10 Differential Equations

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10 Differential Equations

1. Types of Differential Equations . . . 10-1 2. Homogeneous, First-Order Linear

Differential Equations with Constant

Coefficients . . . 10-2 3. First-Order Linear Differential Equations . . 10-2 4. First-Order Separable Differential

Equations . . . 10-2 5. First-Order Exact Differential Equations . . 10-3 6. Homogeneous, Second-Order Linear

Differential Equations with Constant

Coefficients . . . 10-3 7. Nonhomogeneous Differential Equations . . . 10-3 8. Named Differential Equations . . . 10-5 9. Laplace Transforms . . . 10-5 10. Step and Impulse Functions . . . 10-6 11. Algebra of Laplace Transforms . . . 10-6 12. Convolution Integral . . . 10-6 13. Using Laplace Transforms . . . 10-7 14. Third- and Higher-Order Linear Differential

Equations with Constant Coefficients . . . 10-8 15. Application: Engineering Systems . . . 10-8 16. Application: Mixing . . . 10-8 17. Application: Exponential Growth and

Decay . . . 10-9 18. Application: Epidemics . . . 10-10 19. Application: Surface Temperature . . . 10-10 20. Application: Evaporation . . . 10-10

1. TYPES OF DIFFERENTIAL EQUATIONS A differential equation is a mathematical expression combining a functionðe:g:; y ¼ f ðxÞÞ and one or more of its derivatives. The order of a differential equation is the highest derivative in it. First-order differential equa-tions contain only first derivatives of the function, second-order differential equations contain second derivatives (and may contain first derivatives as well), and so on.

A linear differential equation can be written as a sum of products of multipliers of the function and its derivatives.

If the multipliers are scalars, the differential equation is said to have constant coefficients. If the function or one of its derivatives is raised to some power (other than one) or is embedded in another function (e.g., y embedded in sin y or e^y), the equation is said to be nonlinear.

Each term of a homogeneous differential equation contains either the function (y) or one of its deriva-tives; that is, the sum of derivative terms is equal to zero. In a nonhomogeneous differential equation, the

sum of derivative terms is equal to a nonzero forcing function of the independent variable (e.g., g(x)). In order to solve a nonhomogeneous equation, it is often necessary to solve the homogeneous equation first. The homogeneous equation corresponding to a nonhomoge-neous equation is known as a reduced equation or com-plementary equation.

The following examples illustrate the types of differen-tial equations.

y⁰ 7y ¼ 0 homogeneous, first-order linear, with constant coefficients

y⁰⁰ 2y⁰þ 8y ¼ sin 2x nonhomogeneous, second-order linear, with constant coefficients

y⁰⁰ ðx² 1Þy²¼ sin 4x nonhomogeneous, second-order, nonlinear

An auxiliary equation (also called the characteristic equation) can be written for a homogeneous linear dif-ferential equation with constant coefficients, regardless of order. This auxiliary equation is simply the polyno-mial formed by replacing all derivatives with variables raised to the power of their respective derivatives.

The purpose of solving a differential equation is to derive an expression for the function in terms of the independent variable. The expression does not need to be explicit in the function, but there can be no deriva-tives in the expression. Since, in the simplest cases, solving a differential equation is equivalent to finding an indefinite integral, it is not surprising that constants of integration must be evaluated from knowledge of how the system behaves. Additional data are known as initial values, and any problem that includes them is known as an initial value problem.¹

Most differential equations require lengthy solutions and are not efficiently solved by hand. However, several types are fairly simple and are presented in this chapter.

1The term initial implies that time is the independent variable. While this may explain the origin of the term, initial value problems are not limited to the time domain. A boundary value problem is similar, except that the data come from different points. For example, addi-tional data in the form yðx0Þ and y⁰ðx0Þ or yðx0Þ and y⁰ðx1Þ that need to be simultaneously satisfied constitute an initial value problem. Data of the form yðx0Þ and yðx1Þ constitute a boundary value problem. Until solved, it is difficult to know whether a boundary value problem has zero, one, or more than one solution.

