Advanced math

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ADVANCED
ENGINEERING
MATHEMATICS
International Student Edition

PETER V. O’NEIL
University of Alabama
at Birmingham

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Advanced Engineering Mathematics, International Student Edition
by Peter V. O’Neil

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PRELIMINARY CONCEPTS SEPARABLE EQUATIONS
HOMOGENEOUS, BERNOULLI, AND RICCATI EQUA-
CHAPTER 1 TIONS APPLICATIONS TO MECHANICS, ELECTRICAL
CIRCUITS, AND ORTHOGONAL TRAJECTORIES EXI

First-Order
Differential
Equations

1.1 Preliminary Concepts
Before developing techniques for solving various kinds of differential equations, we will develop
some terminology and geometric insight.

1.1.1 General and Particular Solutions
A first-order differential equation is any equation involving a first derivative, but no higher
derivative. In its most general form, it has the appearance
F x y y =0 (1.1)
in which y x is the function of interest and x is the independent variable. Examples are
y − y 2 − ey = 0
y −2 = 0
and
y − cos x = 0
Note that y must be present for an equation to qualify as a first-order differential equation, but
x and/or y need not occur explicitly.
A solution of equation (1.1) on an interval I is a function that satisfies the equation for
all x in I. That is,
Fx x x =0 for all x in I.
For example,
x = 2 + ke−x
3



4 CHAPTER 1 First-Order Differential Equations

is a solution of
y +y = 2
for all real x, and for any number k. Here I can be chosen as the entire real line. And
x = x ln x + cx
is a solution of
y
y = +1
x
for all x > 0, and for any number c.
In both of these examples, the solution contained an arbitrary constant. This is a symbol
independent of x and y that can be assigned any numerical value. Such a solution is called the
general solution of the differential equation. Thus
x = 2 + ke−x
is the general solution of y + y = 2.
Each choice of the constant in the general solution yields a particular solution. For example,
f x = 2 + e−x g x = 2 − e−x
and
√
h x = 2 − 53e−x
are all particular solutions of y + y = 2, obtained by choosing, respectively, k = 1, −1 and
√
− 53 in the general solution.

1.1.2 Implicitly Defined Solutions
Sometimes we can write a solution explicitly giving y as a function of x. For example,
y = ke−x
is the general solution of
y = −y
as can be verified by substitution. This general solution is explicit, with y isolated on one side
of an equation, and a function of x on the other.
By contrast, consider
2xy3 + 2
y =−
3x2 y2 + 8e4y
We claim that the general solution is the function y x implicitly defined by the equation
x2 y3 + 2x + 2e4y = k (1.2)
in which k can be any number. To verify this, implicitly differentiate equation (1.2) with respect
to x, remembering that y is a function of x. We obtain
2xy3 + 3x2 y2 y + 2 + 8e4y y = 0
and solving for y yields the differential equation.
In this example we are unable to solve equation (1.2) explicitly for y as a function of x,
isolating y on one side. Equation (1.2), implicitly defining the general solution, was obtained
by a technique we will develop shortly, but this technique cannot guarantee an explicit solution.



1.1 Preliminary Concepts 5

1.1.3 Integral Curves
A graph of a solution of a first-order differential equation is called an integral curve of the
equation. If we know the general solution, we obtain an infinite family of integral curves, one
for each choice of the arbitrary constant.

EXAMPLE 1.1

We have seen that the general solution of
y +y = 2
is
y = 2 + ke−x
for all x. The integral curves of y + y = 2 are graphs of y = 2 + ke−x for different choices of
k. Some of these are shown in Figure 1.1.

y

30
kϭ6

20

kϭ3
10

k ϭ 0 ( y ϭ 2)
x
Ϫ2 Ϫ1 0 1 2 3 4 5 6

k ϭ Ϫ3
Ϫ10

k ϭ Ϫ6
Ϫ20

FIGURE 1.1 Integral curves of y + y = 2 for k = 0 3 −3 6, and −6.

EXAMPLE 1.2

It is routine to verify that the general solution of
y
y + = ex
x
is
1
y= xex − ex + c
x




for x = 0. Graphs of some of these integral curves, obtained by making choices for c, are shown
in Figure 1.2.

y

40

c ϭ 20
30

20

10 cϭ5

cϭ0 x
0.5 1.0 1.5 2.0 2.5 3.0
Ϫ10
c ϭ Ϫ6
Ϫ20 c ϭ Ϫ10

FIGURE 1.2 Integral curves of y + x y = ex for c = 0 5 20 −6, and
1

−10.

