DIFERENCIAL RESOLUÇAO.pdf

Solutions Manual to Accompany Differential Equations for Engineers and Scientists

By Y. Cengel and W. Palm III

Solutions to Problems in Chapter One

© Solutions Manual Copyright 2012 The McGraw-Hill Companies. All rights reserved. No part of this manual may be displayed, reproduced, or distributed in any form or by any means without the written permission of the publisher or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation. Any other reproduction or translation of this work is unlawful. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher's consent is unlawful.

©2012 McGraw-Hill. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher's consent is unlawful.

CHAPTER 1

Section Review Problems

1-1C Mathematical modeling of physical problems encountered in various fields of science and

engineering requires the application of physical principles and laws by representing the rate of

changes in the dependent variables as derivatives. Therefore the mathematical formulations of

such problems result in differential equations. Consequently, we often use differential equations

instead of algebraic in order to investigate a wide variety of problems in sciences and

engineering.

1-2C Mathematical modeling of physical problems involves two significant steps: (i) All

variables affecting the physical phenomena are identified, reasonable assumptions and

approximations are made, and interdependence of these variables is studied. (ii) The relevant

physical principles and laws are applied, and the problem is formulated mathematically, usually

in the form of a differential equation.

1-3C In order to obtain an accurate model in the form of a differential equation for a given

problem, we should have an adequate knowledge about the nature of the problem, the variables

involved, appropriate simplifying assumptions, and applicable physical laws and principles

involved, as well as a careful analysis.

1-4C The simplifying assumptions are often used to make the physical model easier to solve.

However, a model solution that is not quite consistent with the observed nature of the problem

indicates that that the mathematical model used is too crude. In such cases, a more realistic

model should be prepared by eliminating one or more of the questionable assumptions imposed

in the simplifying the problem.

1-5 Let be vertical position of the parachute varying with time . We take the upward direction

to be positive. Applying Newton’s second law in the -direction we obtain , where is

the net force acting on the parachute, is the total mass of the parachute, and is the local

acceleration of the parachute. Since the parachute cruises at a constant velocity, the net force on

it has to be zero, i.e.,

, where is the drag force by air on

the parachute. Therefore the desired differential equation

with an initial

condition of

1-6 The rate of change of money with a constant interest rate was found in Eq. (1-6) to be

. In this case, the person withdraws money from his account at a constant rate of at the


same time. Therefore we should modify the above equation to account for this withdrawal such

that

.

1-7C A variable is a quantity which may assume various values during a study. If its value

remains unchanged during a study, it is called a constant. A variable whose value can be changed

arbitrarily is called an independent variable. On the other hand, a variable whose value depends

on the values of other variables and thus cannot be varied independently is called dependent

variable. For example, the density ( ) of an ideal gas (e.g., air) depends on its absolute

temperature ( ), absolute pressure , and the gas constant ( ). Therefore we can express this

relation mathematically as , where dependent variable or function, and and

are independent variables or arguments.

1-8C A function of is said to be continuous at a number if (i) the function is defined at that

number, (ii) the limit exists, and (iii) this limit is equal to the values of the function

at . That is, a function is continuous at if . A function that does not meet

these three conditions is said to be discontinuous.

1-9C In mathematics, partial derivative term is used for multivariable functions which are

depend on two or more independent variables. Partial derivative of a function represents the

derivative of that function with respect to one of its independent variable while holding the

other variables constant. Ordinary derivative term, on the other hand, is used to describe the

derivative of a function, which depends on only one independent variable.

1-10C In general, th derivative of a function is denoted by , where is a positive integer

and called order of the derivative. For example is second order derivative of . The term

“order” should not be confused with the degree of a derivative, which is power of the highest

order of derivative. For example is the fourth degree of the first order derivative of .

1-11C Yes it is. In other words,

⁄. Set , then , hence

, and

.

1-12 (a) is a continuous function for all in . (b) Since is continuous on , has to be positive so that be defined. Therefore is continuous in .