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Example 10.1

Write the complementary differential equation for the following nonhomogeneous differential equation.

y⁰⁰þ 6y⁰þ 9y ¼ e^14xsin 5x Solution

The complementary equation is found by eliminating the forcing function, e^14xsin 5x.

y⁰⁰þ 6y⁰þ 9y ¼ 0

Example 10.2

Write the auxiliary equation to the following differen-tial equation.

y⁰⁰þ 4y⁰þ y ¼ 0 Solution

Replacing each derivative with a polynomial term whose degree equals the original order, the auxiliary equation is

r²þ 4r þ 1 ¼ 0

2. HOMOGENEOUS, FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

A homogeneous, first-order linear differential equation with constant coefficients will have the general form of Eq. 10.1.

y⁰þ ky ¼ 0 ^10:1

The auxiliary equation is r + k = 0 and it has a root of r =k. Equation 10.2 is the solution.

y¼ Ae^rx¼ Ae^{kx 10:2} If the initial condition is known to be yð0Þ ¼ y0, the solution is

y¼ y0e^kx 10:3

3. FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS

A first-order linear differential equation has the general form of Eq. 10.4. p(x) and g(x) can be constants or any function of x (but not of y). However, if p(x) is a con-stant and g(x) is zero, it is easier to solve the equation as shown in Sec. 10.2.

y⁰þ pðxÞy ¼ gðxÞ 10:4

The integrating factor (which is usually a function) to this differential equation is

uðxÞ ¼ exp Z

pðxÞdx

 

10:5

The closed-form solution to Eq. 10.4 is y¼ 1

uðxÞ Z

uðxÞgðxÞdx þ C

 

10:6

For the special case where p(x) and g(x) are both con-stants, Eq. 10.4 becomes

y⁰þ ay ¼ b 10:7

If the initial condition is y(0) = y₀, then the solution to Eq. 10.7 is

y¼b

að1  e^axÞ þ y0e^ax 10:8

Example 10.3

Find a solution to the following differential equation.

y⁰ y ¼ 2xe^2x yð0Þ ¼ 1 Solution

This is a first-order linear equation with p(x) =1 and g(x) = 2xe^2x. The integrating factor is

uðxÞ ¼ exp Z

pðxÞdx

 

¼ exp Z

1dx

 

¼ e^x

The solution is given by Eq. 10.6.

y¼ 1 uðxÞ

Z

uðxÞgðxÞdx þ C

 

¼ 1 e^x

Z

e^x2xe^2xdxþ C

 

¼ e^x 2 Z

xe^xdxþ C

 

¼ e^xð2xe^x 2e^xþ CÞ

¼ e^xð2e^xðx  1Þ þ CÞ From the initial condition,

yð0Þ ¼ 1 e⁰

ð

ð2Þðe⁰Þð0  1Þ þ C

Þ

¼ 1 1

ð

ð2Þð1Þð1Þ þ C

Þ

¼ 1 Therefore, C = 3. The complete solution is

y¼ e^x

ð

^2exðx  1Þ þ 3

Þ

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4. FIRST-ORDER SEPARABLE DIFFERENTIAL EQUATIONS

First-order separable differential equations can be placed in the form of Eq. 10.9. For clarity and conve-nience, y⁰ is written as dy/dx.

mðxÞ þ nðyÞdy

dx¼ 0 ^10:9

Equation 10.9 can be placed in the form of Eq. 10.10, both sides of which are easily integrated. An initial value will establish the constant of integration.

mðxÞdx ¼ nðyÞdy 10:10

5. FIRST-ORDER EXACT DIFFERENTIAL EQUATIONS

A first-order exact differential equation has the form f_xðx; yÞ þ fyðx; yÞy⁰¼ 0 10:11

f_xðx; yÞ is the exact derivative of f ðx; yÞ with respect to x, and f_yðx; yÞ is the exact derivative of f ðx; yÞ with respect to y. The solution is

fðx; yÞ  C ¼ 0 10:12

6. HOMOGENEOUS, SECOND-ORDER

LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

Homogeneous second-order linear differential equations with constant coefficients have the form of Eq. 10.13.

