**Unformatted text preview: **—————————————————————————— CHAPTER 1. —— Chapter One
Section 1.1
1. For
, the slopes are negative, and hence the solutions decrease. For
, the
slopes are positive, and hence the solutions increase. The equilibrium solution appears to
be
, to which all other solutions converge.
3. For
, the slopes are
tive, and hence the solutions increase. For
, the slopes are negative, and hence the solutions decrease. All solutions appear to
diverge away from the equilibrium solution
.
5. For , the slopes are
tive, and hence the solutions increase. For
, the slopes are negative, and hence the solutions decrease. All solutions
diverge away from
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page 1 —————————————————————————— CHAPTER 1. ——
the equilibrium solution . 6. For
, the slopes are
tive, and hence the solutions increase. For
,
the slopes are negative, and hence the solutions decrease. All solutions diverge away
from
the equilibrium solution
.
8. For all solutions to approach the equilibrium solution
, we must have
for
, and
for
. The required rates are satisfied by the
differential equation
.
9. For solutions other than
to diverge from
function for
, and a decreasing function for
equation
whose solutions satisfy these criteria is
.
10. For solutions other than
for
, and
for
equation
. ,
must be an increasing
. The simplest differential to diverge from
, we must have
. The required rates are satisfied by the differential 12. Note that
for
and
. The two equilibrium solutions are
and
. Based on the direction field,
for
; thus solutions with initial
values greater than diverge from the solution
. For
, the slopes are
negative, and hence solutions with initial values between and all decrease toward the ________________________________________________________________________
page 2 —————————————————————————— CHAPTER 1. ——
solution
values
less than . For , the slopes are all positive; thus solutions with initial approach the solution . 14. Observe that
for
and
. The two equilibrium solutions are
and
. Based on the direction field,
for
; thus solutions with initial
values greater than diverge from
. For
, the slopes are also
positive, and hence solutions with initial values between and all increase toward the
solution
. For
, the slopes are all negative; thus solutions with initial
values less than diverge from the solution
.
16.
Let
any
given time be the total amount of the drug in milligrams in the patient's body at
. The drug is administered into the body at a constant rate of The rate at which the drug leaves the bloodstream is given by
accumulation rate of the drug is described by the differential equation Hence the Based on the direction field, the amount of drug in the bloodstream approaches the
equilibrium level of
18. Following the discussion in the text, the differential equation is ________________________________________________________________________
page 3 —————————————————————————— CHAPTER 1. —— or equivalently, After a long time, Using the relation Hence the object attains a terminal velocity given by , the required drag coefficient is 19. All solutions appear to approach a linear asymptote
verify that
is a solution. . It is easy to 20. ________________________________________________________________________
page 4 —————————————————————————— CHAPTER 1. —— All solutions approach the equilibrium solution
23. All solutions appear to diverge from the sinusoid
which is also a solution corresponding to the initial value ,
. 25. All solutions appear to converge to
. First, the rate of change is small. The
slopes
eventually increase very rapidly in magnitude.