(c) is continuous for all except . Therefore it is continuous on .

(d)

is continuous for all except , and since the function is

not defined for these values.

1-13 (a) Given:

Solution:


(b) Given:

Solution:

c) Given:

Solution:

d) Given: √

Solution:

√ ( √ ) ( √ )

√

( √ )

1-14 (a) Given:

Solution: ∫

(b) Given:

;

Solution: ∫

[ ]

1-15C The main difference between an algebraic equation and a differential equation is that a differential equation involves derivatives of the dependent variables. An algebraic equation, on the other hand, is simply a mathematical relation between dependent and independent variables. 1-16C A differential equation which involves only ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation, whereas a differential equation which involves partial derivatives of one or more dependent variables with respect to one or more independent variables is called partial differential equations. 1-17C The order of highest derivative in a differential equation indicates the order of the differential equation. 1-18C A differential equation is said to be linear if the dependent variable and all of its derivatives are of the first degree, and their coefficients depend on the independent variable only.

1-19 (a) (Linear, constant coefficient) (b) (Nonlinear, variable coefficient) (c) (Linear, variable coefficient) (d) (Linear, variable coefficient) (e) , dividing both sides by we obtain

, which is a third order nonlinear nonhomogeneous differential equation with constant coefficient.


1-20C In order to obtain a unique solution to a problem, more than just the differential equation is

required. We need to specify some conditions (such as value of the function or its derivatives at

some value of the independent variable) so that forcing the solution to satisfy these conditions at

specified points will result in a unique solution. These conditions are called boundary conditions if

they are specified at two or more values of the independent variable.

1-21C In order to obtain a unique solution to a problem, more than just the differential equation is

required. We need to specify some conditions (such as value of the function or its derivatives at

some value of the independent variable) so that forcing the solution to satisfy these conditions at

specified points will result in a unique solution. These conditions are called initial conditions if all of

them are specified at same values of the independent variable. A differential equation accompanied

by a set of initial conditions is called an initial value problem, whereas a differential equation

accompanied by a set of boundary conditions is called a boundary value problem. For example

, is an initial value problem since both conditions are specified at the

same value of . , , , on the other hand, is a boundary value

problem since both conditions are specified at different value of , and . It should be noted

that the differential equation remained the same while the type of problem was changed.

1-22 Given: , and Solution: The first and second derivatives of are

, and . Then we have .

Therefore is a solution of the differential equation.

The first and second derivatives of is , and

. Then we have .

Therefore is a solution of the differential equation

1-23 Given: , and Solution: The first derivative of are

. Then we have

. Therefore is a solution of the differential equation.

The first derivative of is . Then we have



, and . Then we have


The first and second derivatives of are , and

. Then we have

. Therefore is a solution of the differential equation


, and . Then we

have . Therefore is a solution of the

differential equation.



. Then we have

. Therefore is a solution of the

differential equation

1-26C Some of the linear differential equations, have a single term which involve derivatives, and no terms which involve the unknown function as a factor, can be solved by direct integration. 1-27C Since the order of the differential equation is three, there will be three arbitrary constant in the solution. 1-28 (a) can be solved by direct integration. The solution is , where is an arbitrary constant. (b) can be solved by direct integration. The solution is ∫ ∫

, where is an arbitrary constant.

(c) cannot be solved by direct integration since it involves the integral of the unknown function ∫ . (d) can be solved by direct integration by rewriting it as . The

solution is ∫ ∫

, where is an arbitrary constant.

(e) can be solved by direct integration by rewriting it as ( )

. The solution is ∫ ∫

, √ ,

where is an arbitrary constant. 1-29C No. Not all functions are integrable. For example, is not integrable in closed form in

terms of basic functions.

1-30C With symbolic processing, a computer program solves equations and returns the solution in

symbolic form; that is, as a formula.

1-31C No. For example, using direct integration, the resulting integral might not be integrable.