We will see shortly how these general solutions are obtained. For the moment, we simply
want to illustrate integral curves.
Although in simple cases integral curves can be sketched by hand, generally we need
computer assistance. Computer packages such as MAPLE, MATHEMATICA and MATLAB
are widely available. Here is an example in which the need for computing assistance is clear.

EXAMPLE 1.3

The differential equation
y + xy = 2
has general solution
x
y x = e−x d + ke−x
2 /2 2 /2 2 /2
2e
0

Figure 1.3 shows computer-generated integral curves corresponding to k = 0, 4, 13, −7, −15
and −11.

1.1.4 The Initial Value Problem
The general solution of a first-order differential equation F x y y = 0 contains an arbitrary
constant, hence there is an infinite family of integral curves, one for each choice of the constant.
If we specify that a solution is to pass through a particular point x0 y0 , then we must find that
particular integral curve (or curves) passing through this point. This is called an initial value
problem. Thus, a first order initial value problem has the form
F x y y =0 y x0 = y0
in which x0 and y0 are given numbers. The condition y x0 = y0 is called an initial condition.




y

k ϭ 13
10

5 kϭ4

Ϫ2 kϭ0
0 x
Ϫ4 k ϭ Ϫ7 2 4

Ϫ5

Ϫ10 k ϭ Ϫ11

k ϭ Ϫ15
Ϫ15

FIGURE 1.3 Integral curves of y + xy = 2 for k = 0 4 13 −7 −15, and
−11.

EXAMPLE 1.4

Consider the initial value problem
y +y = 2 y 1 = −5
From Example 1.1, the general solution of y + y = 2 is
y = 2 + ke−x
Graphs of this equation are the integral curves. We want the one passing through 1 −5 . Solve
for k so that
y 1 = 2 + ke−1 = −5
obtaining
k = −7e
The solution of this initial value problem is
y = 2 − 7ee−x = 2 − 7e− x−1
As a check, y 1 = 2 − 7 = −5

The effect of the initial condition in this example was to pick out one special integral curve
as the solution sought. This suggests that an initial value problem may be expected to have a
unique solution. We will see later that this is the case, under mild conditions on the coefficients
in the differential equation.

1.1.5 Direction Fields
Imagine a curve, as in Figure 1.4. If we choose some points on the curve and, at each point,
draw a segment of the tangent to the curve there, then these segments give a rough outline of
the shape of the curve. This simple observation is the key to a powerful device for envisioning
integral curves of a differential equation.




y

x

FIGURE 1.4 Short tangent
segments suggest the shape
of the curve.

The general first-order differential equation has the form

F x y y =0

Suppose we can solve for y and write the differential equation as

y =f x y

Here f is a known function. Suppose f x y is defined for all points x y in some region
R of the plane. The slope of the integral curve through a given point x0 y0 of R is y x0 ,
which equals f x0 y0 . If we compute f x y at selected points in R, and draw a small line
segment having slope f x y at each x y , we obtain a collection of segments which trace out
the shapes of the integral curves. This enables us to obtain important insight into the behavior
of the solutions (such as where solutions are increasing or decreasing, limits they might have
at various points, or behavior as x increases).
A drawing of the plane, with short line segments of slope f x y drawn at selected points
x y , is called a direction field of the differential equation y = f x y . The name derives from
the fact that at each point the line segment gives the direction of the integral curve through that
point. The line segments are called lineal elements.

EXAMPLE 1.5

Consider the equation

y = y2

Here f x y = y2 , so the slope of the integral curve through x y is y2 . Select some points and,
through each, draw a short line segment having slope y2 . A computer generated direction field
is shown in Figure 1.5(a). The lineal elements form a profile of some integral curves and give
us some insight into the behavior of solutions, at least in this part of the plane. Figure 1.5(b)
reproduces this direction field, with graphs of the integral curves through 0 1 , 0 2 , 0 3 ,
0 −1 , 0 −2 and 0 −3 .
By a method we will develop, the general solution of y = y2 is

1
y=−
x+k

so the integral curves form a family of hyperbolas, as suggested by the curves sketched in
Figure 1.5(b).




y

4

2

x
Ϫ4 Ϫ2 0 2 4

Ϫ2

Ϫ4

FIGURE 1.5(a) A direction field for y = y2 .

y

4

2

2 4
x
Ϫ4 Ϫ2 0

Ϫ2

Ϫ4

FIGURE 1.5(b) Direction field for y = y2 and integral curves through 0 1 , 0 2 ,
0 3 0 −1 , 0 −2 , and 0 −3 .




