Advance Engineering Mathematic chaper 1

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A D V A N C E D

E N G I N E E R I N G M A T H E M A T I C S

  International Student Edition PETER V. O’NEIL University of Alabama at Birmingham

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  Licensed to: iChapters User

  Licensed to: iChapters User PRELIMINARY CONCEPTS SEPARABLE EQUATIONS HOMOGENEOUS, BERNOULLI, AND RICCATI EQUA-

TIONS APPLICATIONS TO MECHANICS, ELECTRICAL

  C H A P T E R

1 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

  ′

  (1.1) F x y y

  = 0 in which y x is the function of interest and x is the independent variable. Examples are

  2 y ′

  y − y − e = 0

  ′

  y − 2 = 0 and

  ′

  y − cos x = 0

  ′

  Note that must be present for an equation to qualify as a first-order differential equation, but y 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,

  ′

  for all F x x x in I. x = 0

  For example,

  −x

  x = 2 + ke

  3 Licensed to: iChapters User

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

  x = 2 + ke

  ′

  is the general solution of y

• y = 2. Each choice of the constant in the general solution yields a particular solution. For example,

  −x −x

  f x = 2 + e g x = 2 − e and √

  −x

  53 e h x = 2 −

  ′

  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,

  −x

  y = ke 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

  3

  2 xy

• 2

  

′

  y = −

  2

  2 4 y

  3 x y

• 8e We claim that the general solution is the function y x implicitly defined by the equation

  

2

  3 4 y

  (1.2) x y

• 2x + 2e = k 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

  y

  3

  2 2 ′ 4 ′

  2 xy y y y

• 3x + 2 + 8e = 0

  ′ and solving for yields the differential equation.

  y 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. Licensed to: iChapters User

  1.1 Preliminary Concepts

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

  −x

  y = 2 + ke

  ′ −x

  for all for different choices of x. The integral curves of y

• y = 2 are graphs of y = 2 + ke k. Some of these are shown in Figure 1.1.

  y

  30

  k ⫽ 6

  20

  k ⫽ 3

  10

  k ⫽ 0 ( y ⫽ 2) x

  ⫺2 ⫺1

  1

  2

  3

  4

  5

  6

  k ⫽ ⫺3

  ⫺10

  k ⫽ ⫺6

  ⫺20

  

′

Integral curves of

FIGURE 1.1 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

  x ′

• y

  = e x is

  1

  x x

  xe y = − e + c x Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  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 ⫽ 20

  30

  20

  c ⫽ 5

  10

  c ⫽ 0 x

  0.5

  1.0

  1.5

  2.5

  3.0

  2.0 ⫺10

  c ⫽ ⫺6

  ⫺20

  c ⫽ ⫺10 1 x ′

  Integral curves of for

FIGURE 1.2 y

• y = e c = 0 5 20 −6, and

  x −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
₂ x _{2 2} /2 /2 /2

  −x −x

  2 e y x = e d + ke

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 x y , 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 y x = 0 = y in which and are given numbers. The condition is called an initial condition. x y y x

  = y Licensed to: iChapters User

  1.1 Preliminary Concepts y k ⫽ 13

  10

  k ⫽ 4

  5 ⫺2

  k ⫽ 0 x

  4 ⫺4

  k ⫽ ⫺7

  2 ⫺5

  k ⫽ ⫺11

  ⫺10

  k ⫽ ⫺15

  ⫺15

  ′ Integral curves of

FIGURE 1.3 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

  −x

  y = 2 + ke Graphs of this equation are the integral curves. We want the one passing through 1 −5 . Solve for k so that

  −1

  y 1 = 2 + ke = −5 obtaining k = −7e

  The solution of this initial value problem is

  −x − x−1

  y = 2 − 7ee = 2 − 7e 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. Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  y x Short tangent

  FIGURE 1.4 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 and write the differential equation as y

  ′

  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 x y of R is y x , which equals f x y . 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

  ′

  2

  y = y

  2

  2 Here , so the slope of the integral curve through x y is y . Select some points and,

  f x y = y

  2

  through each, draw a short line segment having slope y . 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 .