End-of-Chapter Problems

1.1. Differential Equations in Sciences and Engineering

1-32 An analyst often finds himself/herself in a position to make a choice between a very accurate

but complex model, and a simple but less accurate model. The right choice is usually the simplest

model, which yield adequate results. Construction of very accurate but complicated models are

often not so difficult, however such models are usually difficult to solve and not preferable. At the

minimum, the mathematical model should reflect the essential features of the physical problem that

it represents.


1.2. How the Differential Equations Arise?

1-33 Let be vertical position of the rock varying with time . We take the upward direction to be

positive. Applying Newton’s second law in the -direction we obtain , where is the net

force acting on the rock, is the mass of the rock, and is the local acceleration of the rock. Hence,

we write

, where is the drag force by air on the rock. The desired

differential equation becomes

with an initial condition of

1-34 Let be vertical position of the rock varying with time . We take the downward direction to

be positive. Applying Newton’s second law in the -direction we obtain , where is the

net force acting on the object, is the mass of the object, and is the local acceleration of the

object. The net force on the object is , where is the spring constant. By Newton’s law

of motion, we obtain

or

with an initial condition of .

1-35 The conservation of energy principle requires that the decrease in energy content of the metal

object during a time period of equals to the total heat transfer from the metal object surface to

the surrounding fluid during the same time interval. Therefore we write or

dividing by

. Taking the limit as yields

. This is the desired

differential equation since it describes the variation of temperature with time. The given initial

condition can be expressed mathematically as .

1-36 The rate of change of is given by

.

1-37 The rate of change of is given by

, . Note that decreases with time as

.

1.3. A Brief Review of Basic Concepts

1-38C An equation can involve more than one independent variable. For example, the velocity

field of a fluid depends in general both spatial coordinates and time . Therefore we can

express the velocity field mathematically ), where and are independent

variables, and is the dependent variables. Similarly, an equation can involve more than one

dependent variable. For example, the conservation of mass principle applied to an

incompressible two-dimensional steady flow can be given by


where and are the velocity component in the - and -directions, respectively. Since this is a

two-dimensional flow, the dependent variables and are a function of independent variables

and , that is and .

1-39C The first derivative of a function is represented by and gives the slope of the

function, that is the “rate” at which the function is changing. The second derivative , on the

other hand, gives the rate at which the slope ( ) is changing, and the point of inflection.

1-40C The derivative of a function gives the slope of the function, that is the “rate” at which the

function is changing. Since the tangent line is parallel to the -axis at , its slope has to be

zero.

1-41C The derivative of a function gives the slope of the function. For the present case, the

tangent line makes an angle of with the -axis at . Since , we can say that

the slope at the given point is .

1-42C The derivative of a function gives the slope of the function, that is the “rate” at which the

function is changing. Since the tangent line is perpendicular to the -axis at , its slope is

infinity.

1-43C By differentiating (

) (

) , and dividing both sides by we

obtain

(

)

(

)

For an equality between

and (

)

, it is evident from above equation that the second term in

the right hand side should be zero. This can be provided if (

) or

, or both are

zero.

1-44C For a given function ,

may or may not be a function of depending functional

nature of the function. Set , then

. The first derivative of the given function is

not a function .

1-45C The ideal gas relation is given. (a) By differentiating both sides we obtain

By realizing that along the constant lines we get

constant. This can be

possible only if constant lines in the - diagram are straight lines. (b) It is clear from


the larger pressures will yield larger

ratios. Since

is the slope of , we can

conclude that high pressure lines are steeper than the low pressure lines.

1-46C Integral can be regarded as the inverse process of differentiation. Integral is widely used

in solving differential equations since solving a differential equation is essentially a process of

removing the derivatives from the equation.

