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602
Perhaps the most important of all the applications of calculus is to differential equa- tions. When physical scientists or social scientists use calculus, more often than not it is to analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying. Although it is often impossible to find an explicit formula for the solution of a differential equation, we will see that graphical and numer- ical approaches provide the needed information.
Direction fields enable us to sketch solutions of differential equations with- out an explicit formula.
DIFFERENTIAL
EQUATIONS
MODELING WITH DIFFERENTIAL EQUATIONS
In describing the process of modeling in Section 1.2, we talked about formulating a math- ematical model of a real-world problem either through intuitive reasoning about the phe- nomenon or from a physical law based on evidence from experiments. The mathematical model often takes the form of a differential equation, that is, an equation that contains an unknown function and some of its derivatives. This is not surprising because in a real- world problem we often notice that changes occur and we want to predict future behavior on the basis of how current values change. Let’s begin by examining several examples of how differential equations arise when we model physical phenomena.
MODELS OF POPULATION GROWTH One model for the growth of a population is based on the assumption that the population grows at a rate proportional to the size of the population. That is a reasonable assumption for a population of bacteria or animals under ideal conditions (unlimited environment, ade- quate nutrition, absence of predators, immunity from disease). Let’s identify and name the variables in this model:
The rate of growth of the population is the derivative. So our assumption that the rate of growth of the population is proportional to the population size is written as the equation
where is the proportionality constant. Equation 1 is our first model for population growth; it is a differential equation because it contains an unknown function P and its derivative. Having formulated a model, let’s look at its consequences. If we rule out a population of 0, then for all t. So, if , then Equation 1 shows that for all t. This means that the population is always increasing. In fact, as increases, Equation 1 shows that becomes larger. In other words, the growth rate increases as the popula- tion increases. Equation 1 asks us to find a function whose derivative is a constant multiple of itself. We know from Chapter 3 that exponential functions have that property. In fact, if we let , then
Thus any exponential function of the form is a solution of Equation 1. In Sec- tion 10.4 we will see that there is no other solution. Allowing C to vary through all the real numbers, we get the family of solutions whose graphs are shown in Figure 1. But populations have only positive values and so we are interested only in the solutions with C 0. And we are probably con-
P t Ce kt
P t Ce kt
P t C ke kt^ k Ce kt^ kP t
P t Ce kt
dP dt
P t
P t 0 k 0 P t 0
dP dt
k
dP dt
(^1) kP
dP dt
P the number of individuals in the population the dependent variable
t time the independent variable
603
N Now is a good time to read (or reread) the dis- cussion of mathematical modeling on page 24.
t
P
FIGURE 1 The family of solutions of dP/dt=kP
mass times acceleration), we have
This is an example of what is called a second-order differential equation because it involves second derivatives. Let’s see what we can guess about the form of the solution directly from the equation. We can rewrite Equation 3 in the form
which says that the second derivative of x is proportional to x but has the opposite sign. We know two functions with this property, the sine and cosine functions. In fact, it turns out that all solutions of Equation 3 can be written as combinations of certain sine and cosine functions (see Exercise 4). This is not surprising; we expect the spring to oscil- late about its equilibrium position and so it is natural to think that trigonometric functions are involved.
GENERAL DIFFERENTIAL EQUATIONS In general, a differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the order of the high- est derivative that occurs in the equation. Thus Equations 1 and 2 are first-order equations and Equation 3 is a second-order equation. In all three of those equations the independent variable is called t and represents time, but in general the independent variable doesn’t have to represent time. For example, when we consider the differential equation
it is understood that y is an unknown function of x. A function is called a solution of a differential equation if the equation is satisfied when and its derivatives are substituted into the equation. Thus is a solution of Equation 4 if
for all values of x in some interval. When we are asked to solve a differential equation we are expected to find all possible solutions of the equation. We have already solved some particularly simple differential equations, namely, those of the form
For instance, we know that the general solution of the differential equation
is given by
where C is an arbitrary constant. But, in general, solving a differential equation is not an easy matter. There is no sys- tematic technique that enables us to solve all differential equations. In Section 10.2, how- ever, we will see how to draw rough graphs of solutions even when we have no explicit formula. We will also learn how to find numerical approximations to solutions.