EXAMPLE 1.6

Figure 1.6 shows a direction field for
y = sin xy
together with the integral curves through 0 1 , 0 2 , 0 3 , 0 −1 , 0 −2 and 0 −3 . In
this case, we cannot write a simple expression for the general solution, and the direction field
provides information about the behavior of solutions that is not otherwise readily apparent.

y

4

2

x
Ϫ4 Ϫ2 0 2 4

Ϫ2

Ϫ4

FIGURE 1.6 Direction field for y = sin xy and integral curves through 0 1 , 0 2 ,
0 3 , 0 −1 , 0 −2 , and 0 −3 .

With this as background, we will begin a program of identifying special classes of first-
order differential equations for which there are techniques for writing the general solution. This
will occupy the next five sections.

SECTION 1.1 PROBLEMS

In each of Problems 1 through 6, determine whether the x2 − 3
5. xy = x − y x = for x = 0
given function is a solution of the differential equation. 2x
√ 6. y + y = 1 x = 1 + Ce−x
1. 2yy = 1 x = x − 1 for x > 1
2. y + y = 0 x = Ce−x In each of Problems 7 through 11, verify by implicit
differentiation that the given equation implicitly defines a
2y + e x
C −e x
3. y = − for x > 0 x = solution of the differential equation.
2x 2x
2xy √ C 7. y2 + xy − 2x2 − 3x − 2y = C
4. y = for x = ± 2 x =
2−x 2 x 2 −2 y − 4x − 2 + x + 2y − 2 y = 0



1.2 Separable Equations 11

8. xy3 − y = C y3 + 3xy2 − 1 y = 0 17. y = x + y y 2 = 2
9. y − 4x + e = C 8x − ye − 2y + xe
2 2 xy xy xy
y =0 18. y = x − xy y 0 = −1
10. 8 ln x − 2y + 4 − 2x + 6y = C; 19. y = xy y 0 = 2
x − 2y 20. y = x − y + 1 y 0 = 1
y =
3x − 6y + 4
In each of Problems 21 through 26, generate a direction
−1
2x3 + 2xy2 − y x field and some integral curves for the differential equation.
11. tan y/x + x = C2
+ 2 y =0
x2 + y 2 x + y2 Also draw the integral curve representing the solution of
the initial value problem. These problems should be done
In each of Problems 12 through 16, solve the initial value
by a software package.
problem and graph the solution. Hint: Each of these dif-
ferential equations can be solved by direct integration. Use 21. y = sin y y 1 = /2
the initial condition to solve for the constant of integration.
22. y = x cos 2x − y y 1 = 0
12. y = 2x y 2 = 1 23. y = y sin x − 3x2 y 0 = 1
−x
13. y = e y 0 =2 24. y = ex − y y −2 = 1
14. y = 2x + 2 y −1 = 1 25. y − y cos x = 1 − x2 y 2 = 2
15. y = 4 cos x sin x y /2 = 0 26. y = 2y + 3 y 0 = 1
16. y = 8x + cos 2x y 0 = −3 27. Show that, for the differential equation y + p x y =
In each of Problems 17 through 20 draw some lineal q x , the lineal elements on any vertical line x = x0 ,
elements of the differential equation for −4 ≤ x ≤ 4, with p x0 = 0, all pass through the single point
−4 ≤ y ≤ 4. Use the resulting direction field to sketch a , where
graph of the solution of the initial value problem. (These 1 q x0
problems can be done by hand.) = x0 + and =
p x0 p x0

1.2 Separable Equations

DEFINITION 1.1 Separable Differential Equation
A differential equation is called separable if it can be written
y =A x B y

In this event, we can separate the variables and write, in differential form,
1
dy = A x dx
By

wherever B y = 0. We attempt to integrate this equation, writing

1
dy = A x dx
By

This yields an equation in x, y, and a constant of integration. This equation implicitly defines
the general solution y x . It may or may not be possible to solve explicitly for y x .




EXAMPLE 1.7

y = y2 e−x is separable. Write

dy
= y2 e−x
dx

as

1
dx = e−x dx
y2

for y = 0. Integrate this equation to obtain

1
− = −e−x + k
y

an equation that implicitly defines the general solution. In this example we can explicitly solve
for y, obtaining the general solution

1
y=
e−x − k

Now recall that we required that y = 0 in order to separate the variables by dividing by
y2 . In fact, the zero function y x = 0 is a solution of y = y2 ex , although it cannot be obtained
from the general solution by any choice of k. For this reason, y x = 0 is called a singular
solution of this equation.
Figure 1.7 shows graphs of particular solutions obtained by choosing k as 0, 3, −3, 6 and
−6.

y
3

2
kϭ0

1

k ϭ Ϫ3
k ϭ Ϫ6
x
Ϫ2 Ϫ1 1 2
kϭ6
kϭ3 Ϫ1

FIGURE 1.7 Integral curves of y = y2 e−x for
k = 0 3 −3 6, and −6.