In Problems 1-47 and 1-48, we are to determine the interval in which the given function is continuous. 1-47 (a) is a continuous function on

(b) √ is defined on [ and continuous in that interval

(c)

is continuous for all except for

(d)

is continuous on { }, where denotes set of reel numbers

1-48 (a) is continuous on (b) is continuous on

(c)

is continuous for all except for

(d)

. The condition will make the function discontinuous. Therefore

the function is continuous on { } 1-49 Given: van der Waals equation of state

Solution: From the given equation

[ ] constant

1-50 (a) satisfies the given condition

(b)

satisfies the given condition

(c) No elementary function can satisfy the given condition.


1-51 (a) satisfies the given condition

(b) satisfies the given condition (c) No elementary function can satisfy the given condition.

1-52 Given: Solution: Since and are independent variables both derivatives are equal to each other.

1-53 (a) Given:

Solution:

(b) Given:

Solution:

(c) Given:

Solution:

1-54 (a) Given:

Solution:

(b) Given:

Solution:

(c) Given:

Solution:


In Problems 1-55 and 1-56, we are to determine the derivative of the given function

1-55 (a) Given:

Solution:

(b) Given: (there is misprint in the problem statement)

Solution:

c) Given:

Solution:

d) Given:

Solution:

1-56 (a) Given:

Solution:

(b) Given:

Solution:

c) Given:

Solution:

( ( ) )

d) Given:

√

Solution:

√

√

√

√ (

)

In Problems 1-57 through 1-59, we are to perform the given integration

1-57 (a) Given:

Solution: ∫ [

]]

[ ]

(b) Given: Solution: ∫[ ] ∫ ∫ ∫

∫


1-58 (a) Given:

Solution: ∫ [

]

∫

∫ ∫

(b) Given:

Solution: ∫[ ] ∫ ∫ ∫

[

]

c) Given:

Solution: ∫[ ] ∫

∫

d) Given: Solution: ∫[ ] ∫ ⏟

[ ]

∫

[ ]

1-59 (a) Given: Solution: ∫[ ] ∫ ∫ ∫

(b) Given:

Solution: ∫ [

]

∫

∫

∫

[ ] [

]

[

]

[ ]

[ ]

(c) Given:

Solution: ∫ [ ]

∫

∫ ∫

∫

[

]

[

]

[ ]

[ ]

[ ]


(d) Given:

Solution: ∫ [

]

∫ ⏟ [ ]

∫ ∫

1.4. Classification of Differential Equations

1-60C A linear differential equation of order can be expressed in the most general form as

A linear differential equation is said to be homogeneous as well if . For example is a linear homogeneous differential equation. 1-61C A differential equation is said to have constant coefficients if the coefficients of all terms which involve dependent variable or its derivatives are constant. However, if any of the terms with the dependent variable or its derivatives involve the independent variables as a coefficient, that equation is said to have variable coefficients. For example is a linear nonhomogeneous differential equation with constant coefficients since the coefficients of and are constant. 1-62C A linear differential equation of order can be expressed in the most general form as

Therefore the given relation, , is a first order, linear, and nonhomogeneous ordinary differential equation with constant coefficient. In Problems 1-63 and 64, we are to determine the order of the differential equation below, whether it is linear or nonlinear, and whether it has constant or variable coefficients 1-63 (a) (Linear, constant coefficient) (b) (Linear, variable coefficient) (c) (Linear, variable coefficient)

(d)

(Linear, constant coefficient)

(e) (Nonlinear, variable coefficient)

1-64 (a) or

(Linear, constant coefficient)

(b) (Linear, variable coefficient) (c) (Linear, variable coefficient)


(d) , dividing both sides by we obtain , which is a second order linear nonhomogeneous differential equation with variable coefficient.

(e) (Nonlinear, variable coefficient)

1.5. Solution of Differential Equations

1-65C In algebra, we usually seek discrete values that satisfy an algebraic equation such as

. When dealing with the differential equations, however, we seek functions that satisfy

the equation in a specified interval. For example satisfies the algebraic equation .

But the differential equation is satisfied by the function for any value of .

1-66C Any function which satisfies a differential equation on an interval is called a solution of the

differential equation. A solution which involves one or more arbitrary constants represents a family

of functions, which satisfy the differential equation is called a general solution of the equation.

