Yunus Cengel Differential Equations for Engineers and Scientists Ch01solutions

Solutions Manual to Accompany Differential Equations for Engineers and Scientists By Y. Cengel and W. Palm III Solutions

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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.

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., the parachute.

, where

Therefore the desired differential equation

is the drag force by air on 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

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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 depend on two or more independent variables. Partial derivative of a function represents derivative of that function with respect to one of its independent variable while holding other variables constant. Ordinary derivative term, on the other hand, is used to describe derivative of a function, which depends on only one independent variable.

are the the the

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 not defined for these values.

except

, and

since the function is

1-13 (a) Given: Solution:

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(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) (b) (c) (d) (e)

(Linear, constant coefficient) (Nonlinear, variable coefficient) (Linear, variable coefficient) (Linear, variable coefficient) , dividing both sides by we obtain , which is a third order nonlinear nonhomogeneous differential equation with constant coefficient.

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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 Therefore is a solution of the differential equation. The first and second derivatives of is Therefore is a solution of the differential equation

, and , and

. Then we have

.

. Then we have

.

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 . Therefore

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

1-24 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 . Therefore

1-25 Given: , Solution: The first and second derivatives of have differential equation.

, and . Then we have is a solution of the differential equation

and are

, and . Therefore

. Then we is a solution of the

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The first and second derivatives of

are

, and . Therefore

. Then we have 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) constant. (b)

can be solved by direct integration. The solution is

, where

is an arbitrary

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 ∫ (e) where

, where



is an arbitrary constant.

can be solved by direct integration by rewriting it as

. The solution is ∫ is an arbitrary constant.



1-29C No. Not all functions are integrable. For example, terms of basic functions.

,

(

)



is not integrable in closed form in

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.

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,

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

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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

nature of the function. Set

, then

depending functional

. The first derivative of the given function is

not a function . 1-45C The ideal gas relation

By realizing that possible only if

along the

is given. (a) By differentiating both sides we obtain

constant lines we get

constant. This can be

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

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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) (b) (c) (d) 1-48

(a) (b) (c)

is a continuous function on and continuous in that interval √ is defined on [ is continuous for all except for is continuous on

{

}, where

denotes set of reel numbers

is continuous on is continuous on is continuous for all except for

(d) . The the function is continuous on

condition will make the function discontinuous. Therefore { }

1-49

Given: van der Waals equation of state [ ] constant Solution: From the given equation

1-50

(a)

(b)

satisfies the given condition

satisfies the given condition

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

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1-51

(a)

(b)

satisfies the given condition

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:

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

(a) Given: Solution: (b) Given: Solution:

(there is misprint in the problem statement)

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:

∫[ ∫

]

] ∫





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1-58

(a) Given: Solution:

∫[

]



(b) Given: Solution:





∫[ ∫

] ∫



[ c) Given: Solution:

d) Given: Solution:

]

∫[

(a) Given: Solution:





∫[ ∫⏟

] ∫ [

[

1-59

]

]

]

∫[ ∫

] ∫



(b) Given: Solution:

∫ [

]





[

]



[

]

[

]

[ (c) Given: Solution:

]

∫ [

[

]

]

∫ ∫

[





[

] ]

[

[

] ]

[

]

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(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 is a linear homogeneous differential equation.

. For example

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) (b) (c) (d) (e)

1-64

(a) (b) (c)

(Linear, constant coefficient) (Linear, variable coefficient) (Linear, variable coefficient) (Linear, constant coefficient) (Nonlinear, variable coefficient) or

(Linear, constant coefficient) (Linear, variable coefficient) (Linear, variable coefficient)

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(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 differential equation. 1-68 Given: , Solution: The first and second derivatives of ⁄ have differential equation. The first and second derivatives of

and are

are

, and . Then we . Therefore is a solution of the

, and . Then we have . Therefore is a solution of the differential

equation 1-69

Given:

,

and

Solution: The first and second derivatives of we have . Therefore The first and second derivatives of have 1-70

Given:

are , and . Then is a solution of the differential equation.

are , and . Then we . Therefore is a solution of the differential equation. ,

and

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Solution: The first and second derivatives of ( )

(

)

are

( )

, and

. Therefore

. Then we have

is a solution of the differential

equation. The first and second derivatives of have

are

(

)

, and

(

)

(

. Then we

)

. Therefore

is a

solution of the differential equation 1-71 Given: , , and Solution: The first and second derivatives of are , and . Therefore is a solution of the differential equation. The first and second derivatives of . Therefore

are , and . Then we have is a solution of the differential equation.