y
x^4 4
C
y x^3
y f x
f x x f x
y f x f
f
(^4) y xy
d^2 x dt^2
k m
x
m
d^2 x dt^2
(^3) kx
SECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS | | | | 605
FIGURE 4
m
x
0
x (^) m
equilibrium position
EXAMPLE 1 Show that every member of the family of functions
is a solution of the differential equation. SOLUTION We use the Quotient Rule to differentiate the expression for y :
The right side of the differential equation becomes
Therefore, for every value of c , the given function is a solution of the differential equation. M
When applying differential equations, we are usually not as interested in finding a fam- ily of solutions (the general solution ) as we are in finding a solution that satisfies some additional requirement. In many physical problems we need to find the particular solution that satisfies a condition of the form. This is called an initial condition , and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem. Geometrically, when we impose an initial condition, we look at the family of solution curves and pick the one that passes through the point. Physically, this corresponds to measuring the state of a system at time and using the solution of the initial-value prob- lem to predict the future behavior of the system.
EXAMPLE 2 Find a solution of the differential equation that satisfies the initial condition. SOLUTION Substituting the values and into the formula
from Example 1, we get
Solving this equation for c , we get , which gives. So the solution of the initial-value problem is
y M
1 13 e t 1 13 e t^
3 e t 3 e t
2 2 c 1 c c (^13)
1 ce^0 1 ce^0
1 c 1 c
y
1 ce t 1 ce t
t 0 y 2
y 0 2
V y 12 y^2 1
t 0
t 0 , y 0
y t 0 y 0
4 ce t 1 ce t^ ^2
2 ce t 1 ce t^ ^2
1 2 y
1 ce t 1 ce t^
2 1
1 ce t^ ^2 1 ce t^ ^2 1 ce t^ ^2
ce t^ c^2 e^2 t^ ce t^ c^2 e^2 t 1 ce t^ ^2
2 ce t 1 ce t^ ^2
y
1 ce t^ ce t^ 1 ce t^ ce t^ 1 ce t^ ^2
y 12 y^2 1
y
1 ce t 1 ce t
V
606 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
N Figure 5 shows graphs of seven members of the family in Example 1. The differential equation shows that if , then. That is borne out by the flatness of the graphs near y 1 and y 1.
y 1 y 0
5
_
_5 5
FIGURE 5
an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton’s Law of Cooling for this particular situ- ation. What is the initial condition? In view of your answer to part (a), do you think this differential equation is an appropriate model for cooling? (c) Make a rough sketch of the graph of the solution of the initial-value problem in part (b).
(c) Make a rough sketch of a possible solution of this differen- tial equation.
14. Suppose you have just poured a cup of freshly brewed coffee with temperature in a room where the temperature is. (a) When do you think the coffee cools most quickly? What happens to the rate of cooling as time goes by? Explain. (b) Newton’s Law of Cooling states that the rate of cooling of
20 C
95 C
608 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
DIRECTION FIELDS AND EULER’S METHOD
Unfortunately, it’s impossible to solve most differential equations in the sense of obtain- ing an explicit formula for the solution. In this section we show that, despite the absence of an explicit solution, we can still learn a lot about the solution through a graphical approach (direction fields) or a numerical approach (Euler’s method).
DIRECTION FIELDS Suppose we are asked to sketch the graph of the solution of the initial-value problem
We don’t know a formula for the solution, so how can we possibly sketch its graph? Let’s think about what the differential equation means. The equation tells us that the slope at any point on the graph (called the solution curve ) is equal to the sum of the x - and y -coordinates of the point (see Figure 1). In particular, because the curve passes through the point , its slope there must be. So a small portion of the solu- tion curve near the point looks like a short line segment through with slope 1. (See Figure 2.) As a guide to sketching the rest of the curve, let’s draw short line segments at a num- ber of points with slope. The result is called a direction field and is shown in Figure 3. For instance, the line segment at the point has slope. The direc- tion field allows us to visualize the general shape of the solution curves by indicating the direction in which the curves proceed at each point.
