chapter 2 linearity and nonlinearity - florida gulf coast...

CHAPTER 2 Linearity andNonlinearity

2.1 Linear Equations: The Nature of TheirSolutions

!!!! Classification

1. First-order, nonlinear

2. First-order, linear, nonhomogeneous, variable coefficients

3. Second-order, linear, homogeneous, variable coefficients

4. Second-order, linear, nonhomogeneous, variable coefficients

5. Third-order, linear, homogeneous, constant coefficients

6. Third-order, linear, nonhomogeneous, constant coefficients

7. Second-order, linear, nonhomogeneous, variable coefficients

8. Second-order, nonlinear

9. Second-order, linear, homogeneous, variable coefficients

10. Second-order, nonlinear

!!!! Pop Quiz

11. ′ + = ⇒ ( ) = +−y y y t ce t2 2 12. ′ + = ⇒ ( ) = +−y y y t ce t2 1 12

2

13. ′ − = ⇒ ( ) = −y y y t ce t3 5 53

3 14. ′ − = ⇒ ( ) = −y y y t ce t008 100 12500 08. .

15. ′ + =y y2 4 , y y t ce t0 1 22( ) = ⇒ ( ) = +− , y c0 1 1( ) = ⇒ = − . Hence, y t e t( ) = − −2 2 .

16. ′ + =y y5 1, y y t ce t1 0 15

5( ) = ⇒ ( ) = +− , y c e1 0 15

5( ) = ⇒ = − . Hence, y t e t( ) = − − −( )15

1 5 1b g .

69

70 CHAPTER 2 Linearity and Nonlinearity

!!!! Nonhomogeneous Principle I

17. The homogeneous equation is 2 0x y− = , which

has solutions

Xh = ( )α α α, arbitrary2k pand describes the line y x= 2 in the xy-plane. A

particular solution of the nonhomogeneous equa-tion 2 5x y− = is X p = −( )0 5, . Hence the solu-

tion of 2 5x y− = consists of

X = −( ) + ( ){ }0 5 1 2, , α

2 4–2

–5

homogeneoussolution

generalsolution

XpXh

y x= 2

y x= −2 5

X X X= +h p

where α is any real number; this solution is the parametric form of the line y x= −2 5 in the xy-

plane.

!!!! Superposition Principle

18. If y1 and y2 are solutions of ′ + ( ) =y p t y 0, then

′ + ( ) =′ + ( ) =

y p t yy p t y

1 1

2 2

00.

Adding these equations gives

′ + ′ + ( ) + ( ) =y y p t y p t y1 2 1 2 0

or

y y p t y y1 2 1 2 0+ ′ + ( ) + =a f a f ,

which shows that y y1 2+ is also a solution of the given equation.

If y1 is a solution, we have

′ + ( ) =y p t y1 1 0

and multiplying by c we getc y p t ycy cp t y

cy p t cy

′ + ( ) =′ + ( ) =

′ + ( ) =

1 1

1 1

1 1

00

0

a f

a f a f ,

which shows that cy1 is also a solution of the equation.

SECTION 2.1 Linear Equations: The Nature of Their Solutions 71

!!!! Second-Order Superposition Principle

19. If y1 and y2 are solutions of

′′ + ( ) ′ + ( ) =y p t y q t y 0,

we have

′′+ ( ) ′ + ( ) =′′+ ( ) ′ + ( ) =y p t y q t y

y p t y q t y1 1 1

2 2 2

00.

Multiplying these equations by c1 and c2 respectively, then adding and using properties of the

derivative, we arrive at

c y c y p t c y c y q t c y c y1 1 2 2 1 1 2 2 1 1 2 2 0+ ′′ + ( ) + ′ + ( ) + =a f a f a f ,

which shows that c y c y1 1 2 2+ is also a solution.

!!!! Linear and Nonlinear Operations

20. L y y y( ) = ′ + 2

L cy cy cy cy cyc y y cL y

( ) = ( )′ + ( ) = ′ += ′ +( ) = ( )

2 22

Hence, L is a linear operator.