26. ________________________________________________________________________
page 5 ————————————————————————— CHAPTER 1. —— The direction field is rather complicated. Nevertheless, the collection of points at which
the slope field is zero, is given by the implicit equation
The graph of
these points is shown below: The y-intercepts of these curves are at
,
. It follows that for solutions with
initial values
, all solutions increase without bound. For solutions with initial
values in the range
and
, the slopes remain negative, and
hence
these solutions decrease without bound. Solutions with initial conditions in the range
initially increase. Once the solutions reach the critical value, given by
the equation
, the slopes become negative and remain negative. These
solutions eventually decrease without bound. ________________________________________________________________________
page 6 —————————————————————————— CHAPTER 1. ——
Section 1.2
1 The differential equation can be rewritten as Integrating both sides of this equation results in
. Applying the initial condition
the constant as
. Hence the solution is , or equivalently,
results in the specification of All solutions appear to converge to the equilibrium solution
1 Rewrite the differential equation as Integrating both sides of this equation results in
equivalently,
. Applying the initial condition
the constant as
. Hence the solution is , or
results in the specification of All solutions appear to converge to the equilibrium solution
than in Problem 1a
2 , but at a faster rate The differential equation can be rewritten as ________________________________________________________________________
page 7 —————————————————————————— CHAPTER 1. —— Integrating both sides of this equation results in
. Applying the initial condition
the constant as
. Hence the solution is , or equivalently,
results in the specification of All solutions appear to diverge from the equilibrium solution
2 Rewrite the differential equation as Integrating both sides of this equation results in
. Applying the initial condition
the constant as
. Hence the solution is , or equivalently,
results in the specification of All solutions appear to diverge from the equilibrium solution
2 . . . The differential equation can be rewritten as Integrating both sides of this equation results in
. Applying the initial condition
the constant as
. Hence the solution is , or equivalently,
results in the specification of ________________________________________________________________________
page 8 —————————————————————————— CHAPTER 1. —— All solutions appear to diverge from the equilibrium solution
3 . . Rewrite the differential equation as
, which is valid for
equivalently,
Note that if . Integrating both sides results in
. Hence the general solution is
, then
, and
is an equilibrium solution. , or As increases, the equilibrium solution gets closer to
, from above.
Furthermore, the convergence rate of all solutions, that is, , also increases.
As increases, then the equilibrium solution
also becomes larger. In
this case, the convergence rate remains the same.
If and both increase but
constant , then the equilibrium solution
remains the same, but the convergence rate of all solutions increases.
5 . Consider the simpler equation
write the equation as . As in the previous solutions, re- Integrating both sides results in
Now set
, and substitute into the original differential equation. We
find that ________________________________________________________________________
page 9 —————————————————————————— CHAPTER 1. ——
.
That is,
, and hence
.
. The general solution of the differential equation is
exactly the form given by Eq.
in the text. Invoking an initial condition
the solution may also be expressed as This is
, 6 . The general solution is
, that is,
.
With
, the specific solution becomes
. This solution is a
decreasing exponential, and hence the time of extinction is equal to the number of
months
it takes, say , for the population to reach zero. Solving
, we find that
months.
The solution,
, is a decreasing exponential as long as
. Hence
has only one root, given by . The answer in part is a general equation relating time of extinction to the value of
the initial population. Setting months , the equation may be written as
, which has solution
answer is
mice . . Since is the initial population, the appropriate 7 . The general solution is
. Based on the discussion in the text, time is
measured in months . Assuming month
days , the hypothesis can be expressed as
. Solving for the rate constant,
, with units of per month .
. days months . The hypothesis is stated mathematically as N/30 .
It follows that
, and hence the rate constant is given by
The units are understood to be per month .
9 . Assuming no air resistance, with the positive direction taken as downward,
Newton's
Second Law can be expressed as in which is the gravitational constant measured in appropriate units. The equation can
be
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page 10 —————————————————————————— CHAPTER 1. ——
written as
initial
velocity . , with solution The object is released with an . Suppose that the object is released from a height of units above the ground. Using the
fact that
, in which is the downward displacement of the object, we obtain
the
differential equation for the displacement as
With the origin placed at
the point of release, direct integration results in
. Based on the
chosen
coordinate system, the object reaches the ground when
. Let
be the time
that it takes the object to reach the ground. Then
. Using the
quadratic
formula to solve for , The positive answer corresponds to the time it takes for the object to fall to the ground.
The
negative answer represents a previous instant at which the object could have been
launched
upward with the same impact speed , only to ultimately fall downward with speed ,
from a height of units above the ground.
. The impact speed is calculated by substituting
. into in part That is, 10 ,b . The general solution of the differential equation is
Given that
mg , the value of the constant is given by
. Hence the amount of
thorium-234 present at any time is given by
. Furthermore, based on the
hypothesis, setting
results in
Solving for the rate constant, we
find that
/week or
/day .
. Let be the time that it takes the isotope to decay to one-half of its original
amount.