0 1 2 x
y
FIGURE 3 Direction field for yª=x+y
0 1 2 x
y
FIGURE 4 The solution curve through (0, 1)
(0, 1)
x , y x y
x , y
y x y
y x y y 0 1
Slope at (¤, fi) is ¤+fi.
Slope at (⁄, ›) is ⁄+›.
0 x
y
FIGURE 1 A solution of yª=x+y
0 x
y
(0, 1) Slope at^ (0, 1) is 0+1=1.
FIGURE 2 Beginning of the solution curve through (0, 1)
Now we can sketch the solution curve through the point by following the direc- tion field as in Figure 4. Notice that we have drawn the curve so that it is parallel to near- by line segments. In general, suppose we have a first-order differential equation of the form
where is some expression in and. The differential equation says that the slope of a solution curve at a point on the curve is. If we draw short line segments with slope at several points , the result is called a direction field (or slope field ). These line segments indicate the direction in which a solution curve is heading, so the direction field helps us visualize the general shape of these curves.
EXAMPLE 1 (a) Sketch the direction field for the differential equation. (b) Use part (a) to sketch the solution curve that passes through the origin. SOLUTION (a) We start by computing the slope at several points in the following chart:
Now we draw short line segments with these slopes at these points. The result is the direction field shown in Figure 5. (b) We start at the origin and move to the right in the direction of the line segment (which has slope ). We continue to draw the solution curve so that it moves parallel to the nearby line segments. The resulting solution curve is shown in Figure 6. Returning to the origin, we draw the solution curve to the left as well. M
The more line segments we draw in a direction field, the clearer the picture becomes. Of course, it’s tedious to compute slopes and draw line segments for a huge number of points by hand, but computers are well suited for this task. Figure 7 shows a more detailed, computer-drawn direction field for the differential equation in Example 1. It enables us to draw, with reasonable accuracy, the solution curves shown in Figure 8 with -intercepts , , , , and.
FIGURE 7
3
_
_3 3
FIGURE 8
3
_
_3 3
y
y x^2 y^2 1
V
F x , y x , y
x , y F x , y
F x , y x y
y F x , y
SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD | | | | 609
x 2 1 0 1 2 2 1 0 1 2... y 0 0 0 0 0 1 1 1 1 1... y x^2 y^2 1 3 0 1 0 3 4 1 0 1 4...
0 x
y
_2 _1 1
1
2
_
FIGURE 5
2
Module 10.2A shows direction fields and solution curves for a variety of differential equations.
TEC
0 x
y
_2 _1 1 2
1
2
_
FIGURE 6
(d) We use the direction field to sketch the solution curve that passes through , as shown in red in Figure 11.
M
Notice from Figure 10 that the line segments along any horizontal line are parallel. That is because the independent variable t does not occur on the right side of the equation
. In general, a differential equation of the form
in which the independent variable is missing from the right side, is called autonomous. For such an equation, the slopes corresponding to two different points with the same y -coordinate must be equal. This means that if we know one solution to an autonomous differential equation, then we can obtain infinitely many others just by shifting the graph of the known solution to the right or left. In Figure 11 we have shown the solutions that result from shifting the solution curve of Example 2 one and two time units (namely, seconds) to the right. They correspond to closing the switch when or.
EULER’S METHOD The basic idea behind direction fields can be used to find numerical approximations to solutions of differential equations. We illustrate the method on the initial-value problem that we used to introduce direction fields:
The differential equation tells us that , so the solution curve has slope 1 at the point. As a first approximation to the solution we could use the linear approx- imation. In other words, we could use the tangent line at as a rough approximation to the solution curve (see Figure 12). Euler’s idea was to improve on this approximation by proceeding only a short distance along this tangent line and then making a midcourse correction by changing direction as indicated by the direction field. Figure 13 shows what happens if we start out along the tangent line but stop when. (This horizontal distance traveled is called the step size. ) Since , we have and we take as the starting point for a new line segment. The differential equation tells us that , so we use the linear function
y 1.5 2 x 0.5 2 x 0.
y 0.5 0.5 1.5 2
L 0.5 1.5 y 0.5 1.5 0.5, 1.5
x 0.