They are most easily solved by finding the two roots of the auxiliary equation, Eq. 10.14.

y⁰⁰þ k¹y⁰þ k²y¼ 0 10:13

r²þ k¹rþ k²¼ 0 ^10:14 There are three cases. If the two roots of Eq. 10.14 are real and different, the solution is

y¼ A¹e^r1xþ A²e^r2x _10:15 If the two roots are real and the same, the solution is

y¼ A1e^rxþ A2xe^rx _10:16 r¼ k1

2 ^10:17

If the two roots are imaginary, they will be of the form ( + i !) and (  i !), and the solution is

y¼ A1e^xcos!x þ A2e^xsin!x 10:18

In all three cases, A₁and A₂must be found from the two initial conditions.

Example 10.4

Solve the following differential equation.

y⁰⁰þ 6y⁰þ 9y ¼ 0 yð0Þ ¼ 0 y⁰ð0Þ ¼ 1 Solution

The auxiliary equation is

r²þ 6r þ 9 ¼ 0 ðr þ 3Þðr þ 3Þ ¼ 0

The roots to the auxiliary equation are r₁= r₂ =3.

Therefore, the solution has the form of Eq. 10.16.

y¼ A¹e^3xþ A²xe^3x The first initial condition is

yð0Þ ¼ 0 A1e⁰þ A²ð0Þe⁰¼ 0 A₁þ 0 ¼ 0 A1¼ 0

To use the second initial condition, the derivative of the equation is needed. Making use of the known fact that A₁= 0,

y⁰¼ d

dxA2xe^3x ¼ 3A²xe^3xþ A²e^3x Using the second initial condition,

y⁰ð0Þ ¼ 1

3A²ð0Þe⁰þ A²e⁰¼ 1 0þ A²¼ 1 A₂¼ 1 The solution is

y¼ xe^3x

7. NONHOMOGENEOUS DIFFERENTIAL EQUATIONS

A nonhomogeneous equation has the form of Eq. 10.19.

f (x) is known as the forcing function.

y⁰⁰þ pðxÞy⁰þ qðxÞy ¼ f ðxÞ 10:19

The solution to Eq. 10.19 is the sum of two equations.

The complementary solution, y_c, solves the complemen-tary (i.e., homogeneous) problem. The particular solu-tion, y_p, is any specific solution to the nonhomogeneous Eq. 10.19 that is known or can be found. Initial values are used to evaluate any unknown coefficients in the

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complementary solution after y_cand y_phave been com-bined. (The particular solution will not have any unknown coefficients.)

y¼ ycþ yp 10:20

Two methods are available for finding a particular solu-tion. The method of undetermined coefficients, as pre-sented here, can be used only when p(x) and q(x) are constant coefficients and f (x) takes on one of the forms in Table 10.1.

The particular solution can be read from Table 10.1 if the forcing function is of one of the forms given. Of course, the coefficients A_iand B_iare not known—these are the undetermined coefficients. The exponent s is the smallest nonnegative number (and will be 0, 1, or 2), which ensures that no term in the particular solution, y_p, is also a solution to the complementary equation, y_c. s must be determined prior to proceeding with the solu-tion procedure.

Once y_p (including s) is known, it is differentiated to obtain y_p⁰ and y_p⁰⁰, and all three functions are substituted into the original nonhomogeneous equation. The result-ing equation is rearranged to match the forcresult-ing function, f (x), and the unknown coefficients are determined, usu-ally by solving simultaneous equations.

If the forcing function, f (x), is more complex than the forms shown in Table 10.1, or if either p(x) or q(x) is a function of x, the method of variation of parameters should be used. This complex and time-consuming method is not covered in this book.

Example 10.5

Solve the following nonhomogeneous differential equation.

y⁰⁰þ 2y⁰þ y ¼ e^xcos x

Solution

step 1: Find the solution to the complementary (homo-geneous) differential equation.

y⁰⁰þ 2y⁰þ y ¼ 0

Since this is a differential equation with constant coefficients, write the auxiliary equation.

r²þ 2r þ 1 ¼ 0

The auxiliary equation factors in (r + 1)² = 0 with two identical roots at r = 1. Therefore, the solution to the homogeneous differential equation is

y_cðxÞ ¼ C1e^xþ C2xe^x

step 2: Use Table 10.1 to determine the form of a par-ticular solution. Since the forcing function has the form P_n(x)e^xcos!x with Pn(x) = 1 (equiva-lent to n = 0), = 1, and ! = 1, the particular solution has the form

y_pðxÞ ¼ x^sðAe^xcos xþ Be^xsin xÞ step 3: Determine the value of s. Check to see if any of

the terms in y_p(x) will themselves solve the homogeneous equation. Try Ae^xcos x first.

