A Single Differential Equation

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Algebraic operations such as addition and multiplication are performed on numbers while the calculus operations of differentiation and integration are performed on functions. Thus algebraic equations (such as $x^2=9$ ) are solved for numbers ( $x=\pm 3$ ) while differential (and integral) equations are solved for functions.

In Chapter ?? we discussed how to solve systems of linear equations, such as

$\begin{eqnarray*} x_1 + x_2 & = & 2 \\ x_1 - x_2 & = & 4 \end{eqnarray*}$

for numbers $x_1=3 \quad \mbox { and } \quad x_2=-1,$ while in this chapter we discuss how to solve some systems of differential equations for functions.

Solving a single linear equation in one unknown $x$ is a simple task. For example, solve $2x = 4$ for $x=2$ . Solving a single differential equation in one unknown function $x(t)$ is far from trivial.

Integral Calculus as a Differential Equation

Mathematically, the simplest type of differential equation is:

$\begin{equation} \label{e:intcalc} \dps\frac{dx}{dt}(t) = f(t) \end{equation}$ where $f$ is some continuous function. In words, this equation asks us to find all functions $x(t)$ whose derivative is $f(t)$ . The fundamental theorem of calculus tells us the answer: $x(t)$ is an antiderivative of $f(t)$ . Thus to find all solutions, we just integrate both sides of (??) with respect to $t$ . Formally, using indefinite integrals, $\begin{equation} \label{E:integrate} \int \frac{dx}{dt}(t)dt = \int f(t)dt + C, \end{equation}$ where $C$ is an arbitrary constant. (It is tempting to put a constant of integration on both sides of (??), but two constants are not needed, as we can just combine both constants on the right hand side of this equation.) Since the indefinite integral of $dx/dt$ is just the function $x(t)$ , we have $\begin{equation} \label{e:intcalcsoln} x(t) = \int f(\tau) d\tau + C. \end{equation}$ In particular, finding closed form solutions to differential equations of the type (??) is equivalent to finding all definite integrals of the function $f(t)$ . Indeed, to find closed form solutions to differential equations like (??) we need to know all of the techniques of integration from integral calculus.

Initial Conditions and the Role of the Integration Constant $C$

Equation (??) tells us that there are an infinite number of solutions to the differential equation (??), each one corresponding to a different choice of the constant $C$ . To understand how to interpret the constant $C$ , consider the example $\dps \frac {dx}{dt}(t) = \cos t.$ Using (??) we see that the answer is $x(t) = \int \cos \tau d\tau + C = \sin t + C.$ Note that $x(0) = \sin (0) + C = C.$ Thus, the constant $C$ represents an initial condition for the differential equation. We will return to the discussion of initial conditions several times in this chapter.

See Exercise ?? for a more interesting example of this type of differential equation.

Solutions to Differential Equations are Functions

Consider the differential equation

$\begin{equation} \label{E:verify} \frac{dx}{dt}(t) = tx(t). \end{equation}$ Are the functions $x_1(t) = t^2 \AND x_2(t) = e^{t^2/2}$ solutions to the differential equation (??)?

To test whether or not the function $x_1(t)$ is a solution, we compute the left and right hand sides of (??):

$\begin{eqnarray*} {\rm LHS:} & \frac{d}{dt}x_1(t) & = 2t\\ {\rm RHS:} & tx_1(t) & = t^3. \end{eqnarray*}$

Since the left and right hand sides are unequal, the function $x_1(t)$ is not a solution to (??).

To test whether or not the function $x_2(t)$ is a solution, we again compute the left and right hand sides of (??):

$\begin{eqnarray*} {\rm LHS:} & \frac{d}{dt}x_2(t) & = te^{t^2/2}\\ {\rm RHS:} & tx_2(t) & = te^{t^2/2}. \end{eqnarray*}$

Since the left and right hand sides are equal, the function $x_2(t)$ is a solution to (??).

Note that we have not discussed how we knew that the function $x_2(t)$ is a solution to (??). For the most part, the issue of how one finds solutions to a differential equation will be discussed in later chapters, though we do determine solutions to a very important equation next.

The Linear Differential Equation of Growth and Decay

The real subject of differential equations begins when the function $f$ on the right hand side of (??) depends explicitly on the function $x$ , and the simplest such differential equation is: $\frac {dx}{dt}(t) = x(t).$ Using results from calculus, we can solve this equation; indeed, we can solve the slightly more complicated equation

$\begin{equation} \label{lin1} \frac{dx}{dt}(t) = \lambda x(t), \end{equation}$ where $\lambda \in \R$ is a constant. The differential equation (??) is linear since $x(t)$ appears by itself on the right hand side. Moreover, (??) is homogeneous since the constant function $x(t)=0$ is a solution.