L x x 1 0, 1
y 0 0 1 1
y x y y 0 1
t 1 t 2
y f y
I 15 3 I
FIGURE 11
0 1 t
I
2 3
2
4
6
SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD | | | | 611
y
0 1 x
1 y=L(x)
solution curve
FIGURE 12 First Euler approximation
y
0 1 x
1
FIGURE 13 Euler approximation with step size 0.
as an approximation to the solution for (the orange segment in Figure 13). If we decrease the step size from to , we get the better Euler approximation shown in Figure 14. In general, Euler’s method says to start at the point given by the initial value and pro- ceed in the direction indicated by the direction field. Stop after a short time, look at the slope at the new location, and proceed in that direction. Keep stopping and changing direc- tion according to the direction field. Euler’s method does not produce the exact solution to an initial-value problem—it gives approximations. But by decreasing the step size (and therefore increasing the number of midcourse corrections), we obtain successively better approximations to the exact solution. (Compare Figures 12, 13, and 14.) For the general first-order initial-value problem , , our aim is to find approximate values for the solution at equally spaced numbers , , ,... , where is the step size. The differential equation tells us that the slope at is , so Figure 15 shows that the approximate value of the solution when is
Similarly,
In general,
EXAMPLE 3 Use Euler’s method with step size to construct a table of approximate values for the solution of the initial-value problem
SOLUTION We are given that , , , and. So we have
This means that if is the exact solution, then. Proceeding with similar calculations, we get the values in the table:
M
For a more accurate table of values in Example 3 we could decrease the step size. But for a large number of small steps the amount of computation is considerable and so we need to program a calculator or computer to carry out these calculations. The following table shows the results of applying Euler’s method with decreasing step size to the initial- value problem of Example 3.
y x y 0.3 1.
y 3 y 2 hF x 2 , y 2 1.22 0.10.2 1.22 1.
y 2 y 1 hF x 1 , y 1 1.1 0.10.1 1.1 1.
y 1 y 0 hF x 0 , y 0 1 0.1 0 1 1.
h 0.1 x 0 0 y 0 1 F x , y x y
y x y y 0 1
yn yn 1 hF x (^) n 1 , yn 1
y 2 y 1 hF x 1, y 1
y 1 y 0 hF x 0 , y 0
x x 1
x 0 , y 0 y F x 0 , y 0
x 2 x 1 h h
x 0 x 1 x 0 h
y F x , y y x 0 y 0
x 0.
612 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
n n 1 0.1 1.100000 6 0.6 1. 2 0.2 1.220000 7 0.7 2. 3 0.3 1.362000 8 0.8 2. 4 0.4 1.528200 9 0.9 2. 5 0.5 1.721020 10 1.0 3.
xn yn xn yn
y
(^0) x¸ ⁄ x
y¸
h
h F(x¸, y¸)
(⁄, ›)
slope=F(x¸, y¸)
FIGURE 15
y
0 1 x
1
FIGURE 14 Euler approximation with step size 0.
Module 10.2B shows how Euler’s method works numerically and visually for a variety of differential equations and step sizes.
TEC
614 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
5. 6. 7. Use the direction field labeled II (above) to sketch the graphs of the solutions that satisfy the given initial conditions. (a) (b) (c) 8. Use the direction field labeled IV (above) to sketch the graphs of the solutions that satisfy the given initial conditions. (a) (b) (c)
9–10 Sketch a direction field for the differential equation. Then use it to sketch three solution curves.
9. 10.