1-67C Any function which satisfies a differential equation and involves the unknown function and

the independent variables only (no derivatives) is a solution of the differential equation. If the

unknown function can be expressed in terms of independent variable only, then the solution is

called explicit; otherwise, the solution the solution is said to be implicit. For example

is an explicit solution. The solution , on the other hand, is an implicit solution

since unknown function in this case cannot be expressed in terms of only.

In Problems 1-68 through 1-78, we are to show that the given functions are solutions of the given



, and . Then we

have ⁄ . Therefore is a solution of the



. Then we have

. Therefore is a solution of the differential

equation

1-69 Given: , and

Solution: The first and second derivatives of are , and

. Then

we have . Therefore is a solution of the differential equation.


. Then we

have . Therefore is a solution of the differential equation.

1-70 Given: ,

and


Solution: The first and second derivatives of

are

, and

. Then we have

(

) (

) (

) . Therefore is a solution of the differential

equation.

The first and second derivatives of

are

, and

. Then we

have (

) (

) (

) . Therefore is a

solution of the differential equation

1-71 Given: , , and Solution: The first and second derivatives of are




. Then we have . Therefore is a solution of the differential equation. The first and second derivatives of are

, and . Then we


1-72 Given: ,

,

and

Solution: The first and second derivatives of

are

, and

. Then we have

(

) (

) (

) . Therefore is a solution of the differential

equation.

The first and second derivatives of

are

, and

. Then we

have (

) (

) (

) . Therefore is a solution

of the differential equation

Since

, the differential equation is already satisfied by .

1-73 Given: , , and Solution: The first and second derivatives of are

, . Then we have


equation.

The first and second derivatives of are ,

. Then we

have ( ) ( ) . Therefore is a solution of

the differential equation.


The first and second derivatives of are ,

. Then we have


equation.

1-74 Given: , √ and [ (√ ) √ ]

Solution: The first and second derivatives of √ are [ (√ )

√ √ ] , [ (√ ) √ √ ] . Then we have

( [ (√ ) √ √ ]) ( [ (√ ) √ √ ])

( √ ) . Therefore is a solution of the differential equation.

After laborious manipulations, the first and second derivatives of [ (√ ) √ ]

are found as

[( √ ) (√ ) ( √ ) √ ]

[( √ ) (√ ) ( √ ) √ ]

Then we have

( [( √ ) (√ ) ( √ ) (√ )])

( [( √ ) (√ ) ( √ ) (√ )])

( [ (√ ) √ ])

Therefore is a solution of the differential equation. 1-75 Given: , , and Solution: The first and second derivatives of are

, and . Then we



. Then we have . Therefore is a solution of the differential equation. The first and second derivatives of are


. Therefore is a solution of the differential equation 1-76 The differential equation for this problem was determined in Example 1-1 to be , where is vertical distance of the rock from the ground. The appropriate initial conditions for this problem will be and . 1-77 The parachute cruises at a constant velocity of . By realizing that the net force acting on it is zero, the differential equation for this problem is , where is vertical distance of the rock from the ground. The appropriate boundary conditions for this problem will be and .


1.6. Solving Differential Equations by Direct Integration

1-78C Since the order of the differential equation is five, there will be five arbitrary constant in the solution. In Problems 1-79 through 1-81, we are to determine if the given differential equation can be solved by direct integration. We also are to obtain the general solution of those that can. 1-79 (a) can be solved by direct integration. The solution, by performing successive two integrals, is , where and are arbitrary constants. (b) cannot be solved by direct integration since it involves a single term with the unknown function ; ∫ .

(c) can be solved by direct integration. ∫ ∫

. Integrating one more time, we obtain ∫ ∫(

) or

, where and are arbitrary constants.

(d) cannot be solved by direct integration since it involves a single term with the unknown function ; ∫ .

(e) can be solved by direct integration by rewriting it as

.