Whenever we use separation of variables, we must be alert to solutions potentially lost
through conditions imposed by the algebra used to make the separation.




EXAMPLE 1.8

x2 y = 1 + y is separable, and we can write
1 1
dy = 2 dx
1+y x
The algebra of separation has required that x = 0 and y = −1, even though we can put x = 0
and y = −1 into the differential equation to obtain the correct equation 0 = 0.
Now integrate the separated equation to obtain
1
ln 1 + y = − + k
x
This implicitly defines the general solution. In this case, we can solve for y x explicitly. Begin
by taking the exponential of both sides to obtain
1 + y = ek e−1/x = Ae−1/x
in which we have written A = ek . Since k could be any number, A can be any positive number.
Then
1 + y = ±Ae−1/x = Be−1/x
in which B = ±A can be any nonzero number. The general solution is
y = −1 + Be−1/x
in which B is any nonzero number.
Now revisit the assumption that x = 0 and y = −1. In the general solution, we actually
obtain y = −1 if we allow B = 0. Further, the constant function y x = −1 does satisfy
x2 y = 1 + y. Thus, by allowing B to be any number, including 0, the general solution
y x = −1 + Be−1/x
contains all the solutions we have found. In this example, y = −1 is a solution, but not a
singular solution, since it occurs as a special case of the general solution.
Figure 1.8 shows graphs of solutions corresponding to B = −8 −5 0 4 and 7.

y
Bϭ7
4
Bϭ4
2
x
0 1 2 3 4 5
Ϫ2 Bϭ0

Ϫ4 B ϭ Ϫ5
Ϫ6
B ϭ Ϫ8
Ϫ8

FIGURE 1.8 Integral curves of x2 y = 1 + y for
B = 0 4 7 −5, and −8.

We often solve an initial value problem by finding the general solution of the differential
equation, then solving for the appropriate choice of the constant.




EXAMPLE 1.9

Solve the initial value problem

y = y2 e−x y 1 = 4

We know from Example 1.7 that the general solution of y = y2 e−x is
1
yx =
e−x − k
Now we need to choose k so that
1
y1 = =4
e−1 − k
from which we get
1
k = e−1 −
4
The solution of the initial value problem is
1
yx =
e−x + 4 − e−1
1

EXAMPLE 1.10

The general solution of
x−1 2
y =y
y+3
is implicitly defined by
1
y + 3 ln y = x−1 3 +k (1.3)
3
To obtain the solution satisfying y 3 = −1, put x = 3 and y = −1 into equation (1.3) to obtain
1 3
−1 = 2 +k
3
hence
11
k=−
3
The solution of this initial value problem is implicitly defined by
1 11
y + 3 ln y = x−1 3 −
3 3

1.2.1 Some Applications of Separable Differential Equations
Separable equations arise in many contexts, of which we will discuss three.




EXAMPLE 1.11

(The Mathematical Policewoman) A murder victim is discovered, and a lieutenant from the
forensic science laboratory is summoned to estimate the time of death.
The body is located in a room that is kept at a constant 68 degrees Fahrenheit. For some time
after the death, the body will radiate heat into the cooler room, causing the body’s temperature
to decrease. Assuming (for want of better information) that the victim’s temperature was a
“normal” 98 6 at the time of death, the lieutenant will try to estimate this time by observing the
body’s current temperature and calculating how long it would have had to lose heat to reach
this point.
According to Newton’s law of cooling, the body will radiate heat energy into the room at
a rate proportional to the difference in temperature between the body and the room. If T t is
the body temperature at time t, then for some constant of proportionality k,
T t = k T t − 68
The lieutenant recognizes this as a separable differential equation and writes
1
dT = k dt
T − 68
Upon integrating, she gets
ln T − 68 = kt + C
Taking exponentials, she gets
T − 68 = ekt+C = Aekt
in which A = eC . Then
T − 68 = ±Aekt = Bekt
Then
T t = 68 + Bekt
Now the constants k and B must be determined, and this requires information. The lieutenant
arrived at 9:40 p.m. and immediately measured the body temperature, obtaining 94 4 degrees.
Letting 9:40 be time zero for convenience, this means that
T 0 = 94 4 = 68 + B
and so B = 26 4. Thus far,
T t = 68 + 26 4ekt
To determine k, the lieutenant makes another measurement. At 11:00 she finds that the
body temperature is 89 2 degrees. Since 11:00 is 80 minutes past 9:40, this means that
T 80 = 89 2 = 68 + 26 4e80k
Then
21 2
e80k =
26 4
so
21 2
80k = ln
26 4