  2 ′

  By a method we will develop, the general solution of is y = y

  1 y = − x + k so the integral curves form a family of hyperbolas, as suggested by the curves sketched in

  Figure 1.5(b). Licensed to: iChapters User

  1.1 Preliminary Concepts y

  4

  2

  x

  ⫺4 ⫺2

  2

  4 ⫺2 ⫺4

  2 ′ A direction field for .

  FIGURE 1.5(a) y = y y

  4

  2

  2

  4

  x

  ⫺4 ⫺2 ⫺2

  ⫺4

  ′

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

  0 3 0 −1 , 0 −2 , and 0 −3 .

CHAPTER 1 First-Order Differential Equations

  

y

′

  2 x for x = 0

  6. y

  ′

  −x In each of Problems 7 through 11, verify by implicit differentiation that the given equation implicitly defines a solution of the differential equation. 7. y

  Licensed to: iChapters User

  S E C T I O N 1 . 1 PROBLEMS In each of Problems 1 through 6, determine whether the given function is a solution of the differential equation.

  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.

  = sin xy and integral curves through 0 1 , 0 2 , 0 3 , 0 −1 , 0 −2 , and 0 −3 .

  4 FIGURE 1.6 Direction field for

  ′ = x − y x = x

  2

  4

  2

  ⫺4 ⫺2 ⫺2 ⫺4

  y x

  = 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.

  ′

Figure 1.6 shows a direction field for y

  EXAMPLE 1.6

  2 − 3

• y = 1 x = 1 + Ce
• y = 0 x = Ce

  2 x for x > 0 x =

  −x 3. y

  1.

  ′ = 0

  2 − 3x − 2y = C y − 4x − 2 + x + 2y − 2 y

  2

  2 yy ′

  = 1 x = √ x − 1 for x > 1 2. y

  ′

  ′ = −

  C − e x

  2 y + e x

  2 − 2 5. xy

  2 x = C x

  √

  2 for x = ±

  2 xy 2 − x

  ′ =

  2 x 4. y

• xy − 2x
• 3xy
• e
• 2xy
• x x
• y
• y

  ′ = 8x + cos 2x y 0 = −3 In each of Problems 17 through 20 draw some lineal elements of the differential equation for

−4 ≤ x ≤ 4,

−4 ≤ y ≤ 4. Use the resulting direction field to sketch a graph of the solution of the initial value problem. (These problems can be done by hand.) 17. y

  ′ = x + y y 2 = 2 18. y

  ′ = x − xy y 0 = −1 19. y

  ′ = xy y 0 = 2 20. y

  ′ = x − y + 1 y 0 = 1 In each of Problems 21 through 26, generate a direction field and some integral curves for the differential equation.

  Also draw the integral curve representing the solution of the initial value problem. These problems should be done by a software package. 21. y

  ′ = sin y y 1 = /2 22. y

  ′ = x cos 2x − y y 1 = 0 23. y

  ′ = y sin x − 3x

  2 y 0 = 1 24. y

  ′ = e x

  − y y −2 = 1 25. y

  2 y 2 = 2 26. y

  ′ − y cos x = 1 − x

  ′ = 2y + 3 y 0 = 1

  27. Show that, for the differential equation y ′

  1 p x and = q x p x

  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 B y dy = A x dx wherever

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

  1 B y dy = A x dx

  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 .

  Licensed to: iChapters User

  ′ = 4 cos x sin x y /2 = 0 16. y

  −x y 0 = 2 14. y

  ′ = 2x + 2 y −1 = 1 15. y

  ′ = x − 2y

  1.2 Separable Equations

  8. xy

  3 − y = C y

  3

  2 − 1 y

  ′ = 0 9. y

  2 − 4x

  2

  xy = C 8x − ye xy

  − 2y + xe xy y

  ′

= 0

10. 8 ln x − 2y + 4 − 2x + 6y = C; y

  3 x − 6y + 4 11. tan

  ′ = e

  −1 y/x + x

  2 = C

  2 x

  3

  2 − y x

  2

  2

  2

  

2

y

  ′ = 0 In each of Problems 12 through 16, solve the initial value problem and graph the solution. Hint: Each of these dif- ferential equations can be solved by direct integration. Use the initial condition to solve for the constant of integration.