The solution is ∫ ∫ , √ . It is not possible to obtain

the unknown function since the integral ∫√ cannot be found in terms of elementary functions.

1-80 (a) can be solved by direct integration. The solution, by performing successive three integrals, is

, where , and are arbitrary constants. (b) cannot be solved by direct integration since it involves the integral of the unknown function ∫ . (c) can be solved by direct integration. The solution, by performing

successive three integrals, is

, where , and are arbitrary

constants (d) By performing successive two integrals we are left with ,

where and are arbitrary constants. The resulting differential equation cannot be solved by direct integration since it involves the integral of the unknown function ∫ .


. The solution is ∫ ∫ , √ . However the required two more integrals are not so easy to perform. Maple gives the following results:

> f:=sqrt(C1-4*exp(-2*x));

> int(f,x);

> int(f,x,x);


1-81 (a) can be solved by direct integration. The solution is

, where is

an arbitrary constant. (b) cannot be solved by direct integration since it involves the integral of the unknown function ∫ . (c) can be solved by direct integration. By performing successive two

integrals, we are left , and

, where and

are arbitrary constants. (d) cannot be solved by direct integration since it involves a single term with the unknown function ; ∫ .


.

The solution is ∫ ∫ , √ . It is not possible to

obtain the unknown function since the integral ∫√ cannot be found in terms of elementary functions. 1.7. Introduction to Computer Methods

1-82C Suppose we have a differential equation model of a vibratory system and we need to

determine the period of the oscillations. We can determine this from a plot of the solution by

measuring the time between peaks.

1-83C The error is

It can be calculated and plotted in MATLAB as follows: x=[0:0.001:0.6]; y=((x-sin(x))./sin(x))*100; plot(x,y,x,5*ones(size(x))),grid, ylabel('Error (%)'),… xlabel('Theta (rad)'),ginput(1) The plot is shown below. The ginput function gives the result rad.


1-84 The function can be plotted in MATLAB as follows: t=[0:1:300]; V=(6-0.0215*t).^2; plot(t,V),xlabel('Time (s)'),… ylabel('Volume, V (cups)'),ginput(1) The plot is shown below. The volume reaches zero at approximately s.


1-85 The MATLAB code is x=[0:0.001:1]; y=-2*x.*exp(2*x)+2*x-3; plot(x,y),xlabel('x'),ylabel('y')

1-86 All three curves have the same shape, simply shifted up by the value of . It is best to

display them separately to show the shape of the plots. The following MATLAB code illustrates the

use of the subplot function.

xlim=5; x=[0:0.001:xlim]; y1=1/4-(1/4)*exp(-2*x).*(1+2*x+2*x.^2); y2=1/4-(1/4)*exp(-2*x).*(1+2*x+2*x.^2)+5; y3=1/4-(1/4)*exp(-2*x).*(1+2*x+2*x.^2)+10; subplot(3,1,1),plot(x,y1),xlabel('x'),ylabel('y'),gtext('y(0)=0') subplot(3,1,2),plot(x,y2),xlabel('x'),ylabel('y'),gtext('y(0)=5') subplot(3,1,3),plot(x,y3),xlabel('x'),ylabel('y'),gtext('y(0)=10')

The plot is shown below. Note how the MATLAB autoscale feature selects the best range of values on the ordinate axis.


1-87 We can compute the integral numerically using the MATLAB quad function as follows: sin_sq=@(x)sin(x.^2); I=quad(sin_sq,0,sqrt(2*pi)) The result is I = 0.4304. To evaluate the integral symbolically, we can use the MuPad as follows: int(sin(x^2),x=0..sqrt(2*PI)): float(%) The result is 0.4304077247.

1-88 We can compute the integral numerically using the MATLAB quad function as follows: integrand=@(x)sqrt(x.^4+5); I=quad(integrand,0,1) The result is I = 2.2796 . To evaluate the integral symbolically, we can use MuPad as follows: int(sqrt(x.^4+5),x=0..1): float(%) The result is 2.279625687.