11–14 Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. , 12. ,
, 14. ,
15–16 Use a computer algebra system to draw a direction field for the given differential equation. Get a printout and sketch on it the solution curve that passes through. Then use the CAS to draw the solution curve and compare it with your sketch.
15. 16. 17. Use a computer algebra system to draw a direction field for the differential equation y y^3 4 y. Get a printout and
CAS
y x^2 sin y y x y^2 4
0, 1
CAS
13. y y x y 0, 1 y x x y 1, 0 11. y y 2 x 1, 0 y 1 x y 0, 0
y 1 y y x^2 y^2
y 0 1 y 0 0 y 0 1
y 0 1 y 0 2 y 0 1
y
0 x
4
_2 2
2
y
_2^0 2 x
2
_
y
0 x
4
_2 2
2
y
_2^0 2 x
2
_
I II
III IV
1. A direction field for the differential equation y ^ ^ x^ ^ y^ ^1^ y ^ ^ sin^ x^ sin^ y is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) (ii) (iii) (iv) (b) Find all the equilibrium solutions. 2. A direction field for the differential equation is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) (ii) (iii) (iv) (v) (b) Find all the equilibrium solutions.
3–6 Match the differential equation with its direction field (labeled I–IV). Give reasons for your answer.
3. y 2 y 4. y x 2 y
y
0 x
3
_3 3
5
4
_2 _1 1 2
1
2
y 0 4 y 0 5
y 0 1 y 0 2 y 0
y x sin y
y
0 x
3
_3 3
_
_2 _1 1 2
1
2
_
_
y 0 3 y 0 3
y 0 1 y 0 1
y y (1 14 y^2 )
10.2 E X E R C I S E S
22. Use Euler’s method with step size to estimate , where is the solution of the initial-value problem , . Use Euler’s method with step size to estimate , where is the solution of the initial-value problem ,. 24. (a) Use Euler’s method with step size to estimate , where is the solution of the initial-value problem ,. (b) Repeat part (a) with step size. ; 25.^ (a) Program a calculator or computer to use Euler’s method to compute , where is the solution of the initial- value problem
(i) (ii) (iii) (iv) (b) Verify that is the exact solution of the differential equation. (c) Find the errors in using Euler’s method to compute with the step sizes in part (a). What happens to the error when the step size is divided by 10?
26. (a) Program your computer algebra system, using Euler’s method with step size 0.01, to calculate , where is the solution of the initial-value problem
(b) Check your work by using the CAS to draw the solution curve.
27. The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of farads (F), and a resistor with a resistance of ohms ( ). The voltage drop across the capacitor is , where is the charge (in coulombs), so in this case Kirchhoff’s Law gives
But , so we have
Suppose the resistance is , the capacitance is F, and a battery gives a constant voltage of 60 V. (a) Draw a direction field for this differential equation. (b) What is the limiting value of the charge?
C
E R
5 0.
R dQ dt 1 C Q E t
I dQ dt
RI
Q C E t
Q C Q
R
C
y x^3 y^3 y 0 1
y 2 y
CAS
y 1
y 2 e x 3
h 0.01 h 0.
h 1 h 0.
y 0 3 dy dx 3 x^2 y 6 x^2
y 1 y x
y x x y y 1 0
y x
0.2 y 1.4
y y x y y 0 1
y x
23. 0.1 y 0.5
y 0 0
y x y 1 x y
sketch on it solutions that satisfy the initial condition 0.2 y 1 for various values of. For what values of does exist? What are the possible values for this limit? Make a rough sketch of a direction field for the autonomous differential equation , where the graph of is as shown. How does the limiting behavior of solutions depend on the value of?
(a) Use Euler’s method with each of the following step sizes to estimate the value of , where is the solution of the initial-value problem. (i) (ii) (iii) (b) We know that the exact solution of the initial-value problem in part (a) is. Draw, as accurately as you can, the graph of , together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figures 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler’s method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler’s method to estimate the true value of , namely. What happens to the error each time the step size is halved?