In words (??) asks: For which functions $x(t)$ is the derivative of $x(t)$ a scalar multiple of $x(t)$ . The function $x(t)=e^{\lambda t}$ is such a function, since $\frac {dx}{dt}(t) = \frac {d}{dt}e^{\lambda t} = \lambda e^{\lambda t} = \lambda x(t).$ More generally, the function

$\begin{equation} \label{soln1} x(t) = K e^{\lambda t} \end{equation}$ is a solution to (??) for any real constant $K$ . We claim that the functions (??) list all (differentiable) functions that solve (??).

To verify this claim, we let $x(t)$ be a solution to (??) and show that the ratio $\frac {x(t)}{e^{\lambda t}} = x(t)e^{-\lambda t}$ is a constant (independent of $t$ ). Using the product rule and (??), compute

$\begin{eqnarray*} \frac{d}{dt}\left[x(t)e^{-\lambda t}\right] & = & \frac{d}{dt}\left(x(t)\right) e^{-\lambda t} + x(t)\frac{d}{dt}\left(e^{-\lambda t}\right) \\ & = & (\lambda x(t)) e^{-\lambda t} + x(t)(-\lambda e^{-\lambda t}) \\ & = & 0. \end{eqnarray*}$

Now recall that the only functions whose derivatives are identically zero are the constant functions. Thus, $x(t) e^{-\lambda t} = K$ for some constant $K\in \R$ . Hence $x(t)$ has the form (??), as claimed.

Next, we discuss the role of the constant $K$ . We have written the function as $x(t)$ , and we have meant the reader to think of the variable $t$ as time. Thus $x(0)$ is the initial value of the function $x(t)$ at time $t=0$ ; we say that $x(0)$ is the initial value of $x(t)$ . From (??) we see that $x(0) = K,$ and that $K$ is the initial value of the solution of (??). Henceforth, we write $K$ as $x_0$ so that the notation calls attention to the special meaning of this constant.

By deriving (??) we have proved:

There is a unique solution to the initial value problem $\begin{equation} \label{ivp1} \begin{array}{rcl} \dps \frac{dx}{dt}(t) & = & \lambda x(t) \\ x(0) & = & x_0. \end{array} \end{equation}$ That solution is $x(t) = x_0e^{\lambda t}.$

As a consequence of Theorem ?? we see that there is a qualitative difference in the behavior of solutions to (??) depending on whether $\lambda >0$ or $\lambda <0$ . Suppose that $x_0>0$ . Then

$\begin{equation} \label{explimits} \lim_{t\to\infty} x(t) = \lim_{t\to\infty} x_0e^{\lambda t} =\left\{ \begin{array}{rl} +\infty & \quad\lambda>0 \\ 0 & \quad\lambda<0 . \end{array} \right. \end{equation}$ When $\lambda >0$ we say that the solution has exponential growth and when $\lambda < 0$ we say that the solution has exponential decay . In either case, however, the number $\lambda$ is called the growth rate. We can visualize this discussion by graphing the solutions in MATLAB.

Suppose we set $x_0=1$ and $\lambda =\pm 0.5$ . Type

x0 = 1;
 
lambda = 0.5;
 
t = linspace(-1,4,100);
 
x = x0*exp(lambda*t);
 
plot(t,x)
 
hold on
 
xlabel(’t’)
 
ylabel(’x’)
 
lambda = -0.5;
 
x = x0*exp(lambda*t);
 
plot(t,x)

The result of this calculation is shown in Figure ??. In this way we can actually see the difference between exponential growth ( $\lambda =0.5$ ) and exponential decay ( $\lambda =-0.5$ ), as discussed in (??).

Figure 1: Solutions of (??) for $t\in [-1,4]$ , $x_0=1$ and $\lambda =\pm 0.5$ .

The Inhomogeneous Linear Differential Equation

It follows from (??) that solutions to the linear homogeneous differential equation (??) are either unbounded as $t\to \infty$ or they approach zero. Now we consider an inhomogeneous differential equation and show that solutions can approach fixed values for increasing $t$ that are neither zero nor infinity.