1-89 Table 1-2 shows how to evaluate a definite integral. To evaluate an indefinite integral, modify

the commands given in Table 1-2 by deleting the limits 0, x or 0..x, but leaving the integration

variable x.


(a) The answer is

[ ]

if , and

If .

(b) The answer is

1-90 Table 1-2 shows how to evaluate a definite integral. To evaluate an indefinite integral, modify


variable x.

(a) The answer is

(b) The answer is

(c) The answer is

{[ ] } (d) The answer is

[ ] 1-91 Table 1-2 shows how to evaluate a definite integral. To evaluate an indefinite integral, modify


variable x.

(a) The answer is


The symbolic processing programs often do not simplify their ansers, partly because it depends on

individual preferences. For, example, some would prefer the expression to

. Also, some

would prefer to combine the exponentials into a hyberbolic sine, as

(b) The answer is

[ ]

[ ]

Another form returned is

[ ]

(c) The answer is

[ ]

[ ]

(d) The answer is

Another form returned is

[ ]

1-92 If these problems can be solved by direct integration, the required integral will be an indefinite

integral, since no initial conditions are given. Table 1-2 shows how to evaluate a definite integral. To

evaluate an indefinite integral, modify the commands given in Table 1-2 by deleting the limits 0, or

0.. , but leaving the integration variable .

(a) Using direct integration,

∫ ∫


∫ ∫

The answer is

(b) Rearrange the equation as

This equation cannot be solved by direct integration since it is not possible to integrate the left side

without knowing . However, the more advanced methods of Chapter 5 give the solution in

terms of the special functions called Bessel’s functions.

(c) There is a misprint in the first printing of the text. The equation should be

This equation can be solved by direct integration as follows.

∫ ∫

Integrating one more time, we obtain

∫ ∫(

)

or

where and are arbitrary constants. (d) Rearrange the equation as


without knowing . However, the more advanced methods of Chapter 5 give the solution in

terms of the special functions called Airy’s functions.

(e) This equation can be solved partly by direct integration by rewriting it as

. The

solution is ∫ ∫ , √ . It is not possible to obtain the

unknown function in terms of elementary functions since the integral ∫√ cannot be found in terms of elementary functions.


1-93 (a) The answer is



without knowing . However, the more advanced computer methods of later chapters give the

following solution.

√

√

(√

)

√

(√

)

(c) Rearrange the equation as

∫ ∫

The answer is

(d) By performing successive two integrations we are left with , where and

are arbitrary constants. The resulting differential equation cannot be solved by direct integration

since it involves the integral of the unknown function ∫ . However, the more advanced computer

methods of later chapters give the following solution.

(e) The equation can be solved by direct integration by rewriting it as

The solution is

∫ ∫

√

However the required two more integrations are not so easy to perform. Maple gives the following

results:


> f:=sqrt(C1-4*exp(-2*x));

> int(f,x);

> int(f,x,x);

1-94 (a) Integrate both sides to obtain

∫ ∫

and

∫ ∫(

)

The answer is

These integrations can be easily performed with the standard symbolic programs.



without knowing .

(c) Integrate both sides to obtain

∫ ∫

and

∫ ∫[ ]


The answer is

(d) Rearrange the equation as


without knowing .

(e) This equation can be solved by direct integration by rewriting it as

The solution is

∫ ∫

√

It is not possible to obtain the unknown function since the integral ∫√ cannot be found in terms of elementary functions. Review Problems In Problems 1-95 through 1-103, we are to determine the values of for which the given differential equation has a solution of the form 1-95 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since, , . Solving for gives and . (b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , Solving for gives .

(c) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since, , . Solving for gives and .


1-96 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these

derivatives in the given differential equation results in ( ) . Since,

, . Solving for gives and . (b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , Solving for gives .

(c) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these

derivatives in the given differential equation results in ( ) . Since,

, . Solving for gives and . 1-97 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives and . (b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , Solving for gives .