  12. y

  ′ = 2x y 2 = 1 13. y

• p x y = q x , the lineal elements on any vertical line x = x , with p x = 0, all pass through the single point , where = x +

1.2 Separable Equations

  Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  EXAMPLE 1.7

  2 ′ −x

  is separable. Write y e = y dy

  2 −x

  e = y dx as

  1

  −x

  dx dx = e

  2

  y for y = 0. Integrate this equation to obtain

  1

  −x

  − = −e + 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 =

  −x

  e − k

  Now recall that we required that y = 0 in order to separate the variables by dividing by

  2 2 x ′

  . In fact, the zero function , although it cannot be obtained y e y x = 0 is a solution of y = y 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

  k ⫽ 0

  1

  k ⫽ ⫺3 k ⫽ ⫺6 x

  ⫺2 ⫺1

  2

  1

  k ⫽ 6 k ⫽ 3

  ⫺1

  ′ 2 −x

FIGURE 1.7 Integral curves of for

  y e = y 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. Licensed to: iChapters User

  1.2 Separable Equations EXAMPLE 1.8

  2 ′

  x y = 1 + y is separable, and we can write

  1

  1 dx dy =

  2

  1 x

• y 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

  k −1/x −1/x

  e 1 + y = e = Ae

  k

  in which we have written . Since k could be any number, A can be any positive number.

  A = e Then

  −1/x −1/x

  1

• y = ±Ae = Be in which

  B = ±A can be any nonzero number. The general solution is

  −1/x

  y = −1 + Be 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

  2 ′

  x y = 1 + y. Thus, by allowing B to be any number, including 0, the general solution

  −1/x

  y x = −1 + Be 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 B ⫽ 7

  4 B ⫽ 4

  2

  x

  1

  2

  3

  4

  5 B ⫽ 0 ⫺2 ⫺4

  B ⫽ ⫺5

  ⫺6

  B ⫽ ⫺8

  ⫺8

  2 ′

FIGURE 1.8 Integral curves of x 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. Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  EXAMPLE 1.9

  Solve the initial value problem

  2 ′ −x

  y e = y y 1 = 4

  2 ′ −x

  We know from Example 1.7 that the general solution of is y e = y

  1 y x =

  −x

  e − k

  Now we need to choose k so that 1 y 1 = = 4

  −1

  e − k from which we get

  1

  −1

  k = e −

  4 The solution of the initial value problem is

  1 y x =

  1 −x −1

  e − e +

  4 EXAMPLE 1.10

  The general solution of

  2

  x − 1

  ′

  y = y y + 3 is implicitly defined by

  1

  3

  (1.3) y + 3 ln y = x − 1 + k

  3 To obtain the solution satisfying y 3 = −1, put x = 3 and y = −1 into equation (1.3) to obtain

  1

  3

  2 −1 = + k

  3 hence 11 k = −

  3 The solution of this initial value problem is implicitly defined by

  1

  11

  3

  y + 3 ln y = x − 1 −

  3

  3

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

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  1.2 Separable Equations 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

  kt kt+C

  T − 68 = e = Ae

  C

  in which . Then A = e

  kt kt

  T − 68 = ±Ae = Be Then

  kt

  T t = 68 + Be 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,

  kt

  T t = 68 + 26 4e 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

  k

  80 T 80 = 89 2 = 68 + 26 4e

  Then

  21

  2

  80 k

  e =

  26

  4 so 21 2

  80 k = ln

  26

  4 Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  and

  1 21 2 ln k =

  80

  26

  4 The lieutenant now has the temperature function:

  ln 21 2/26 4 t/80

  T t = 68 + 26 4e In order to find when last time when the body was 98 6 (presumably the time of death), solve for the time in