d

dxðAe^xcos xÞ ¼ Ae^xcos x Ae^xsin x d²

dx²ðAe^xcos xÞ ¼ 2Ae^xsin x

Substitute these quantities into the homoge-neous equation.

y⁰⁰þ 2y⁰þ y ¼ 0

2Ae^xsin xþ 2Ae^xcos x

 2Ae^xsin xþ Ae^xcos x¼ 0 3Ae^xcos x 4Ae^xsin x¼ 0 Disregarding the trivial (i.e., A = 0) solution, Ae^xcos x does not solve the homogeneous equation.

Table 10.1 Particular Solutions*

form of fðxÞ form of y_p

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Substitute these quantities into the homoge-neous equation.

y⁰⁰þ 2y⁰þ y ¼ 0 2Be^xcos xþ 2Be^xcos x

þ 2Be^xsin xþ Be^xsin x¼ 0 3Be^xsin xþ 4Be^xcos x¼ 0 Disregarding the trivial (B = 0) case, Be^xsin x does not solve the homogeneous equation.

Since none of the terms in y_p(x) solve the homo-geneous equation, s = 0, and a particular solu-tion has the form

y_pðxÞ ¼ Ae^xcos xþ Be^xsin x

step 4: Use the method of unknown coefficients to deter-mine A and B in the particular solution. Draw-ing on the previous steps, substitute the quantities derived from the particular solution into the nonhomogeneous equation.

y⁰⁰þ 2y⁰þ y ¼ e^xcos x

2Ae^xsin xþ 2Be^xcos x þ 2Ae^xcos x 2Ae^xsin x

þ 2Be^xcos xþ 2Be^xsin x

þAe^xcos xþ Be^xsin x¼ e^xcos x Combining terms,

ð4A þ 3BÞe^xsin xþ ð3A þ 4BÞe^xcos x

¼ e^xcos x

Equating the coefficients of like terms on either side of the equal sign results in the following simultaneous equations.

4A þ 3B ¼ 0 3Aþ 4B ¼ 1 The solution to these equations is

A¼ 3 25 B¼ 4

25 A particular solution is

y_pðxÞ ¼ 25³

 ðe^xcos xÞ þ 25⁴

 ðe^xsin xÞ

step 5: Write the general solution.

yðxÞ ¼ ycðxÞ þ ypðxÞ

¼ C¹e^xþ C²xe^xþ 25³

 ðe^xcos xÞ þ 25⁴

 ðe^xsin xÞ

The values of C₁and C₂would be determined at this time if initial conditions were known.

8. NAMED DIFFERENTIAL EQUATIONS Some differential equations with specific forms are named after the individuals who developed solution techniques for them.

. Bessel equation of order 

x²y⁰⁰þ xy⁰þ xð ² ²Þy ¼ 0 ^10:21 . Cauchy equation

a0xⁿdⁿy

dxⁿþ a¹x^n1d^n1y dx^n1þ    þ aⁿ1xdy

dxþ aⁿy¼ f ðxÞ 10:22

. Euler’s equation

x²y⁰⁰þ xy⁰þ y ¼ 0 ^10:23 . Gauss’ hypergeometric equation

xð1  xÞy⁰⁰þ

c ða þ b þ 1Þx

y⁰ aby ¼ 0 10:24

. Legendre equation of order 

ð1  x²Þy⁰⁰ 2xy⁰þ ð þ 1Þy ¼ 0 ½1<^x<¹

10:25

9. LAPLACE TRANSFORMS

Traditional methods of solving nonhomogeneous differ-ential equations by hand are usually difficult and/or time consuming. Laplace transforms can be used to reduce many solution procedures to simple algebra.

Every mathematical function, fðtÞ, for which Eq. 10.26 exists has a Laplace transform, written asLðf Þ or F (s).

The transform is written in the s-domain, regardless of the independent variable in the original function.²(The variable s is equivalent to a derivative operator, although it may be handled in the equations as a simple

2It is traditional to write the original function as a function of the independent variable t rather than x. However, Laplace transforms are not limited to functions of time.

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variable.) Equation 10.26 converts a function into a Laplace transform.