From part
, it follows that
, in which
/week . Taking the
natural logarithm of both sides, we find that
weeks or
s.
11. The general solution of the differential equation
is
,
in which
is the initial amount of the substance. Let be the time that it takes
the substance to decay to one-half of its original amount ,
. Setting
in the
solution,
we have
. Taking the natural logarithm of both sides, it follows that
or
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page 11 —————————————————————————— CHAPTER 1. —— 12. The differential equation governing the amount of radium-226 is
,
with solution
Using the result in Problem 11, and the fact that the
half-life
years , the decay rate is given by
per year . The
amount of radium-226, after years, is therefore
Let be
the time that it takes the isotope to decay to
of its original amount. Then setting
,
and
, we obtain
Solving for the decay
time, it follows that
or
years .
13. The solution of the differential equation, with
As , is , the exponential term vanishes, and hence the limiting value is . 14 . The accumulation rate of the chemical is
grams per hour . At any
given time , the concentration of the chemical in the pond is
grams per gallon
.
Consequently, the chemical leaves the pond at a rate of
grams per hour .
Hence, the rate of change of the chemical is given by
gm/hr .
Since the pond is initially free of the chemical, . . The differential equation can be rewritten as Integrating both sides of the equation results in
.
Taking
the natural logarithm of both sides gives
. Since
, the
value of the constant is
. Hence the amount of chemical in the pond at any
time
is
grams . Note that year
hours . Setting
, the amount of chemical present after one year is
grams ,
that is,
kilograms .
. With the accumulation rate now equal to zero, the governing equation becomes
gm/hr . Resetting the time variable, we now assign the new
initial value as
grams .
. The solution of the differential equation in Part
is
Hence, one year after the source is removed, the amount of chemical in the pond is
grams .
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page 12 —————————————————————————— CHAPTER 1. ——
. Letting be the amount of time after the source is removed, we obtain the equation
Taking the natural logarithm of both sides,
or
hours
years . 15 . It is assumed that dye is no longer entering the pool. In fact, the rate at which the
dye leaves the pool is
kg/min
gm per hour
.
Hence the equation that governs the amount of dye in the pool is
gm/hr .
The initial amount of dye in the pool is grams . . The solution of the governing differential equation, with the specified initial value,
is
. The amount of dye in the pool after four hours is obtained by setting
. That is,
grams . Since size of the pool is
gallons , the
concentration of the dye is
grams/gallon .
. Let be the time that it takes to reduce the concentration level of the dye to
grams/gallon . At that time, the amount of dye in the pool is
grams . Using
the answer in part , we have
. Taking the natural logarithm of
both sides of the equation results in the required time
hours .
. Note that . Consider the differential equation
. Here the parameter corresponds to the flow rate, measured in gallons per minute .
Using the same initial value, the solution is given by
In order
to determine the appropriate flow rate, set
and
. (Recall that
gm of
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page 13 —————————————————————————— CHAPTER 1. ——
dye has a concentration of
gm/gal ). We obtain the equation
Taking the natural logarithm of both sides of the equation results in the required flow rate
gallons per minute . ________________________________________________________________________
page 14 —————————————————————————— CHAPTER 1. ——
Section 1.3
1. The differential equation is second order, since the highest derivative in the equation
is of order two. The equation is linear, since the left hand side is a linear function of
and
its derivatives.
3. The differential equation is fourth order, since the highest derivative of the function
is of order four. The equation is also linear, since the terms containing the dependent
variable is linear in and its derivatives.
4. The differential equation is first order, since the only derivative is of order one. The
dependent variable is squared, hence the equation is nonlinear.
5. The differential equation is second order. Furthermore, the equation is nonlinear,
since the dependent variable is an argument of the sine function, which is not a linear
function.
7.
Also, . Hence
and 9. . Substituting into the differential equation, we have
. Hence the given function is a solution. . Thus 10.
and
Clearly,
is
a solution. Likewise,
,
,
,
. Substituting into the left hand side of the equation, we
find that
. Hence both
functions are solutions of the differential equation.