As an example, consider the linear differential equation

$\begin{equation} \label{E:ivp2} \frac{dx}{dt} = -2x-6. \end{equation}$ Observe that $x(t)=0$ is not a solution of (??) and therefore that (??) is inhomogeneous. It is easy to verify, however, that the constant function $x(t)=-3$ is a solution.

Equation (??) can be solved by introducing a new function that transforms (??) into a homogeneous equation. Let $y(t) = x(t) + 3,$ and compute $\frac {dy}{dt} = \frac {dx}{dt} = -2x-6 = -2y.$

Using Theorem ??, it follows that $y(t)$ has the form $y(t) = y_0e^{-2t}.$ Therefore $x(t) = y_0e^{-2t}-3$ is a solution of (??) for every constant $y_0$ . Moreover, $\lim _{t\to \infty } x(t) = \lim _{t\to \infty } y_0e^{-2t} -3=-3,$ which is neither zero nor infinity.

Any equation of the form

$\begin{equation} \label{ivp2} \frac{dx}{dt}(t) = \lambda x(t) +\rho \end{equation}$ can be solved in a similar fashion. Just set $y(t) = x(t) + \frac {\rho }{\lambda }.$ It follows from (??) that $\frac {dy}{dt} = \frac {dx}{dt} = \lambda x(t) +\rho = \lambda \left ( y(t)-\frac {\rho }{\lambda }\right ) +\rho = \lambda y(t).$ Theorem ?? implies that $y(t) = y_0e^{\lambda t}$ for some constant $y_0$ and $x(t) = y_0e^{\lambda t} - \frac {\rho }{\lambda }.$ Note that the limit of $x(t)$ as $t\to \infty$ is $-\rho /\lambda$ when $\lambda <0$ .

Some Examples of (??)

Even though the differential equation (??) is one of the simplest differential equations, it still has some use in applications. We present two here: compound interest and population dynamics.

Compound Interest

Banks pay interest on an account in the following way. At the end of each day, the bank determines the interest rate $r_{day}$ for that day, checks the principal $P$ in the account, and then deposits an additional $r_{day}P$ . So the next day the principal in this account is $(1+r_{day})P$ . Note that if $r$ denotes the interest rate per year, then $r_{day} = r/365$ . Of course, a day is just a convenient measure for elapsed time. Before computers were prevalent, banks paid interest yearly or quarterly or monthly or, in a few cases, even weekly, depending on the particular bank rules.

Observe that the more frequently interest is paid, the more money is earned. For example, if interest is paid only once at the end of a year, then the money in the account at the end of the year is $(1+r)P$ , and the amount $rP$ is called simple interest. But if interest is paid twice a year, then the principal at the end of six months will be $(1+\frac {r}{2})P$ , and the principal at the end of the year will be $(1+\frac {r}{2})^2P$ . Since $\left (1+\frac {r}{2}\right )^2 = 1+r+\frac {1}{4}r^2 > 1+r,$ there is more money in the account at the end of the year if the interest is compounded semiannually rather than annually. But how much is the difference and what is the maximum earning potential?

While making the calculation in the previous paragraph, we implicitly made a number of simplifying assumptions. In particular, we assumed

an initial principal $P_0$ is deposited in the bank on January 1,
the money is not withdrawn for one year,
no new money is deposited in that account during the year,
the yearly interest rate $r$ remains constant throughout the year, and
interest is added to the account $N$ times during the year.

In this model, simple interest corresponds to $N=1$ , compound monthly interest to $N=12$ , and compound daily interest to $N=365$ .

We first answer the question: How much money is in this account after one year? After one time unit of $\frac {1}{N}$ year, the amount of money in the account is $Q_1 = \left (1+\frac {r}{N}\right )P_0.$ The interest rate in each time period is $\frac {r}{N}$ , the yearly rate $r$ divided by the number of time periods $N$ . Here we have used the assumption that the interest rate remains constant throughout the year. After two time units, the principal is: $Q_2 = \left (1+\frac {r}{N}\right )Q_1 = \left (1+\frac {r}{N}\right )^2P_0,$ and at the end of the year (that is, after $N$ time periods)

$\begin{equation} \label{compint} Q_N = \left(1+\frac{r}{N}\right)^N P_0. \end{equation}$ Here we have used the assumption that money is neither deposited nor withdrawn from our account. Note that $Q_N$ is the amount of money in the bank after one year assuming that interest has been compounded $N$ (equally spaced) times during that year, and the effective interest rate when compounding $N$ times is: $\left (1+\frac {r}{N}\right )^N - 1.$

For the curious, we can write a program in MATLAB to compute (??). Suppose we assume that the initial deposit $P_0=\$1,000$ , the simple interest rate is $6\$ per year, and the interest payments are made monthly. In MATLAB type

                                                                  

                                                                  
N  = 12;
 
P0 = 1000;
 
r  = 0.06;
 
QN = (1 + r/N)^N*P0

The answer is $QN=\$1,061.68$ , and the effective interest rate for monthly payments is $6.16778\$ . For daily interest payments $N=365$ , the answer is $QN=\$1,061.83$ , and the effective interest rate is $6.18313\$ .