21. L y y y( ) = ′ + 2

L cy cy cy cy c yc y ycL y

( ) = ( )′ + ( ) = ′ +

≠ ′ +

= ( )

2 2 2

2b g

Hence, L is not a linear operator.

22. L y y ty( ) = ′ + 2

L y y y y t y yy ty y ty

L y L y

L cy cy t cy c y tycL y

1 2 1 2 1 2

1 1 2 2

1 2

22 2

2 2

+ = + ′ + += ′ + + ′ += +

( ) = ( )′ + ( ) = ′ +( )= ( )

a f a f a fa f a fa f a f

Hence, L is a linear operator. This problem illustrates the fact that the coefficients of a DE can befunctions of t and the operator will still be linear.

23. L y y e yt( ) = ′ −

L y y y y e y yy e y y e y

L y L y

L cy cy e cy c y e ycL y

t

t t

t t

1 2 1 2 1 2

1 1 2 2

1 2

+ = + ′ − +

= ′ − + ′ −

= +

( ) = ( )′ − ( ) = ′ −

= ( )

a f a f a fb g b ga f a f

b g


Hence, L is a linear operator. This problem illustrates the fact that a linear operator need not havecoefficients that are linear functions of t.

24. L y y t y( ) = ′′ + ( )sin

L y y y y t y yy t y y t y

L y L y

L cy cy t cyc y t ycL y

1 2 1 2 1 2

1 1 2 2

1 2

+ = + ′′ + ( ) += ′′+ ( ) + ′′ + ( )

= +

( ) = ( )′′ + ( )( )= ′′ + ( ){ }= ( )

a f a f a fk p k pa f a f

sinsin sin

sinsin

Hence, L is a linear operator. This problem illustrates the fact that a linear operator need not havecoefficients that are linear functions of t.

25. L y y y y y( ) = ′′ + − ′ +1 2b gL cy cy cy y cy

c y y y ycL y

( ) = ( )′′ + − ( ) ′ + ( )

≠ ′′ + − ′ +

= ( )

1

1

2

2

n sb gl q

Hence, L y( ) is not a linear operator.

!!!! Many from One

26. Because y t t( ) = 2 is a solution of a linear homogeneous equation, we know by parquetry (A) that

ct2 is also a solution for any real number c.

!!!! Guessing Solutions

We can often find a particular solution of a nonhomogeneous DE by inspection (guessing). For the first-order equations given for Problems 27–31 the general solutions come in two parts: solutions to theassociated homogeneous equation (which could be found by separation of variables) plus a particularsolution of the nonhomogeneous equation. For second-order linear equations as Problems 32–35 we canalso sometimes find solutions by inspection.

27. ′ + = ⇒ ( ) = +−y y e y t ce et t t12

28. ′ + = ⇒ ( ) = +− − −y y e y t ce tet t t

29. ′ − = ⇒ ( ) = +y y e y t ce tet t t 30. ′ − = ⇒ ( ) =y ty y t cet0 2 2

31. ′ + = ⇒ ( ) = +y yt

t y t ct

t34

5

32. ′′ − = ⇒ ( ) = + −y a y y t c e c eat at21 20 . An alternative form is y t c at c at( ) = ( ) + ( )1 2sinh cosh .

SECTION 2.1 Linear Equations: The Nature of Their Solutions 73

33. ′′ + = ⇒ ( ) = +y a y y t c at c at21 20 sin cos 34. ′′ + ′ = ⇒ ( ) = + −y y y t c c e t0 1 2

35. ′′ − ′ = ⇒ ( ) = +y y y t c c et0 1 2

!!!! Linear Algebraic Equations

36. The homogeneous system corresponding tox zy z

+ =− =

20

isx zy z

+ =− =

00

The solution of this system is

Xh = −( )α α α α, , arbitraryk p ,which is a line in a three-dimensional space. A particular solution of the nonhomogeneoussystem is

X p x y z= ( ) = ( ), , , 2, 0 0 .

Hence, the general solution of the nonhomogeneous solution is

X X X= + = ( ) + −( )

= −( )h p 2, 0 0

2

arbitrary

arbitrary

, , ,

, , .