  ln 21 2/26 4 t/80

  T t = 98 6 = 68 + 26 4e To do this, the lieutenant writes

  30

  6

  ln 21 2/26 4 t/80

  = e

  26

  4 and takes the logarithm of both sides to obtain 30 6 t 21 2 ln ln

  =

  26

  80

  26

  4

  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

  kt kt+c

  m t = e = Ae in which A can be any positive number. Licensed to: iChapters User

  1.2 Separable Equations

  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

  kt

  m t = Me If at some later time grams, then

  T we find that there are M

  T kT

  m T = M T = Me Then

  M

  T

  ln = kT

  M hence

  1 M

  T

  ln k = T M

  This gives us k and determines the mass at any time:

  ln M /M t/T _T

  m t = Me 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 M

  T = M/2 and M T /M = 1/2. Now the expression for the mass becomes ln 1/2 t/H

  m t = Me or

  − ln 2 t/H

  (1.4) m t = Me 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 t and t

  1 1 + H,

  exactly half of the mass of the element present at t will radiate away. To see this, write

  1 − ln 2 t +H /H ₁

  m t

  1 + H = Me /H − ln 2 t − ln 2 H/H − ln 2 ₁

  e m t = Me = e

  1

  1

  m t =

  1

  2 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

  14

  some of it into radioactive carbon-14, or 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

  14 fraction of C in the atmosphere, compared to regular carbon, has been essentially constant.

  14 This means that living matter (plant or animal) has injested 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

  14

  14

  ceases its intake of

  C, which then begins to decay. By measuring the ratio of C to carbon in an artifact, we can estimate the amount of the decay, and hence the time it took, giving an Licensed to: iChapters User

CHAPTER 1 First-Order Differential Equations

  estimate of the time the organism was alive. This process of estimating the age of an artifact

  14

  is called carbon dating. Of course, in reality the ratio of 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

  5730 H in which

  ≈ means “approximately equal” (not all decimal places are listed). Equation (1.4) becomes

  −0 000120968t

  m t = Me Now suppose we have an artifact, say a piece of fossilized wood, and measurements show that

  14

  the ratio of 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 000120968T

  37 = e We find that ln 0 37

  T = − ≈ 8,219 000120968

  2

  years. This is a little less than one and one-half half-lives, a reasonable estimate if nearly of

  3

  14 the 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

  2

  the kinetic energy, . Therefore, mv

  2

  2 gh t v t = Licensed to: iChapters User

  1.2 Separable Equations

  18 18 ⫺ h

  r h FIGURE 1.9

  Putting these two equations together yields dV 2 (1.5) gh t

  = −kA 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 to t t

  1 = t + t. The volume V of water draining from the tank in this time equals the ∗ ∗

  volume of a disk of thickness between h (the change in depth) and radius r t , for some t and . Therefore t t

  1

  2 ∗

  h V = r t so

  V

  2 h ∗

  = r t t t In the limit as t → 0, dV dh

  2

  = r dt dt Putting this into equation (1.5) yields dh

  2

  2 r gh = −kA dt

  Now V has been eliminated, but at the cost of introducing r t . However, from Figure 1.9,

  2

  2

  2

  2

  r = 18 − 18 − h = 36h − h so _dh

  2

  2 gh 36h − h = −kA dt

  This is a separable differential equation, which we write as

  2

  36 h − h 2 g dt dh = −kA

  1 /2

  h Take g to be 32 feet per second per second. The radius of the circular opening is 3 inches, or

  1

  feet, so its area is A = /16 square feet. For water, and an opening of this shape and size,

  4

  the experiment gives k = 0 8. The last equation becomes √

  1

  

1 /2

3 /2

  36h 64 dt − h dh = − 0 8

  16

CHAPTER 1 First-Order Differential Equations

  2 6. y

  2 y

  ′ =

  ′ = cos y 8. x y y

  2 y 7. x sin y y

  2 − 2y

  ′ = x + 1

  ′

  y + y ′

  ′ = 3x 5. xy

  ′ = sin x + y 4. e x+y y

  ′ = 0 3. cos y y

  2 2. y + xy

  = 4x/y

  3 y ′

  2

  = e x

  S E C T I O N 1 . 2 PROBLEMS In each of Problems 1 through 10, determine if the dif- ferential equation is separable. If it is, find the general solution (perhaps implicitly defined). If it is not separa- ble, do not attempt a solution at this time.