To find the maximum effective interest, we ask the bank to compound interest continuously; that is, we ask the bank to compute $\lim _{N\to \infty } \left (1 + \frac {r}{N}\right )^N.$ We compute this limit using differential equations. The concept of continuous interest is rephrased as follows. Let $P(t)$ be the principal at time $t$ , where $t$ is measured in units of years. Suppose that we assume that interest is compounded $N$ times during the year. The length of time in each compounding period is $\Delta t = \frac {1}{N},$ and the change in principal during that time period is $\Delta P = \frac {r}{N} P = rP\Delta t.$ It follows that $\frac {\Delta P}{\Delta t} = rP,$ and, on taking the limit $\Delta t \to 0$ , we have the differential equation $\frac {dP}{dt}(t) = rP(t).$

Since $P(0)=P_0$ the solution of the initial value problem given in Theorem ?? shows that $P(t) = P_0 e^{rt}.$ After one year ( $t=1$ ) we find that $P(1) = e^r P_0.$ Note that $P(1) = \lim _{N\to \infty } Q_N,$ and we have thus verified that $\lim _{N\to \infty } \left (1 + \frac {r}{N}\right )^N = e^r.$ Thus the maximum effective interest rate is $e^r-1$ . When $r=6\$ the maximum effective interest rate is $6.18365\$ .

An Example from Population Dynamics

To provide a second interpretation of the constant $\lambda$ in (??), we discuss a simplified model for population dynamics. Let $p(t)$ be the size of a population of a certain species at time $t$ and let $r$ be the rate at which the population $p$ is changing at time $t$ . In general, $r$ depends on the time $t$ and is a complicated function of birth and death rates and of immigration and emigration, as well as of other factors. Indeed, the rate $r$ may well depend on the size of the population itself. (Overcrowding can be modeled by assuming that the death rate increases with the size of the population.) These population models assume that the rate of change in the size of the population $dp/dt$ is given by

$\begin{equation} \label{pop_model} \frac{dp}{dt}(t) = r p(t), \end{equation}$ they just differ on the precise form of $r$ . In general, the rate $r$ will depend on the size of the population $p$ as well as the time $t$ , that is, $r$ is a function $r(p,t)$ .

The simplest population model — which we now assume — is the one in which $r$ is assumed to be constant. Then equation (??) is identical to (??) after identifying $p$ with $x$ and $r$ with $\lambda$ . Hence we may interpret $r$ as the growth rate for the population. The form of the solution in (??) shows that the size of a population grows exponentially if $r>0$ and decays exponentially if $r<0$ .

The mathematical description of this simplest population model shows that the assumption of a constant growth rate leads to exponential growth (or exponential decay). Is this realistic? Surely, no population will grow exponentially for all time, and other factors, such as limited living space, have to be taken into account. On the other hand, exponential growth describes well the growth in human population during much of human history. So this model, though surely oversimplified, gives some insight into population growth.

Exercises

In Exercises ?? – ?? find solutions to the given initial value problems.

$\frac {\dps dx}{\dps dt} = \sin (2t),\quad x(\pi ) = 2$ .

$\frac {\dps dx}{\dps dt} = t^2, \quad x(2) = 8$ .

$\frac {\dps dx}{\dps dt} = \frac {1}{t^2},\quad x(1)=1$ .

In Exercises ?? – ?? determine whether or not each of the given functions $x_1(t)$ and $x_2(t)$ is a solution to the given differential equation.

ODE: $\frac {\dps dx}{\dps dt} = \frac {\dps t}{\dps x-1}$ . Functions: $x_1(t)=t+1 \AND x_2(t)= \frac {\dps 1+\sqrt {4t^2+1}}{\dps 2}$ .

ODE: $\frac {\dps dx}{\dps dt} = x + e^t$ . Functions: $x_1(t)=te^t \AND x_2(t)= 2e^t$ .