α α α α

α α α α

k pk p

!!!! Nonhomogeneous Principle II

In these problems, the verification of yp is a straightforward substitution. To find the rest of the solution

we simply add to yp all the homogeneous solutions yh , which we find by inspection or separation of

variables.

37. ′ − =y y et3 has general solution y t y y ce teh pt t( ) = + = + 3

38. ′ + =y y t2 10sin has general solution y t y y ce t th pt( ) = + = + −−2 4 2sin cos

39. ′ − =yt

y y2 2 has general solution y t y y ct th p( ) = + = +2 3

40. ′ ++

=yt

y11

2 has general solution y t y y ct

t tth p( ) = + =

++

++1

21

2

!!!! Suggested Journal Entry

41. Student Project


2.2 Solving the First-Order Linear DifferentialEquation

!!!! General Solutions

The solutions for Problems 1–15 can be found using either the Euler-Lagrange method or the integratingfactor method. For problems where we find a particular solution by inspection (Problems 2, 6, 7) we usethe Euler-Lagrange method. For the other problems we find it more convenient to use the integratingfactor method, which gives both the homogeneous solutions and a particular solution in one swoop. Youcan use the Euler-Lagrange method to get the same results.

1. ′ + =y y2 0

By inspection we have y t ce t( ) = −2 .

2. ′ + =y y et2 3

We find the homogeneous solution by inspection as y ceht= −2 . A particular solution on the

nonhomogeneous equation can also be found by inspection, and we see y ept= . Hence the

general solution is y t ce et t( ) = +−2 .

3. ′ − =y y et3

We multiply each side of the equation by the integrating factor

µ t e e ep t dt dt tb g b g b g

= = =z z − −1

giving

e y yt− ′ −{ } = 3

or simply

ddt

ye t− =l q 3.

Integrating, we find

ye t ct− = +3

or

y t ce tet t( ) = + 3 .

SECTION 2.2 Solving the First-Order Linear Differential Equation 75

4. ′ + =y y tsin

We multiply each side of the equation by the integrating factor µ t et( ) = , giving

e y y e tt t′ +{ } = sin

or

ddt

ye e tt tl q = sin .

Integrating, we get

ye e t t ct t= −( ) +12

sin cos .

Solving for y, we find

y t ce t tt( ) = + −− 12

12

sin cos .

5. ′ + =+

y yet

11

We multiply each side of the equation by the integrating factor µ t et( ) = , giving

e y y ee

ddt

ye ee

tt

t

tt

t

′ +{ } =+

=+

1

1l q .

Integrating, we get

ye e ct t= + +ln 1b g .

Hence,

y t ce e et t tb g d i= + +− − ln 1 .

6. ′ + =y ty t2

In this problem we see that y tp( ) =12

is a solution of the nonhomogeneous equation (there are

other single solutions, but this is the easiest to find). Hence, to find the general solution we solvethe corresponding homogeneous equation

′ + =y ty2 0

by separation of variables getting

dyy

tdt= −2 ,


which has the general solution

y ce t= − 2

where c is any constant. Adding the solutions of the homogeneous equation to the particular

solution yp =12

we get the general solution of the nonhomogeneous equation:

y t ce t( ) = +− 2 12

.

7. ′ + =y t y t3 2 2

In this problem we see that y tp( ) =13

is a solution of the nonhomogeneous equation (there are

other single solutions, but this is the easiest to find). Hence, to find the general solution, we solvethe corresponding homogeneous equation

′ + =y t y3 02

by separation of variables, getting

dyy

t dt= −3 2 ,


y t ce t( ) = − 3

where c is any constant. Adding the solutions of the homogeneous equation to the particular

solution yp =13

, we get the general solution of the nonhomogeneous equation

y t ce t( ) = +− 3 13

.

8. ′ + =yt

yt

1 12 , t ≠( )0


µ t e e tdt t t( ) = z = =ln

giving

t yt

yt

ddt

tyt

′ +RSTUVW =

{ } =

1 1

1.



ty t c= +ln .

Solving for y, we get

y t ct t

t( ) = + FH IK1 ln .