and
1 21 2
k= ln
80 26 4
The lieutenant now has the temperature function:
T t = 68 + 26 4eln 21 2/26 4 t/80
In order to find when last time when the body was 98 6 (presumably the time of death), solve
for the time in
T t = 98 6 = 68 + 26 4eln 21 2/26 4 t/80
To do this, the lieutenant writes
30 6
= eln 21 2/26 4 t/80
26 4
and takes the logarithm of both sides to obtain
30 6 t 21 2
ln = ln
26 4 80 26 4
Therefore the time of death, according to this mathematical model, was
80 ln 30 6/26 4
t=
ln 21 2/26 4
which is approximately −53 8 minutes. Death occurred approximately 53 8 minutes before
(because of the negative sign) the first measurement at 9:40, which was chosen as time zero.
This puts the murder at about 8:46 p.m.

EXAMPLE 1.12

(Radioactive Decay and Carbon Dating) In radioactive decay, mass is converted to energy by
radiation. It has been observed that the rate of change of the mass of a radioactive substance
is proportional to the mass itself. This means that, if m t is the mass at time t, then for some
constant of proportionality k that depends on the substance,
dm
= km
dt
This is a separable differential equation. Write it as
1
dm = k dt
m
and integrate to obtain
ln m = kt + c
Since mass is positive, m = m and
ln m = kt + c
Then
m t = ekt+c = Aekt
in which A can be any positive number.




Determination of A and k for a given element requires two measurements. Suppose at
some time, designated as time zero, there are M grams present. This is called the initial mass.
Then
m 0 =A=M
so
m t = Mekt
If at some later time T we find that there are MT grams, then
m T = MT = MekT
Then
MT
ln = kT
M
hence
1 MT
k= ln
T M
This gives us k and determines the mass at any time:
m t = Meln MT /M t/T
We obtain a more convenient formula for the mass if we choose the time of the second
measurement more carefully. Suppose we make the second measurement at that time T = H
at which exactly half of the mass has radiated away. At this time, half of the mass remains, so
MT = M/2 and MT /M = 1/2. Now the expression for the mass becomes
m t = Meln 1/2 t/H
or
m t = Me− ln 2 t/H (1.4)
This number H is called the half-life of the element. Although we took it to be the time
needed for half of the original amount M to decay, in fact, between any times t1 and t1 + H,
exactly half of the mass of the element present at t1 will radiate away. To see this, write
m t1 + H = Me− ln 2 t1 +H /H

= Me− ln 2 t1 /H e− ln 2 H/H = e− ln 2 m t1
= 2 m t1
1

Equation (1.4) is the basis for an important technique used to estimate the ages of certain
ancient artifacts. The earth’s upper atmosphere is constantly bombarded by high-energy cosmic
rays, producing large numbers of neutrons, which collide with nitrogen in the air, changing
some of it into radioactive carbon-14, or 14 C. This element has a half-life of about 5,730 years.
Over the relatively recent period of the history of this planet in which life has evolved, the
fraction of 14 C in the atmosphere, compared to regular carbon, has been essentially constant.
This means that living matter (plant or animal) has injested 14 C at about the same rate over a
long historical period, and objects living, say, two million years ago would have had the same
ratio of carbon-14 to carbon in their bodies as objects alive today. When an organism dies, it
ceases its intake of 14 C, which then begins to decay. By measuring the ratio of 14 C to carbon
in an artifact, we can estimate the amount of the decay, and hence the time it took, giving an




estimate of the time the organism was alive. This process of estimating the age of an artifact
is called carbon dating. Of course, in reality the ratio of 14 C in the atmosphere has only been
approximately constant, and in addition a sample may have been contaminated by exposure to
other living organisms, or even to the air, so carbon dating is a sensitive process that can lead
to controversial results. Nevertheless, when applied rigorously and combined with other tests
and information, it has proved a valuable tool in historical and archeological studies.
To apply equation (1.4) to carbon dating, use H = 5730 and compute
ln 2 ln 2
= ≈ 0 000120968
H 5730
in which ≈ means “approximately equal” (not all decimal places are listed). Equation (1.4)
becomes
m t = Me−0 000120968t
Now suppose we have an artifact, say a piece of fossilized wood, and measurements show that
the ratio of 14 C to carbon in the sample is 37 percent of the current ratio. If we say that the
wood died at time 0, then we want to compute the time T it would take for one gram of the
radioactive carbon to decay this amount. Thus, solve for T in
0 37 = e−0 000120968T
We find that
ln 0 37
T =− ≈ 8,219
0 000120968
2
years. This is a little less than one and one-half half-lives, a reasonable estimate if nearly 3
of
the 14 C has decayed.

