and \(3\text{.}\)
The population will decrease toward \(P = 3\) with \(P\) always being above \(3\text{.}\)
There's a maximum threshold of \(P = 3\text{.}\)
\(P\) increases for \(1 \lt P \lt 3\) since \(\frac{dP}{dt} \gt 0\) for these \(P\)-values.
\(P\) decreases for \(P \lt 1\) and \(3 \lt P \lt 4\) since \(\frac{dP}{dt} \lt 0\) for these \(P\)-values.
If \(P(0) = 3\text{,}\) \(\frac{dP}{dt}\vert_{P=3} = 0\text{,}\) so \(P\) will not change at all with time since its derivative is \(0\text{.}\)
If the initial population satisfies \(0\lt P(0)\lt 1\text{,}\) we expect that the population will decrease toward \(P = 0\) with \(P\) always being positive. Since \(\frac{dP}{dt}\) is always negative for \(0\lt P \lt 1\text{,}\) \(P\) is always decreasing on this interval. The population can't drop below \(P = 0\) because at that value of \(P\text{,}\) \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
If the initial population satisfies \(1\lt P(0)\lt 3\text{,}\) we expect that the population will increase toward \(P = 3\) with \(P\) always being between \(1\) and \(3\text{.}\) Since \(\frac{dP}{dt}\) is always positive for \(1 \lt P \lt 3\text{,}\) \(P\) is always increasing on this interval. The population can't increase above \(P = 3\) because at that value of \(P\text{,}\) \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
If the initial population satisfies \(3\lt P(0)\text{,}\) we expect that the population will decrease toward \(P = 3\) with \(P\) always being above \(3\text{.}\) Since \(\frac{dP}{dt}\) is always negative for \(P \gt 3\text{,}\) \(P\) is always decreasing on this interval. The population can't drop below \(P = 3\) because at that value of \(P\text{,}\) \(\frac{dP}{dt} = 0\) and at that value the population wouldn't change.
It looks like there are two thresholds for the population: there's a minimum threshold of \(P = 1\text{,}\) since below that population, the population decreases to \(0\text{,}\) and there's a maximum threshold of \(P = 3\text{,}\) where if we are above the minimum but below the maximum, the population works to increase toward the maximum threshold, while if the population is above the maximum, its value will fall back toward it.
6.
In this problem, we test further what it means for a function to be a solution to a given differential equation.
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Consider the differential equation
\begin{equation*} \frac{dy}{dt} = y - t\text{.} \end{equation*}Determine whether the following functions are solutions to the given differential equation.
\(y(t) = t + 1 + 2e^t\)
\(y(t) = t + 1\)
\(y(t) = t + 2\)
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When you weigh bananas in a scale at the grocery store, the height \(h\) of the bananas is described by the differential equation
\begin{equation*} \frac{d^2h}{dt^2} = -kh \end{equation*}where \(k\) is the spring constant, a constant that depends on the properties of the spring in the scale. After you put the bananas in the scale, you (cleverly) observe that the height of the bananas is given by \(h(t) = 4\sin(3t)\text{.}\) What is the value of the spring constant?
\(y(t) = t + 1 + 2e^t\) is a solution to the DE.
\(y(t) = t + 1\) is a solution to the DE.
\(y(t) = t + 2\) is a not solution to the DE.
\(k = 9\text{.}\)
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To see if \(y(t) = t + 1 + 2e^t\) is a solution to \(\frac{dy}{dt} = y - t\text{,}\) we compute \(\frac{dy}{dt}\) and \(y - t\) separately and see if they are equal. First, using \(y(t) = t + 1 + 2e^t\text{,}\)
\begin{equation*} \frac{dy}{dt} = 1 + 2e^t\text{.} \end{equation*}Next,
\begin{equation*} y - t = ( t + 1 + 2e^t ) - t = 1 + 2e^t\text{.} \end{equation*}Thus, \(y(t) = t + 1 + 2e^t\) satisfies the DE and is indeed a solution.
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For \(y(t) = t + 1\text{,}\) observe that
\begin{equation*} \frac{dy}{dt} = 1 \end{equation*}and
\begin{equation*} y - t = 1 \end{equation*}as well, so \(y(t) = t + 1\) is a solution to the DE.
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If \(y(t) = t + 2\text{,}\) then
\begin{equation*} \frac{dy}{dt} = 1 \end{equation*}but
\begin{equation*} y - t = 2 \end{equation*}as well, so \(\frac{dy}{dt} \ne y-t\text{,}\) and therefore \(y(t) = t + 2\) is a not solution to the DE.
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In order for \(h(t) = 4\sin(3t)\) to be a solution to
\begin{equation*} \frac{d^2h}{dt^2} = -kh\text{,} \end{equation*}it must make the equation true. To find \(k\text{,}\) we substitute the formula for \(h\) into the DE and see what is required of \(k\text{.}\) We observe that for \(h(t) = 4\sin(3t)\text{,}\)
\begin{equation*} \frac{dh}{dt} = 12\cos(3t) \end{equation*}and
\begin{equation*} \frac{d^2h}{dt^2} = -36\sin(3t)\text{.} \end{equation*}To satisfy the DE
\begin{equation*} \frac{d^2h}{dt^2} = -kh\text{,} \end{equation*}it follows that we must have \(-36\sin(3t) = -kh = -k \cdot 4\sin(3t)\text{.}\) Thus, we need \(-36 = -4k\text{,}\) and hence \(k = 9\text{.}\)