9. ty y t′ + = 2

We rewrite the equation as

′ + =yt

y1 2

and multiply each side of the equation by the integrating factor

µ t e e tdt t t( ) = z = =ln ,

giving

t yt

y t

ddt

ty t

′ +RSTUVW =

{ } =

1 2

2 .


ty t c= +2 .


y t ct

t( ) = + .

10. cos sint y tyb g ′ + = 1

We rewrite the equation as

′ + =y t y ttan secb gand multiply each side of the equation by the integrating factor

µ t e e e ttdt t t( ) = = = =z − ( ) ( )−tan ln cos ln cos sec1

,

giving

sec tan sec

sec sec .

t y t y t

ddt

t y t

′ + =

=

b gm rb gm r

2

2



sec tant y t cb g = + .


y t c t t( ) = +cos sin .

11. ′ − =yt

y t t2 2 cos , t ≠( )0


µ t e e e tt dt t t( ) = = = =−z − −−2 2 22a f ln ln ,

giving

t yt

y t

ddt

t y t

−

−

′ −RSTUVW =

=

2

2

2 cos

cos .l q


t y t c− = +2 sin .


y t ct t t( ) = +2 2 sin .

12. ′ + =yt

y tt

33

sin , t ≠( )0


µ t e e e tt dt t t( ) = z = = =3 3 33a f b gln ln ,

giving

t yt

y t

ddt

t y t

3

3

3′ +RSTUVW =

=

sin

sin .l q


t y t c3 = − +cos .


y t ct t

t( ) = −3 31 cos .


13. 1 0+ ′ + =e y e yt tb gWe rewrite the equation as

′ ++FHGIKJ =y e

ey

t

t10

and then multiply each side of the equation by the integrating factor

µ t e e ee e dt e tt t t( ) = z = = ++ +1 1 1b g b gln ,

giving

ddt

e yt1 0+ =d io t .


1+ =e y ctb g .

Solving for y, we have

y t cet( ) =

+1.

14. t y ty2 9 0+ ′ + =b gWe rewrite the equation as

′ ++FH IK =y t

ty2 9

0

and then multiply each side of the equation by the integrating factor

µ t e e tt t dt t( ) = z = = ++ +2 29 1 2 9 2 9b g a f b gln ,

giving

ddt

t y2 9 0+ ={ } .


t y c2 9+ = .

Solving for y, we find

y t ct

( ) =+2 9

.


15. ′ ++FH IK =y tt

y t2 1 2 , t ≠( )0


µ t e tet tdt t( ) = z =+( )2 1 2 ,

giving

ddt

te y t et t2 2 22l q = .


te y t e te e ct t t t2 2 2 2 212

= − + + .

Solving for y, we have

y t c et t

tt

( ) = FHGIKJ + + −

−2 12

1 .

!!!! Initial-Value Problems

16. ′ − =y y 1, y 0 1( ) =

By inspection, the homogeneous solutions are y ceht= . A particular solution of the nonho-

mogeneous can also be found by inspection to be yp = −1. Hence, the general solution is

y y y ceh pt= + = − 1.

Plugging in y 0 1( ) = gives c − =1 1 or c = 2. Hence, the solution of the IVP is

y t et( ) = −2 1.

17. ′ + =y ty t2 3, y 1 1( ) =

We can solve the differential equation using either the Euler-Lagrange method or the integratingfactor method to get

y t t ce t( ) = − + −12

12

2 2 .

Plugging in y 1 1( ) = we find ce− =1 1 or c = e. Hence, the solution of the IVP is

y t t e t( ) = − + −12

12

2 1 2 .


18. ′ − FH IK =yt

y t3 3 , y 1 4( ) =

We find the integrating factor to be

µ t e e e tt dt t t( ) = z = = =− − −−3 3 33a f ln ln .

Multiplying the DE by this, we get

ddt

t y− =3 1l q .

Hence,

t y t c− = +3

or

y ct t= +3 4 .

Plugging in y 1 4b g = gives c + =1 4 or c = 3. Hence, the solution of the IVP is

y t t t( ) = +3 3 4.

19. ′ + =y ty t2 , y 0 1( ) =

We solved this differential equation in Problem 6 using the integrating factor method and found

y t ce t( ) = +− 2 12

.