Equation 10.26 is not often needed because tables of transforms are readily available. (Appendix 10.A con-tains some of the most common transforms.)

Extracting a function from its transform is the inverse Laplace transform operation. Although other methods exist, this operation is almost always done by finding the transform in a set of tables.³

fðtÞ ¼ L^1 FðsÞ

10:27

Example 10.6

Find the Laplace transform of the following function.

fðtÞ ¼ e^at ½s > a

10. STEP AND IMPULSE FUNCTIONS

Many forcing functions are sinusoidal or exponential in nature; others, however, can only be represented by a step or impulse function. A unit step function, u_t, is a function describing the disturbance of magnitude 1 that is not present before time t but is suddenly there after time t. A step of magnitude 5 at time t = 3 would be represented as 5u₃. (The notation 5u (t 3) is used in some books.)

The unit impulse function,^t, is a function describing a disturbance of magnitude 1 that is applied and removed so quickly as to be instantaneous. An impulse of magni-tude 5 at time 3 would be represented by 53. (The notation 5(t  3) is used in some books.)

Example 10.7

What is the notation for a forcing function of magnitude 6 that is applied at t = 2 and that is completely removed at t = 7?

Since the Laplace transform is an integral that starts at t = 0, the value of f(t) prior to t = 0 is irrelevant.

11. ALGEBRA OF LAPLACE TRANSFORMS Equations containing Laplace transforms can be simpli-fied by applying the following principles.

. linearity theorem (c is a constant.) L

cfðtÞ

¼ cL fðtÞ

¼ cFðsÞ ^10:28

. superposition theorem (f (t) and g(t) are different functions.)

. time-shifting theorem (delay theorem) L

fðt  bÞub

¼ e^bsFðsÞ 10:30

. Laplace transform of a derivative L

A complex Laplace transform, F (s), will often be recog-nized as the product of two other transforms, F₁(s) and

3Other methods include integration in the complex plane, convolution, and simplification by partial fractions.

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F₂(s), whose corresponding functions f₁(t) and f₂(t) are known. Unfortunately, Laplace transforms cannot be computed with ordinary multiplication. That is, f (t)6¼ f₁(t)f₂(t) even though F (s) = F₁(s)F₂(s).

However, it is possible to extract f (t) from its convolu-tion, h(t), as calculated from either of the convolution integrals in Eq. 10.35. This process is demonstrated in Ex. 10.9. is a dummy variable.

fðtÞ ¼ L^1

F1ðsÞF²ðsÞ

¼Z t 0

f₁ðt  Þf2ðÞd

¼Z t 0

f₁ðÞf2ðt  Þd 10:35

Example 10.9

Use the convolution integral to find the inverse trans-form of

FðsÞ ¼ 3 s²ðs²þ 9Þ Solution

F(s) can be factored as F1ðsÞF²ðsÞ ¼

1 s²

 3

s²þ 9

 

As the inverse transforms of F₁(s) and F₂(s) are f₁(t) = t and f₂(t) = sin 3t, respectively, the convolution integral from Eq. 10.35 is

fðtÞ ¼Z t 0

ðt  Þsin 3d

¼ Z t

0 ðt sin 3   sin 3Þd

¼ tZ t 0

sin 3d Z t 0

 sin 3d

Expand using integration by parts.

fðtÞ ¼ ¹₃t cos 3 þ¹₃ cos 3 ¹₉sin 3^t

0

¼3t sin 3t 9

13. USING LAPLACE TRANSFORMS

Any nonhomogeneous linear differential equation with constant coefficients can be solved with the following procedure, which reduces the solution to simple algebra.

A complete table of transforms simplifies or eliminates step 5.

step 1: Put the differential equation in standard form (i.e., isolate the y⁰⁰term).

y⁰⁰þ k¹y⁰þ k²y¼ f ðtÞ 10:36

step 2: Take the Laplace transform of both sides. Use the linearity and superposition theorems. (See Eq. 10.28 and Eq. 10.29.)

Lðy⁰⁰Þ þ k1Lðy⁰Þ þ k2LðyÞ ¼ L fðtÞ

10:37

step 3: Use Eq. 10.38 and Eq. 10.39 to expand the equa-tion. (These are specific forms of Eq. 10.31.) Use a table to evaluate the transform of the forcing function.