  ′ = 3x

  4 degrees. What is the outside temperature (which is assumed to be constant)? Licensed to: iChapters User

  17. A thermometer is carried outside a house whose am- bient temperature is 70 degrees Fahrenheit. After five minutes the thermometer reads 60 degrees, and fifteen minutes after this, 50

  An object having a temperature of 90 degrees Fahren- heit is placed into an environment kept at 60 degrees. Ten minutes later the object has cooled to 88 degrees. What will be the temperature of the object after it has been in this environment for 20 minutes? How long will it take for the object to cool to 65 degrees?

  ′ = 2x sec 3y y 2/3 = /3 16.

  3 14. 2 yy ′ = e x−y ² y 4 = −2 15. yy

  2 y y 2 = e

  2 y + 2 y 2 = 8 13. ln y x y

  − sin y 10. cos x + y + sin x − y y

  ′ = 3x

  2 = 2 12. y

  ′ = y + 1 y 3e

  2 y

  11. xy

  ′ = cos 2x In each of Problems 11 through 15, solve the initial value problem.

  1.

  . 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.

  or 36h

  2

  60 h

  5 t + c or

  2

  = −

  5 /2

  5 h

  −

  − h

  

3 /2

  24 h

  dh = −0 4 dt A routine integration yields

  3 /2

  − h

  

1

/2

  3 /2

  5 /2

  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

  

3 /2

  28 seconds.

  The tank is empty when h = 0, and this occurs when t = 2268 √ 2 seconds, or about 53 minutes,

  2 − t

  = 2268 √

  5 /2

  − h

  √ 2 and h t is implicitly determined by the equation 60 h

  = −t + k Now h 0 = 18, so

  = k Thus k = 2268

  5 /2

  − 18

  3 /2

  18

  60

• y = y
• 1 x + 1 9.

  1.2 Separable Equations

  −t ² dt = √

  29. Suppose the conical tank of Problem 27, vertex at the bottom, is initially empty and water is added at the constant rate of /10 cubic feet per second. Does the tank ever overflow? 30. (Draining a Sphere) Determine the time it takes to completely drain a spherical tank of radius 18 feet if it is initially full of water and the water drains through a circular hole of radius 3 inches located in the bottom of the tank. Use k = 0 8.

  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. 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 possible to determine the location of the drain hole if we are told the rate of change of the depth and the depth of the water in the tank? Can this be done without knowing the size of the drain opening?

  27. (Draining a Cone) A tank shaped like a right circular cone, with its vertex down, is 9 feet high and has a diameter of 8 feet. It is initially full of water. (a) Determine the time required to drain the tank through a circular hole of diameter 2 inches at the vertex. Take k = 0 6. (b) Determine the time it takes to drain the tank if it is inverted and the drain hole is of the same size and shape as in (a), but now located in the new base. 28. (Drain Hole at Unknown Depth) Determine the rate of change of the depth of water in the tank of Problem

  H dt dh dh (c) Determine how much longer it takes to drain the lower half than the upper half of the tub. Hint: Use the integral suggested in (b), with different limits for the two halves.

  (b) Calculate the time T required to drain the hot tub if it is initially full. Hint: One way to do this is to write T =

  = dh dV dV dt = dV/dt dV/dh

  (a) Assume a value k = 0 6 to determine the rate at which the depth of the water is changing. Here it is useful to write dh dt

  8 inches in diameter located in the base of the tub.

  5

  25. Calculate the time required to empty the hemispher- ical tank of Example 1.13 if the tank is positioned with its flat side down. 26. (Draining a Hot Tub) Consider a cylindrical hot tub with a 5-foot radius and height of 4 feet, placed on one of its circular ends. Water is draining from the tub through a circular hole

  2 .

  24. Derive the fact used in Example 1.13 that v t = 2gh t . Hint: Consider a free-falling particle hav- ing height h t at time t. The work done by gravity in moving the particle from its starting point to a given point is mgh t , and this must equal the change in the kinetic energy, which is 1/2 mv

  /2. Finally, evaluate I 3 .