(c) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in

. Since , . Solving for gives

( √ ).

1-98 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives

(b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in

. Since , Solving for gives

( √ ).


. Since , . Solving for gives √ .

1-99 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives

(b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these


derivatives in the given differential equation results in

. Since , Solving for gives

( √ ).


. Since , . Solving for gives ( √ )

1-100 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since ,

. Solving for gives

( √ )

(b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives .

1-101 (a) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since ,

. Solving for gives

( √ )

(b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since ,

. Solving for gives √ . 1-102 (a) Dividing both sides of the given equation by we get . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives

(b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since ,

. Solving for gives

( √ )

1-103 (a) Dividing both sides of the given equation by we get . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting these derivatives in the given differential equation results in . Since , . Solving for gives

√ (b) . The solution is in the form of . Taking the first and second derivatives of the proposed solution we obtain and . Substituting


these derivatives in the given differential equation results in . Since , . Solving for gives .

1-104 Given:

,

|

m/s, and .

Solution: Integrating the given differential equation twice we obtain

. By

inserting second initial condition,

Therefore we get

. The first initial condition, on the other hand, yields

|

,

then . Substituting the calculated values of and into the general solution, we obtain

. For seconds, the position of the rock will be

m.

1-105 Given:

, and

Solution: Since the parachute cruises at a constant velocity of , we can write

| .

Integrating the differential equation we obtain . The boundary condition requires that . Substituting into the general solution, we obtain . Since m at , m/s.

1-106 Given:

(

)

with

|

(due to thermal symmetry), and .

Solution: The differential equation can be arranged to give

(

)

Integrating the differential equation once we obtain

Thermal symmetry about the midpoint of the spherical body requires that

|

Therefore the differential equation takes the form

Integrating the differential equation one more time we obtain

Introducing the boundary condition yields

Substituting the calculated into the general solution of the differential equation we obtain


The temperature at the center is calculated to be

1-107 Given:

(

)

with

|

(due to thermal symmetry), and .

Solution: Multiplying both sides of the differential equation by and rearranging gives

(

)

Integrating the differential equation once we obtain

Thermal symmetry about the centerline of the spherical body requires that

|

Therefore the differential equation takes the form

Integrating the differential equation one more time we obtain

Introducing the boundary condition yields

Substituting into the general solution of the differential equation we obtain

Which is the desired solution for the temperature distribution in the wire as a function of . The temperature at the center is calculated to be

1-108 Given:

with BCs and


Solution: This is a second order linear homogeneous differential equation whose general solution can be obtained by direct integration twice such that

Applying the boundary conditions gives BC at

BC at

Substituting and into the general solution, the variation of temperature is determined to be

Maple solution: > restart: > with(DEtools): > T1:=80:T2:=10:L:=0.2: > ode := diff(T(x), x, x) = 0: > bcs := T(0) = T1, T(L) = T2: > dsolve({bcs, ode});

1-109 Given:

with

|

, and .

Solution: Rearranging the differential equation and integrating

where Integrating one more time

Applying the boundary conditions:

B.C. at

B.C. at

Substituting the and relations into the general solution of the differential equation give


which is the desired solution for the differential equation. The temperature at the insulated surface is then

Therefore the temperature at the insulated surface is . This problem can also be solved with Maple as follows: Maple solution > restart; > with(DEtools); > p := 0.2e-1; k := 30; To := 30; g := 7*10^4; L := .5: > ode := diff(T(x), x, x) = -g*exp(p*x)/k: > bcs := T(L) = To, (D(T))(0) = 0: > dsolve({bcs, ode}): > TEMP := unapply(dsolve({bcs, ode}), x); > evalf(TEMP(0));

1-110 (a) With ,

Thus

And

Thus

(b) Since

we have

∫

∫

Thus

]

and


(c)

1-111 (a) With ,

Thus

and

and

Thus

and

√

√

(b) Since

we have

∫

∫

Thus

]

So

and

√

which is the desired result.

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