Plugging this in y 0 1( ) = gives c + =12

1 or c =12

. Hence, the solution of the IVP is

y t e t( ) = +−12

12

2 .

20. 1 0+ ′ + =e y e yt tb g , y 0 1( ) =

We solved this DE in Problem 13 and found

y t cet( ) =

+1.

Plugging in y 0 1( ) = gives c2

1= or c = 2. Hence, the solution of the IVP is

y tet( ) =

+2

1.


!!!! Synthesizing Facts

21. (a) y t t tt

t( ) =++

> −( )2 2

11,

(b) y t t t( ) = + > −( )1 1,

(c) The algebraic solution given in Example1 for k = 1 is

y t t tt

tt

( ) =+ +

+=

+( )+

2 22 11

11

.

Hence, when t ≠ −1 we have y t= + 1.

–3

3y

3t

–3

(d) The solution passing through the origin 0 0, ( ) asymptotically approaches the liney t= + 1 as t → ∞ , which is the solution passing through y 0 1( ) = . The entire liney t= + 1 is not a solution of the DE, as the slope is not defined when t = −1. The segmentof the line y t= + 1 for t > −1 is the solution passing through y 0 1( ) = .

On the other hand, if the initial condition were y −( ) = −5 4 , then the solutionwould be the segment of the line y t= + 1 for t less than –1. Notice in the direction fieldthe slope element is not defined at −( )1 0, .

!!!! Using Integrating Factors

In each of the following equations, we first write in the form ′ + ( ) = ( )y p t y f t and then identify p t( ).

22. ′ + =y y2 0

Here p t( ) = 2, therefore

µ t e e ep t dt dt t( ) = z = z =( ) 2 2 .

Multiplying each side of the equation ′ + =y y2 0 by e t2 yields

ddt

ye t2 0l q = .

Integrating gives

ye ct2 = .

Solving for y gives

y t ce t( ) = −2 .


23. ′ + =y y et2 3

Here p t( ) = 2, therefore

µ t e e ep t dt dt t( ) = z = z =( ) 2 2 .

Multiplying each side of the equation

′ + =y y et2 3

by e t2 yields

ddt

ye et t2 33l q = .

Integrating gives

ye e ct t2 3= + .

Solving for y gives

y t ce et t( ) = +−2 .

24. ′ − =y y e t3

Here p t( ) = −1, therefore

µ t e e ep t dt dt t( ) = z = z =( ) − − .


′ − =y y e t3

by e t− yields

ddt

ye et t− =l q 2 .

Integrating gives

ye e ct t− = +12

2 .

Solving for y gives

y t ce et t( ) = +12

3 .

25. ′ + =y y tsin

Here p t( ) = 1 therefore the integrating factor is

µ t e e ep t dt dt t( ) = z = z =( ) .



′ + =y y tsin

by et gives the differential

ddt

ye e tt tl q = sin .

Integrating gives

ye e t t ct t= − +12

sin cosl q .

Solving for y gives

y t t t ce t( ) = −{ } + −12

sin cos .

26. ′ + =+

y yet

11

Here p t( ) = 1 therefore

µ t e e ep t dt dt t( ) = z = z =( ) .


′ + =+

y yet

11

by et yields

ddt

ye ee

tt

tl q =+1

.

Integrating gives

ye e ct t= + +ln 1b g .

Solving for y gives

y t e e cet t t( ) = + +− −ln 1b g .

27. ′ + =y ty t2

Here p t t( ) = 2 , therefore

µ t e e ep t dt tdt t( ) = z = z =( ) 2 2 .


′ + =y ty t2


by et 2 yields

ddt

ye tet t2 2n s = .

Integrating gives

ye e ct t2 212

= + .

Solving for y gives

y t ce t( ) = +− 2 12

.

28. ′ + =y t y t3 2 2

Here p t t( ) = 3 2, therefore

µ t e e ep t dt t dt t( ) = z = z =( ) 3 2 3 .


′ + =y t y t3 2 2

by et3 yields

ddt

ye t et t3 32n s = .