11.
and
hand side of the equation, we have . Substituting into the left Likewise,
and
. Substituting into the left
hand side of the differential equation, we have
. Hence both functions are solutions of the differential equation.
12.
and
. Substituting into the left hand
side of the differential equation, we have
2
. Likewise,
and
. Substituting into the left hand side of the equation, we have
2 2 2 ________________________________________________________________________
page 15 —————————————————————————— CHAPTER 1. ——
2 2 Hence both functions are solutions of the 2 differential equation.
13. and
. Substituting into the left hand side of the differential equation, we have
.
Hence the function is a solution of the differential equation. 15. Let
. Then
results in
. Since
The roots of this equation are
17.
equation, we have
equation , and substitution into the differential equation
, we obtain the algebraic equation and
, that is, . Substituting into the differential
. Since
, we obtain the algebraic
. The roots are
, . 18. Let
. Then
,
and
the derivatives into the differential equation, we have
, we obtain the algebraic equation
that
. Clearly, the roots are
, . Substituting
. Since
By inspection, it follows
and 20.
and
. Substituting the derivatives
into the differential equation, we have
. After
some algebra, it follows that
. For
, we obtain the
algebraic equation
The roots of this equation are
and
21. The order of the partial differential equation is two, since the highest derivative, in
fact each one of the derivatives, is of second order. The equation is linear, since the left
hand side is a linear function of the partial derivatives.
23. The partial differential equation is fourth order, since the highest derivative, and in
fact each of the derivatives, is of order four. The equation is linear, since the left hand
side is a linear function of the partial derivatives.
24. The partial differential equation is second order, since the highest derivative of the
function
is of order two. The equation is nonlinear, due to the product
on
the left hand side of the equation.
25.
It is evident that
derivatives are and
Likewise, given , the second ________________________________________________________________________
page 16 —————————————————————————— CHAPTER 1. —— 2 2 Adding the partial derivatives,
2 2 .
Hence is also a solution of the differential equation. 27. Let . Then the second derivatives are It is easy to see that . Likewise, given Clearly, is also a solution of the partial differential equation. 28. Given the function It follows that
Hence , we have , the partial derivatives are .
is a solution of the partial differential equation. ________________________________________________________________________
page 17 —————————————————————————— CHAPTER 1. ——
29 . The path of the particle is a circle, therefore polar coordinates are intrinsic to the
problem. The variable is radial distance and the angle is measured from the vertical.
Newton's Second Law states that F
a In the tangential direction, the equation of
motion may be expressed as , in which the tangential acceleration, that is, the linear acceleration along the path is
is positive in the direction
of increasing . Since the only force acting in the tangential direction is the component
of weight, the equation of motion is Note that the equation of motion in the radial direction will include the tension in the
rod .
. Rearranging the terms results in the differential equation ________________________________________________________________________
page 18 —————————————————————————— CHAPTER 2. —— Chapter Two
Section 2.1
1 Based on the direction field, all solutions seem to converge to a specific increasing
function.
The integrating factor is
, and hence
It follows that all solutions converge to the function
2 . All slopes eventually become positive, hence all solutions will increase without
bound.
The integrating factor is
, and hence
evident that all solutions increase at an exponential rate. It is 3 ________________________________________________________________________
page 18 —————————————————————————— CHAPTER 2. —— . All solutions seem to converge to the function
The integrating factor is
, and hence
clear that all solutions converge to the specific solution
4 It is
. . . Based on the direction field, the solutions eventually become oscillatory.
The integrating factor is in which
function
5 , and hence the general solution is is an arbitrary constant. As becomes large, all solutions converge to the . ________________________________________________________________________
page 19 —————————————————————————— CHAPTER 2. —— . All slopes eventually become positive, hence all solutions will increase without
bound.
The integrating factor is
can
be written as
, that is,
sides of the equation results in the general solution
all solutions will increase exponentially. The differential equation
Integration of both
It follows that 6 All solutions seem to co...

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- Spring '08
- Doran
- Math, Differential Equations, Equations, Slope