20. A direction field for a differential equation is shown. Draw, with a ruler, the graphs of the Euler approximations to the solution curve that passes through the origin. Use step sizes and. Will the Euler estimates be under- estimates or overestimates? Explain.
Use Euler’s method with step size to compute the approx- imate -values of the solution of the initial- value problem y y 2 x , y 1 0.
y y 1 , y 2 , y 3 , and y 4
21. 0.
y 2
1
0 1 2 x
h 1 h 0.
y 0.4 e 0.
y e x , 0 x 0.
y e x
h 0.4 h 0.2 h 0.
y y , y 0 1
y 0.4 y
19.
_2 _1 0 1 2 y
f(y)
y 0
y f y f
18.
lim t l y t
y 0 c c c
SECTION 10.2 DIRECTION FILEDS AND EULER’S METHOD | | | | 615
EXAMPLE 1
(a) Solve the differential equation.
(b) Find the solution of this equation that satisfies the initial condition. SOLUTION (a) We write the equation in terms of differentials and integrate both sides:
where is an arbitrary constant. (We could have used a constant on the left side and another constant on the right side. But then we could combine these constants by writing .) Solving for , we get
We could leave the solution like this or we could write it in the form
where. (Since is an arbitrary constant, so is .) (b) If we put in the general solution in part (a), we get. To satisfy the initial condition , we must have and so. Thus the solution of the initial-value problem is
M
EXAMPLE 2 Solve the differential equation.
SOLUTION Writing the equation in differential form and integrating both sides, we have
where is a constant. Equation 3 gives the general solution implicitly. In this case it’s impossible to solve the equation to express explicitly as a function of. M
EXAMPLE 3 Solve the equation. SOLUTION First we rewrite the equation using Leibniz notation:
dy dx
x^2 y
y x^2 y
y x
C
(^3) y^2 sin y 2 x^3 C
y ^2 y^ ^ cos^ y dy^ ^ y 6 x (^2) dx
2 y cos y dy 6 x^2 dx
dy dx
6 x^2 2 y cos y
V
y s^3 x^3 8
y 0 2 s^3 K 2 K 8
x 0 y 0 s^3 K
K 3 C C K
y s^3 x^3 K
y s^3 x^3 3 C
y
C C 2 C 1
C 2
C C 1
1 3 y
3 x
3 C
y y^2 dy^ ^ y x^2 dx
y^2 dy x^2 dx
y 0 2
dy dx
x^2 y^2
SECTION 10.3 SEPARABLE EQUATIONS | | | | 617
N Figure 1 shows graphs of several members of the family of solutions of the differential equation in Example 1. The solution of the initial- value problem in part (b) is shown in red.
3
_
_3 3
FIGURE 1
N Some computer algebra systems can plot curves defined by implicit equations. Figure 2 shows the graphs of several members of the family of solutions of the differential equation in Example 2. As we look at the curves from left to right, the values of are , , , , , , and 3.
C 3210 1 2
4
_
_2 2
FIGURE 2
If , we can rewrite it in differential notation and integrate:
This equation defines implicitly as a function of. But in this case we can solve explicitly for as follows:
so
We can easily verify that the function is also a solution of the given differential equation. So we can write the general solution in the form
where is an arbitrary constant ( , or , or ). M
EXAMPLE 4 In Section 10.2 we modeled the current in the electric circuit shown in Figure 5 by the differential equation
Find an expression for the current in a circuit where the resistance is , the induc- tance is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on when
. What is the limiting value of the current? SOLUTION With L 4, R 12, and , the equation becomes
or
dI dt
4 15 3 I
dI dt
12 I 60
E t 60
t 0
L
dI dt
RI E t
V I t
6
_
_2 2
FIGURE 3 FIGURE 4
2
_
0 x
y
_2 _1 1 2
4
6
_
_
A A e C A e C A 0
y Ae x^
(^3) 3
y 0
y e Ce x
(^3) 3
y e ln^ y^ e x
(^3) 3 C e Ce x
(^3) 3
y
y x
ln y
x^3 3
C
y
dy y
(^) y x^2 dx
dy y
x^2 dx y 0
y 0
618 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
N If a solution is a function that satisfies for some , it follows from a uniqueness theorem for solutions of differential equations that y x 0 for all x.
y x 0 x
y
N Figure 3 shows a direction field for the differ- ential equation in Example 3. Compare it with Figure 4, in which we use the equation to graph solutions for several values of. If you use the direction field to sketch solution curves with -intercepts , , , , and , they will resemble the curves in Figure 4.