Integrating gives

ye e ct t3 313

= + .

Solving for y gives

y t ce t( ) = +− 3 13

.

29. ′ + =yt

yt

1 12

Here p tt

( ) =1 , therefore

µ t e e e tp t dt t dt t( ) = z = z = =( ) 1a f ln .


′ + =yt

yt

1 12


by t yields

ddt

tyt

{ } =1 .

Integrating gives

ty t c= +ln .

Solving for y gives

y t ct t

t( ) = +1 1ln .

30. ty y t′ + = 2

Here p tt

( ) =1 , therefore

µ t e e e tp t dt t dt t( ) = z = z = =( ) 1a f ln .


′ + =y yt

2

by t yields

ddt

ty t{ } = 2 .

Integrating gives

ty t c= +2 .

Solving for y gives

y t ct

t( ) = +1 .

!!!! Switch for Linearity dydt t y

=+1 , y −( ) =1 0

31. Flipping both sides of the equation, yields the equivalent linear form dtdy

t y= + , or dtdy

t y− = .

Solving this equation we get

t y ce yy( ) = − − 1.

Using the condition y −( ) =1 0, we find − = −1 10ce , and so c = 0. Thus, we have t y= − − 1 and

solving for y gives

y t t( ) = − −1.


!!!! The Tough Made Easy dydt

ye tyy=

−

2

2

32. We flip both sides of the equation, getting

dtdy

e tyy

y=

− 22

or

dtdy y

t ey

y+ =

22 .

We solve this linear DE for t y( ) getting

t y e cy

y( ) =

+2 .

!!!! A Useful Transformation

33. (a) Letting z y= ln , we have y ez= and dydt

e dzdt

z= . Now the equation

dydt

ay by y+ = ln

can be rewritten as

e dzdt

ae bzez z z+ = .

Dividing by ez gives the simple linear equation

dzdt

bz a− = − .

Solving yields

z ce ab

bt= +

and using z y= ln , the solution becomes

y t e a b cebt( ) = +a f .

(b) If a b= = 1, we have

y t e cet( ) = +1b g.

Note that when c = 0 we have the constant solution

y e= .


!!!! Bernoulli Equation ′ + ( ) = ( )y p t y q t yα , α ≠ 0 , α ≠ 1

34. (a) Let y = −( )υ α1 1 , with yα α αυ= −( )1 . Hence,

dydt

ddt

=−

−( )11

1α

υ υα α .

Plugging everything in the original equation yields

υα

υ υ υα α

α α α1

1 1 11

−( )−( ) −( )

−+ ( ) = ( )

ddt

p t q t .

We can now rewrite

ddt

p t q tυ α υ α+ −( ) ( ) = −( ) ( )1 1 ,

which is a linear DE in υ t( ) .

(b) α = 3, p t( ) = −1, and q t( ) = 1; hence ddtυ υ+ = −2 2 , which has the general solution

υ t ce t( ) = − + −1 2 .

Because υ =12y

, this yields

y t ce t( ) = − + − −1 2 1 2b g .

Note, too, that y = 0 satisfies the given equation.

(c) When α = 0 the Bernoulli equation is

dydt

p t y q t+ ( ) = ( ) ,

which is the general first-order linear equation we solved by the integrating factormethod and the Euler-Lagrange method.

When α = 1 the Bernoulli equation is

dydt

p t y q t y+ ( ) = ( )

or

dydt

p t q t y+ ( ) − ( )( ) = 0,

which can be solved by separation of variables.


!!!! Ricatti Equation ′ = ( ) + ( ) + ( )y p t q t y r t y2

35. (a) Suppose y1 satisfies the DE so that

dydt

p t q t y r t y11 1

2= ( ) + ( ) + ( ) .

If we define a new variable y y= +11υ

, then

dydt

dydt

ddt

= − FHIK1

21

υυ .

Substituting for dydt

1 yields

dydt

p t q t y r t y ddt

= ( ) + ( ) + ( ) − FH IK1 12

21

υυ .