EXAMPLE 1.13

(Torricelli’s Law) Suppose we want to estimate how long it will take for a container to empty
by discharging fluid through a drain hole. This is a simple enough problem for, say, a soda
can, but not quite so easy for a large oil storage tank or chemical facility.
We need two principles from physics. The first is that the rate of discharge of a fluid
flowing through an opening at the bottom of a container is given by
dV
= −kAv
dt
in which V t is the volume of fluid in the container at time t, v t is the discharge velocity
of fluid through the opening, A is the cross sectional area of the opening (assumed constant),
and k is a constant determined by the viscosity of the fluid, the shape of the opening, and the
fact that the cross-sectional area of fluid pouring out of the opening is slightly less than that
of the opening itself. In practice, k must be determined for the particular fluid, container, and
opening, and is a number between 0 and 1.
We also need Torricelli’s law, which states that v t is equal to the velocity of a free-falling
particle released from a height equal to the depth of the fluid at time t. (Free-falling means
that the particle is influenced by gravity only). Now the work done by gravity in moving the
particle from its initial point by a distance h t is mgh t , and this must equal the change in
1
the kinetic energy, 2 mv2 . Therefore,

vt = 2gh t




18 Ϫ h 18
r
h

FIGURE 1.9

Putting these two equations together yields
dV
= −kA 2gh t (1.5)
dt
We will apply equation (1.5) to a specific case to illustrate its use. Suppose we have a
hemispherical tank of water, as in Figure 1.9. The tank has radius 18 feet, and water drains
through a circular hole of radius 3 inches at the bottom. How long will it take the tank to
empty?
Equation (1.5) contains two unknown functions, V t and h t , so one must be eliminated.
Let r t be the radius of the surface of the fluid at time t and consider an interval of time
from t0 to t1 = t0 + t. The volume V of water draining from the tank in this time equals the
volume of a disk of thickness h (the change in depth) and radius r t∗ , for some t∗ between
t0 and t1 . Therefore
V= r t∗ 2
h
so
V h
= r t∗ 2
t t
In the limit as t → 0,
dV dh
= r2
dt dt
Putting this into equation (1.5) yields
dh
r2 = −kA 2gh
dt
Now V has been eliminated, but at the cost of introducing r t . However, from Figure 1.9,
r 2 = 182 − 18 − h 2 = 36h − h2
so
dh
36h − h2 = −kA 2gh
dt
This is a separable differential equation, which we write as
36h − h2
dh = −kA 2g dt
h1/2
Take g to be 32 feet per second per second. The radius of the circular opening is 3 inches, or
1
4
feet, so its area is A = /16 square feet. For water, and an opening of this shape and size,
the experiment gives k = 0 8. The last equation becomes
1 √
36h1/2 − h3/2 dh = − 0 8 64 dt
16




or
36h1/2 − h3/2 dh = −0 4 dt
A routine integration yields
2 2
24h3/2 − h5/2 = − t + c
5 5
or
60h3/2 − h5/2 = −t + k
Now h 0 = 18, so
60 18 3/2 − 18 5/2 = k
√
Thus k = 2268 2 and h t is implicitly determined by the equation
√
60h3/2 − h5/2 = 2268 2 − t
√
The tank is empty when h = 0, and this occurs when t = 2268 2 seconds, or about 53 minutes,
28 seconds.

The last three examples contain an important message. Differential equations can be used
to solve a variety of problems, but a problem usually does not present itself as a differential
equation. Normally we have some event or process, and we must use whatever information
we have about it to derive a differential equation and initial conditions. This process is called
mathematical modeling. The model consists of the differential equation and other relevant
information, such as initial conditions. We look for a function satisfying the differential equation
and the other information, in the hope of being able to predict future behavior, or perhaps better
understand the process being considered.