2
y 521 1
A
y Ae x^^3 ^3
R
E
switch
L
FIGURE 5
members of the family. If we differentiate , we get
This differential equation depends on , but we need an equation that is valid for all values of simultaneously. To eliminate we note that, from the equation of the given general parabola , we have and so the differential equation can be written as
or
This means that the slope of the tangent line at any point on one of the parabolas is
. On an orthogonal trajectory the slope of the tangent line must be the nega- tive reciprocal of this slope. Therefore the orthogonal trajectories must satisfy the differ- ential equation
This differential equation is separable, and we solve it as follows:
where is an arbitrary positive constant. Thus the orthogonal trajectories are the family of ellipses given by Equation 4 and sketched in Figure 9. M
Orthogonal trajectories occur in various branches of physics. For example, in an elec- trostatic field the lines of force are orthogonal to the lines of constant potential. Also, the streamlines in aerodynamics are orthogonal trajectories of the velocity-equipotential curves.
MIXING PROBLEMS A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed solution of some substance, such as salt. A solution of a given concentration enters the tank at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which may differ from the entering rate. If denotes the amount of substance in the tank at time t , then is the rate at which the substance is being added minus the rate at which it is being removed. The mathematical description of this situation often leads to a first-order sepa- rable differential equation. We can use the same type of reasoning to model a variety of phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into the bloodstream.
y t
y t
C
x^2
y^2 2
4 C
y^2 2
x^2 C
y y^ dy^ ^ y 2 x^ dx
dy dx
2 x y
y y 2 x
x , y
dy dx
y 2 x
dy dx
2 ky
x y^2
y
x ky^2 k x y^2
k k
k
dy dx
2 ky
1 2 ky or
dy dx
x ky^2
620 | | | | CHAPTER 10 DIFFERENTIAL EQUATIONS
x
y
FIGURE 9
EXAMPLE 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that con- tains 0.03 kg of salt per liter of water enters the tank at a rate of 25 Lmin. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt remains in the tank after half an hour? SOLUTION Let be the amount of salt (in kilograms) after minutes. We are given that and we want to find. We do this by finding a differential equation satis- fied by. Note that is the rate of change of the amount of salt, so
where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which salt leaves the tank. We have
The tank always contains 5000 L of liquid, so the concentration at time is (measured in kilograms per liter). Since the brine flows out at a rate of 25 Lmin, we have
Thus, from Equation 5, we get
Solving this separable differential equation, we obtain
Since , we have , so
Therefore
Since is continuous and and the right side is never 0, we deduce that is always positive. Thus and so
The amount of salt after 30 min is
y 30 150 130 e ^30 ^200 38.1 kg M
y t 150 130 e t ^200
150 y t 150 y 150 y
y t y 0 20
150 y 130 e t ^200
ln 150 y
t 200
ln 130
y 0 20 ln 130 C
ln 150 y
t 200
C
y
dy 150 y
(^) y
dt 200
dy dt
y t 200
150 y t 200
rate out (^)
y t 5000
kg L ^25
L
min ^
y t 200
kg min
t y t 5000
rate in (^) 0.
kg L ^25
L
min ^ 0.^
kg min
dy dt
(^5) rate in rate out
y t dy dt
y 0 20 y 30
y t t
SECTION 10.3 SEPARABLE EQUATIONS | | | | 621
N Figure 10 shows the graph of the function of Example 6. Notice that, as time goes by, the amount of salt approaches 150 kg.
y t
t
y
(^0 200 )
50
100
150
FIGURE 10