Lðy⁰⁰Þ ¼ s²LðyÞ  syð0Þ  y⁰ð0Þ ^10:38 Lðy⁰Þ ¼ sLðyÞ  yð0Þ 10:39

step 4: Use algebra to solve forLðyÞ.

step 5: If needed, use partial fractions to simplify the expression forLðyÞ.

step 6: Take the inverse transform to find y(t).

yðtÞ ¼ L^1 LðyÞ

10:40

Example 10.10

Find y (t) for the following differential equation.

y⁰⁰þ 2y⁰þ 2y ¼ cos t yð0Þ ¼ 1 y⁰ð0Þ ¼ 0 Solution

step 1: The equation is already in standard form.

step 2: Lðy⁰⁰Þ þ 2Lðy⁰Þ þ 2LðyÞ ¼ Lðcos tÞ

step 3: Use Eq. 10.38 and Eq. 10.39. Use App. 10.A to find the transform of cos t.

s²LðyÞ  syð0Þ  y⁰ð0Þ þ 2sLðyÞ  2yð0Þ þ 2LðyÞ

¼ s

s²þ 1

But, y (0) = 1 and y⁰ð0Þ ¼ 0.

s²LðyÞ  s þ 2sLðyÞ  2 þ 2LðyÞ ¼ s s²þ 1

step 4: Combine terms and solve forLðyÞ.

LðyÞðs²þ 2s þ 2Þ  s  2 ¼ s s²þ 1

LðyÞ ¼ s

s²þ 1þ s þ 2 s²þ 2s þ 2

¼ s³þ 2s²þ 2s þ 2 ðs²þ 1Þðs²þ 2s þ 2Þ

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step 5: Expand the expression for LðyÞ by partial fractions. The following simultaneous equations result.

A1 þ A² ¼ 1

step 6: Refer to App. 10.A and take the inverse trans-forms. The numerator of the second term is rewritten from (4s + 6) to ((4s + 4) + 2).

14. THIRD- AND HIGHER-ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

The solutions of third- and higher-order linear differen-tial equations with constant coefficients are extensions of the solutions for second-order equations of this type.

Specifically, if an equation is homogeneous, the auxiliary equation is written and its roots are found. If the equa-tion is nonhomogeneous, Laplace transforms can be used to simplify the solution.

Consider the following homogeneous differential equa-tion with constant coefficients.

yⁿþ k¹y^n1þ    þ kⁿ1y⁰þ kⁿy¼ 0 ^10:41 The auxiliary equation to Eq. 10.41 is

rⁿþ k¹r^n1þ    þ kⁿ1rþ kⁿ¼ 0 ^10:42 For each real and distinct root r, the solution contains the term

y¼ Ae^{rx 10:43}

For each real root r that repeats m times, the solution contains the term

y¼ ðA1þ A2xþ A3x²þ    þ Amx^m1Þe^rx 10:44

For each pair of complex roots of the form r ¼  ± i!

the solution contains the terms

y¼ e^xðA¹sin!x þ A²cos!xÞ 10:45

15. APPLICATION: ENGINEERING SYSTEMS There is a wide variety of engineering systems (mechan-ical, electr(mechan-ical, fluid flow, heat transfer, and so on) whose behavior is described by linear differential equa-tions with constant coefficients.

16. APPLICATION: MIXING

A typical mixing problem involves a liquid-filled tank.

The liquid may initially be pure or contain some solute.

Liquid (either pure or as a solution) enters the tank at a known rate. A drain may be present to remove thor-oughly mixed liquid. The concentration of the solution (or, equivalently, the amount of solute in the tank) at some given time is generally unknown. (See Fig. 10.1.) If m(t) is the mass of solute in the tank at time t, the rate of solute change will be m⁰ðtÞ. If the solute is being added at the rate of a(t) and being removed at the rate of r (t), the rate of change is

m⁰ðtÞ ¼ rate of addition  rate of removal

¼ aðtÞ  rðtÞ ^10:46

The rate of solute addition a(t) must be known and, in fact, may be constant or zero. However, r(t) depends on the concentration, c(t), of the mixture and volumetric flow rates at time t. If o(t) is the volumetric flow rate out of the tank, then

rðtÞ ¼ cðtÞoðtÞ 10:47

However, the concentration depends on the mass of solute in the tank at time t. Recognizing that the