SECTION 1.2 PROBLEMS

In each of Problems 1 through 10, determine if the dif- In each of Problems 11 through 15, solve the initial value
ferential equation is separable. If it is, find the general problem.
solution (perhaps implicitly defined). If it is not separa-
11. xy2 y = y + 1 y 3e2 = 2
ble, do not attempt a solution at this time.
12. y = 3x2 y + 2 y 2 = 8
1. 3y = 4x/y2 13. ln yx y = 3x2 y y 2 = e3
2
2. y + xy = 0 14. 2yy = ex−y y 4 = −2
3. cos y y = sin x + y 15. yy = 2x sec 3y y 2/3 = /3
4. e x+y
y = 3x 16. An object having a temperature of 90 degrees Fahren-
heit is placed into an environment kept at 60 degrees.
5. xy + y = y 2
Ten minutes later the object has cooled to 88 degrees.
x + 1 2 − 2y What will be the temperature of the object after it has
6. y =
2y been in this environment for 20 minutes? How long
7. x sin y y = cos y will it take for the object to cool to 65 degrees?
x 2y2 + 1 17. A thermometer is carried outside a house whose am-
8. y = bient temperature is 70 degrees Fahrenheit. After five
y x+1
minutes the thermometer reads 60 degrees, and fifteen
9. y + y = ex − sin y minutes after this, 50 4 degrees. What is the outside
10. cos x + y + sin x − y y = cos 2x temperature (which is assumed to be constant)?




18. Assume that the population of bacteria in a petri dish 25. Calculate the time required to empty the hemispher-
changes at a rate proportional to the population at ical tank of Example 1.13 if the tank is positioned
that time. This means that, if P t is the population with its flat side down.
at time t, then 26. (Draining a Hot Tub) Consider a cylindrical hot tub
dP with a 5-foot radius and height of 4 feet, placed on
= kP
dt one of its circular ends. Water is draining from the tub
through a circular hole 5 inches in diameter located
8
for some constant k. A particular culture has a popu-
in the base of the tub.
lation density of 100,000 bacteria per square inch. A
culture that covered an area of 1 square inch at 10:00 (a) Assume a value k = 0 6 to determine the rate at
a.m. on Tuesday was found to have grown to cover 3 which the depth of the water is changing. Here it is
square inches by noon the following Thursday. How useful to write
many bacteria will be present at 3:00 p.m. the follow- dh dh dV dV/dt
ing Sunday? How many will be present on Monday at = =
dt dV dt dV/dh
4:00 p.m.? When will the world be overrun by these
bacteria, assuming that they can live anywhere on the (b) Calculate the time T required to drain the hot tub if
earth’s surface? (Here you need to look up the land it is initially full. Hint: One way to do this is to write
area of the earth.)
0 dt
19. Assume that a sphere of ice melts at a rate pro- T= dh
H dh
portional to its surface area, retaining a spherical
shape. Interpret melting as a reduction of volume with (c) Determine how much longer it takes to drain the
respect to time. Determine an expression for the vol- lower half than the upper half of the tub. Hint: Use
ume of the ice at any time t. the integral suggested in (b), with different limits for
20. A radioactive element has a half-life of ln 2 weeks. the two halves.
If e3 tons are present at a given time, how much will 27. (Draining a Cone) A tank shaped like a right circular
be left 3 weeks later? cone, with its vertex down, is 9 feet high and has a
21. The half-life of uranium-238 is approximately diameter of 8 feet. It is initially full of water.
4 5 · 109 years. How much of a 10-kilogram block (a) Determine the time required to drain the tank
of U-238 will be present 1 billion years from now? through a circular hole of diameter 2 inches at the
22. Given that 12 grams of a radioactive element decays vertex. Take k = 0 6.
to 9 1 grams in 4 minutes, what is the half-life of this (b) Determine the time it takes to drain the tank if it
element? is inverted and the drain hole is of the same size and
23. Evaluate shape as in (a), but now located in the new base.
28. (Drain Hole at Unknown Depth) Determine the rate of
−t2 −9/t2 change of the depth of water in the tank of Problem
e dt
0 27 (vertex at the bottom) if the drain hole is located in
the side of the cone 2 feet above the bottom of the tank.
Hint: Let
What is the rate of change in the depth of the water when
the drain hole is located in the bottom of the tank? Is it
e−t
2− x/t 2
Ix = dt possible to determine the location of the drain hole if we
0
are told the rate of change of the depth and the depth of
the water in the tank? Can this be done without knowing
Calculate I x by differentiating under the integral
the size of the drain opening?
sign, then let u = x/t. Show that I x = −2I x and
solve for I x . Evaluate the constant by using the stan- 29. Suppose the conical tank of Problem 27, vertex at the
√
dard result that 0 e−t dt =
2
/2. Finally, evaluate bottom, is initially empty and water is added at the
I3. constant rate of /10 cubic feet per second. Does the
tank ever overflow?
24. Derive the fact used in Example 1.13 that v t =
2gh t . Hint: Consider a free-falling particle hav- 30. (Draining a Sphere) Determine the time it takes to
ing height h t at time t. The work done by gravity in completely drain a spherical tank of radius 18 feet if
moving the particle from its starting point to a given it is initially full of water and the water drains through
point is mgh t , and this must equal the change in the a circular hole of radius 3 inches located in the bottom
kinetic energy, which is 1/2 mv2 . of the tank. Use k = 0 8.




