Solutions Manual to Accompany Differential Equations for ...€¦ · such problems result in differential equations. Consequently, we often use differential equations instead of algebraic
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Solutions Manual to Accompany Differential Equations for Engineers and Scientists
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.
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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.
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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)
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
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. 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 ∫ ∫ , √ . 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 ∫ .
(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 ∫ ∫ , √ . 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.
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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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1-87 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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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 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 .
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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
( √ ).
(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 √ .
1-99 (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 . 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
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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
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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 . 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
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