The first and second derivatives of have 1-72

Given:

. Then we have

are , and . Then we . Therefore is a solution of the differential equation. ,

,

and

Solution: The first and second derivatives of ( )

(

)

are

( )

, and

. Therefore

. Then we have

is a solution of the differential

equation. The first and second derivatives of have

are

(

)

, and

(

)

(

)

. Then we . Therefore

is a solution

of the differential equation Since

, the differential equation is already satisfied by

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

and are Therefore

.

, . Then we have is a solution of the differential

equation. The first and second derivatives of have ( ) the differential equation.

are (

, )

. Therefore

. Then we is a solution of

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The first and second derivatives of

are . Therefore

, . Then we have is a solution of the differential

equation. 1-74 Given: , and √ Solution: The first and second derivatives of √

,

]



( (

[

(√

)

(√ √

. Therefore

)



[

)







])

(

[

(√

√ .

] [

(√

) are

√ [

Then )



] (√

we



])

(√

)

) have

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: , , Solution: The first and second derivatives of have . Therefore

and are , and . Then we 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. The first and second derivatives of

are , and . Then we have . 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 .

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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 ∫ the unknown function elementary functions.

, ∫ since the integral ∫ √



. It is not possible to obtain cannot be found in terms of

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

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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 ; ∫ . (e) can be solved by direct integration by rewriting it as . The solution is ∫ obtain the unknown function elementary functions. 1.7.

, ∫ since the integral ∫ √



. It is not possible to cannot be found in terms of

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.

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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.

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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.

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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.

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(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 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

(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 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

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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, ∫



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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

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 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.

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1-93 (a) 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 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:

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> 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. (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 . (c) Integrate both sides to obtain ∫



and ∫

∫[

]

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The answer is

(d) Rearrange the equation as

This equation cannot be solved by direct integration since it is not possible to integrate the left side 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 found in terms of elementary functions.

since the integral ∫ √

cannot be

Review Problems In Problems 1-95 through 1-103, we are to determine the values of equation has a solution of the form 1-95 (a) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in , . Solving for gives and (b) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in . Since , Solving for gives (c) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in , . Solving for gives and

and . and

for which the given differential . Taking the first and second . Substituting these . Since, . Taking the first and second . Substituting these

. . Taking the first and second and . Substituting these . Since, .

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1-96 (a) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in , . Solving for gives and (b) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in . Since , Solving for gives (c) . The solution is in the form of derivatives of the proposed solution we obtain derivatives in the given differential equation results in , . Solving for gives and

and

. Taking the first and second . Substituting these . Since, ( )

. and

. Taking the first and second . Substituting these

. . Taking the first and second and . Substituting these . Since, ( ) .

1-97 (a) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , . Solving for gives (b) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , Solving for gives (c) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , . Solving for gives 1-98 (a) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , . Solving for gives (b) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , Solving for gives (c) . The solution is in the form of derivatives of the proposed solution we obtain and derivatives in the given differential equation results in . Since , . Solving for gives

. Taking the first and second . Substituting these and . . Taking the first and second . Substituting these . . Taking the first and second . Substituting these (



).

. Taking the first and second . Substituting these . Taking the first and second . Substituting these ( √ ). . Taking the first and second . Substituting these √

.

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

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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 ( √ ) 1-100 (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 derivatives of the proposed solution we obtain and these derivatives in the given differential equation results in . Solving for gives ( √ ) (b) . The solution is in the form of derivatives of the proposed solution we obtain and these derivatives in the given differential equation results in . Solving for gives √ .

. Taking the first and second . Substituting . Since , . Taking the first and second . Substituting . Since ,

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 derivatives of the proposed solution we obtain and these derivatives in the given differential equation results in . Solving for gives ( √ )

. Taking the first and second . Substituting . Since ,

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 derivatives of the proposed solution we obtain and

. Taking the first and second . Substituting

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these derivatives in the given differential equation results in . Solving for gives . 1-104 Given:

,

m/s, and

|

. Since .

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 Integrating the differential equation we obtain requires that . Substituting . Since m at , 1-106 Given:

(

)

with

, we can write

. The boundary condition into the general solution, we obtain m/s.

(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

Substituting the calculated

yields

into the general solution of the differential equation we obtain

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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

Substituting

yields

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

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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

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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 solved with Maple as follows: Maple solution

. This problem can also be

> 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

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(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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