1.3 Linear Differential Equations

DEFINITION 1.2 Linear Differential Equation
A first-order differential equation is linear if it has the form
y x +p x y = q x

Assume that p and q are continuous on an interval I (possibly the whole real line). Because
of the special form of the linear equation, we can obtain the general solution on I by a clever
observation. Multiply the differential equation by e p x dx to get
e p x dx
y x +p x e p x dx
y=q x e p x dx

p x dx
The left side of this equation is the derivative of the product y x e , enabling us to write

d
yxe p x dx
=q x e p x dx
dx
Now integrate to obtain

yxe p x dx
= qxe p x dx
dx + C

Finally, solve for y x :

y x = e− p x dx
qxe p x dx
dx + Ce− p x dx
(1.6)

The function e p x dx is called an integrating factor for the differential equation, because multi-
plication of the differential equation by this factor results in an equation that can be integrated to
obtain the general solution. We do not recommend memorizing equation (1.6). Instead, recognize
the form of the linear equation and understand the technique of solving it by multiplying by e p x dx .

EXAMPLE 1.14

The equation y + y = sin x is linear. Here p x = 1 and q x = sin x , both continuous for
all x. An integrating factor is
dx
e
or ex . Multiply the differential equation by ex to get
y ex + yex = ex sin x
or
yex = ex sin x
Integrate to get
1
yex = ex sin x dx = ex sin x − cos x + C
2
The general solution is
1
yx = sin x − cos x + Ce−x
2



1.3 Linear Differential Equations 23

EXAMPLE 1.15

Solve the initial value problem
y
y = 3x2 − y 1 =5
x
First recognize that the differential equation can be written in linear form:
1
y + y = 3x2
x

An integrating factor is e 1/x dx
= eln x = x, for x > 0. Multiply the differential equation by x
to get
xy + y = 3x3
or
xy = 3x3
Integrate to get
3
xy = x4 + C
4
Then
3 C
y x = x3 +
4 x
for x > 0. For the initial condition, we need
3
y 1 =5= +C
4
so C = 17/4 and the solution of the initial value problem is
3 17
y x = x3 +
4 4x
for x > 0.

Depending on p and q, it may not be possible to evaluate all of the integrals in the general
solution 1.6 in closed form (as a finite algebraic combination of elementary functions). This
occurs with
y + xy = 2
whose general solution is

y x = 2e−x dx + Ce−x
2 /2 2 /2 2 /2
ex
2
We cannot write ex /2 dx in elementary terms. However, we could still use a software package
to generate a direction field and integral curves, as is done in Figure 1.10. This provides some
idea of the behavior of solutions, at least within the range of the diagram.




y
4

2

x
Ϫ4 Ϫ2 0 2 4

Ϫ2

Ϫ4

FIGURE 1.10 Integral curves of y + xy = 2 passing through 0 2 ,
0 4 , 0 −2 , and 0 −5 .

Linear differential equations arise in many contexts. Example 1.11, involving estimation
of time of death, involved a separable differential equation which is also linear and could have
been solved using an integrating factor.

EXAMPLE 1.16

(A Mixing Problem) Sometimes we want to know how much of a given substance is present in
a container in which various substances are being added, mixed, and removed. Such problems
are called mixing problems, and they are frequently encountered in the chemical industry and
in manufacturing processes.
As an example, suppose a tank contains 200 gallons of brine (salt mixed with water), in
which 100 pounds of salt are dissolved. A mixture consisting of 1 pound of salt per gallon is
8
flowing into the tank at a rate of 3 gallons per minute, and the mixture is continuously stirred.
Meanwhile, brine is allowed to empty out of the tank at the same rate of 3 gallons per minute
(Figure 1.11). How much salt is in the tank at any time?

1
8lb/gal;
3 gal/min

3 gal/min

FIGURE 1.11

Before constructing a mathematical model, notice that the initial ratio of salt to brine in
1
the tank is 100 pounds per 200 gallons, or 2 pound per gallon. Since the mixture pumped in
1
has a constant ratio of 8 pound per gallon, we expect the brine mixture to dilute toward the
incoming ratio, with a “terminal” amount of salt in the tank of 1 pound per gallon, times 200
8
gallons. This leads to the expectation that in the long term (as t → ) the amount of salt in the
tank should approach 25 pounds.


Advanced math

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