ODE: $\frac {\dps dx}{\dps dt} = x^2 + 1$ . Functions: $x_1(t)=-\tan t \AND x_2(t)= \tan t$ .

ODE: $\frac {\dps dx}{\dps dt} = \frac {\dps x}{\dps t}$ . Functions: $x_1(t)=t+1 \AND x_2(t)= 5t$ .

Solve the differential equation $\frac {dx}{dt} = 2x,$ where $x(0)=1$ . At what time $t_1$ will $x(t_1)=2$ ?

Solve the differential equation $\frac {dx}{dt} = -3x.$ At what time $t_1$ will $x(t_1)$ be half of $x(0)$ ?

Bacteria grown in a culture increase at a rate proportional to the number present. If the number of bacteria doubles every $2$ hours, then how many bacteria will be present after $5$ hours? Express your answer in terms of $x_0$ , the initial number of bacteria.

Suppose you deposit $10,000 in a bank at an interest of $7.5\$ compounded continuously. How much money will be in your account a year and a half later? How much would you have if the interest were compounded monthly?

Newton’s law of cooling states that the rate at which a body changes temperature is proportional to the difference between the body temperature and the temperature of the surrounding medium. That is, $\begin{equation} \label{e:Newton} \frac{dT}{dt} = \alpha(T-T_m) \end{equation}$ where $T(t)$ is the temperature of the body at time $t$ , $T_m$ is the constant temperature of the surrounding medium, and $\alpha$ is the constant of proportionality. Suppose the body is in air of temperature $50^\circ$ and the body cools from $100^\circ$ to $75^\circ$ in $20$ minutes. What will the temperature of the body be after one hour? Hint: Rewrite (??) in terms of $U(t) = T(t) - T_m$ .

Let $p(t)$ be the population of group Grk at time $t$ measured in years. Let $r$ be the growth rate of the group Grk. Suppose that the population of Grks changes according to the differential equation (??). Find $r$ so that the population of Grks doubles every $50$ years. How large must $r$ be so that the population doubles every $25$ years?

You deposit $4,000 in a bank at an interest of $5.5\$ but after half a year the bank changes the interest rate to $4.5\$ . Suppose that the interest is compounded continuously. How much money will be in your account after one year?

As an application of (??) answer the following question (posed by R.P. Agnew).

One day it started snowing at a steady rate. A snowplow started at noon and went two miles in the first hour and one mile in the second hour. Assume that the speed of the snowplow times the depth of the snow is constant. At what time did it start to snow?

To set up this problem, let $d(t)$ be the depth of the snow at time $t$ where $t$ is measured in hours and $t=0$ is noon. Since the snow is falling at a constant rate $r$ , $d(t)= r(t-t_0)$ where $t_0$ is the time that it started snowing. Let $x(t)$ be the position of the snowplow along the road. The assumption that speed times the depth equals a constant $k$ means that $\frac {dx}{dt}(t) = \frac {k}{d(t)} = \frac {K}{t-t_0}$ where $K=k/r$ . The information about how far the snowplow goes in the first two hours translates to $x(1) = 2 \AND x(2) =3.$ Now solve the problem.

In Exercises ?? – ?? use MATLAB to graph the given function $f$ on the specified interval.

$f(t) = t^2$ on the interval $t\in [0,2]$ .

$f(t) = e^t-t$ on the interval $t\in [0,3]$ .

$f(t) = \cos (2t)-t$ on the interval $t\in [2,8]$ .

$f(t) = \sin (5t)$ on the interval $t\in [0,6.5]$ .

Hint: Use the fact that the trigonometric functions $\sin$ and $\cos$ can be evaluated in MATLAB in the same way as the exponential function, that is, by using sin and cos instead of exp.

Two banks each pay $7\$ interest per year — one compounds money daily and one compounds money continuously. What is the difference in earnings in one year in an account having $10,000.

There are two banks in town — Intrastate and Statewide. You plan to deposit $5,000 in one of these banks for two years. Statewide Bank’s best savings account pays $8\$ interest per year compounded quarterly and charges $10 to open an account. Intrastate Bank’s best savings account pays $7.75\$ interest compounded daily. Which bank will pay you the most money when you withdraw your money? Would your answer change if you had planned to keep your money in the bank for only one year?

In the beginning of the year 1990 the population of the United States was approximately 250,000,000 people and the growth rate was estimated at $3\$ per year. Assuming that the growth rate does not change, during what year will the population of the United States reach 400,000,000?