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volume, V(t), of the liquid in the tank may be changing with time,

cðtÞ ¼mðtÞ

VðtÞ ^10:48

The differential equation describing this problem is

m⁰ðtÞ ¼ aðtÞ mðtÞoðtÞ

VðtÞ ^10:49

Example 10.11

A tank contains 100 gal of pure water at the beginning of an experiment. Pure water flows into the tank at a rate of 1 gal/min. Brine containing ¹=4 lbm of salt per gallon enters the tank from a second source at a rate of 1 gal/min. A perfectly mixed solution drains from the tank at a rate of 2 gal/min. How much salt is in the tank 8 min after the experiment begins?

brine

pure water

mixture

2 gal/min

1 gal/min 1 gal/min

V(0) ⫽ 100 gal

Solution

Let m(t) represent the mass of salt in the tank at time t.

0.25 lbm of salt enters the tank per minute (that is,

a(t) = 0.25 lbm/min). The salt removal rate depends on the concentration in the tank. That is,

rðtÞ ¼ oðtÞcðtÞ ¼ 2 gal min

 

mðtÞ 100 gal

 

¼ 0:02 1 min

 

mðtÞ

From Eq. 10.46, the rate of change of salt in the tank is m⁰ðtÞ ¼ aðtÞ  rðtÞ

¼ 0:25 lbm

min 0:02 1 min

 

mðtÞ

m⁰ðtÞ þ 0:02 1 min

 

mðtÞ ¼ 0:25 lbm=min This is a first-order linear differential equation of the form of Eq. 10.7. Since the initial condition is m(0) = 0, the solution is

mðtÞ ¼ 0:25lbm min 0:02 1

min 0

B@

1

CA 1  e^{ 0:02} ^min1 

t

 

¼ ð12:5 lbmÞ 1  e^{ 0:02} ^min1 

t

 

At t = 8,

mðtÞ ¼ ð12:5 lbmÞ 1  e^{ 0:02} ^min1 

ð8 minÞ

 

¼ ð12:5 lbmÞð1  0:852Þ

¼ 1:85 lbm

17. APPLICATION: EXPONENTIAL GROWTH AND DECAY

Equation 10.50 describes the behavior of a substance whose quantity, m(t), changes at a rate proportional to the quantity present. The constant of proportional-ity, k, will be negative for decay (e.g., radioactive decay) and positive for growth (e.g., compound interest).

m⁰ðtÞ ¼ kmðtÞ ^10:50

m⁰ðtÞ  kmðtÞ ¼ 0 ^10:51 If the initial quantity of substance is m(0) = m₀, then Eq. 10.51 has the solution

mðtÞ ¼ m⁰e^kt 10:52

If m(t) is known for some time t, the constant of pro-portionality is

k¼1

tln mðtÞ m0

 

10:53 Figure 10.1 Fluid Mixture Problem

mixing paddle

solvent

solute

mixture drain

i(t) a(t)

V(t) m(t)

o(t) r(t)

Backgroundand Support

...

...

...

For the case of a decay, the half-life, t₁₌₂, is the time at which only half of the substance remains. The relation-ship between k and t1=2is

kt₁₌₂¼ ln¹2¼ 0:693 10:54

18. APPLICATION: EPIDEMICS

During an epidemic in a population of n people, the density of sick (contaminated, contagious, affected, etc.) individuals is s(t) = s(t)/n, where s(t) is the number of sick individuals at a given time, t. Similarly, the density of well (uncontaminated, unaffected, suscep-tible, etc.) individuals is w(t) = w(t)/n, where w(t) is the number of well individuals. Assuming there is no quarantine, the population size is constant, individuals move about freely, and sickness does not limit the

During an epidemic in a population of n people, the density of sick (contaminated, contagious, affected, etc.) individuals is s(t) = s(t)/n, where s(t) is the number of sick individuals at a given time, t. Similarly, the density of well (uncontaminated, unaffected, suscep-tible, etc.) individuals is w(t) = w(t)/n, where w(t) is the number of well individuals. Assuming there is no quarantine, the population size is constant, individuals move about freely, and sickness does not limit the

In document Civil Engineering- Reference PE (Page 125-135)