Now, if we require, as suggested, that υ satisfies the linear equation

ddt

q t r t y r tυ υ= − ( ) + ( ) − ( )2 1a f ,

then substituting in the previous equation gives

dydt

p t q t y r t y q t r t y r t= ( ) + ( ) + ( ) +

( )+

( )+

( )1 1

2 12

2υ υ υ

,

which simplifies to

dydt

p t q t y r t y y p t q t y r t y= + +FHGIKJ + + FHG

IKJ +

FHG

IKJ = + +b g b g b g b g b g b g1 1

2 12

21 2 1υ υ υ

.

Hence, y y= +11υ

satisfies the Ricatti equation as well, as long as υ satisfies its given

equation.

(b) ′ = − + −y y y1 2 2

Let y1 1= so ′ =y1 0, and substitution in the DE gives

0 1 2 1 2 1 01 12= − + − = − + − =y y .

Hence, y1 satisfies the given equation. To find υ and then y, note that

p t( ) = −1, q t( ) = 2 , r t( ) = −1.

Now find υ from the assumed requirement that

ddtυ υ= − + −( )( )( ) − −( )2 2 1 1 1 ,


which reduces to ddtυ

= 1. This gives υ t t c( ) = + , hence

y t yt c

( ) = + = ++1

1 1 1υ

.

!!!! Computer Visuals

36. (a) ′ + =y y t2

(b) y t ceht( ) = −2 , y tp = −

12

14

The general solution is

y t y y ce th pt( ) = + = + −−2 1

214

.

The curves in the figure in part (a) are labeled for different values of c.

(c) The homogeneous solution yh is transient because yh → 0 as t → ∞ . However, althoughall solutions are attracted to yp , we would not call yp a steady-state solution because itis neither constant nor periodic; yp → ∞ as t → ∞ .

–5

5y

5–5t

y x= −05 0 25. .

37. (a) ′ − =y y e t3

(b) y t ceht( ) = , y ep

t=12

3 . The general solution is

y t y y ce eh pt

y

t

yh

p

( ) = + = +1 :12

3 .


(c) There is no steady-state solution (includ-ing both yh and yp ) go to ∞ as t → ∞ .

The c values are approximate:

{0.5; –0.8; –1.5; –2; –2.5; –3.1}

as counted from the top-most curve tothe bottom-most one.

3–3t

–3

3y

c = –2

38. (a) ′ + =y y tsin

6–6t

–2

2y

c=-0.002

(b) y t ceht( ) = − , y t tp = −

12

12

sin cos .


y t y y ce t th pt( ) = + = + −− 1

212

sin cos .

The curves in the figure in part (a) are labeled for different values of c.

(c) The sinusoidal steady-state solution

y t tp = −12

12

sin cos

occurs when c = 0 . Note that the other solutions approach this solution as t → ∞ .


39. (a) ′ + =y y tsin2

6–6t

–2

2y

(b) y t ceht( ) = − , y t tp = −( )

15

2 2 2sin cos .


y t ce t tt

yy

h

p

( ) = +−−3 sin cos2 2 25! "## $##

.

(c) The steady-state solution is yp , which attracts all other solutions. The transient solution

is yh .

40. (a) ′ + =y ty2 0

–2

2y

c = –2

c = 2

c = –1

c = 1

c = 0 2–2t

(b) This equation is homogeneous.The general solution is

y t ceht( ) = − 2 .

(c) The equation has steady-statesolution y = 0. All solutions tend

towards zero as t → ∞ .


41. (a) ′ + =y ty2 1

–4

4y

4–4t

The approximate c values correspondingto the curves in the center counted fromtop to bottom, are

{1; –1; 2; –2}

and approximately 50,000 (left curve) and–50,000 (right curve) for the side curves.

(b) y t ceht( ) = − 2 , y e e dtp

t t= − z2 2 .


y t ce e e dtt

y

t t

yhp

( ) = +− − z2 2 2; ! "# $# .

(c) The steady-state solution is y t( ) = 0, which is not equal to yp . Both yh and yp are

transient, but as t → ∞ , all solutions approach 0.

!!!! Computer Numerics

42. ′ + =y y t2 , y 0 1( ) =

(a) Using step size h = 01. and h = 0 01. and Euler’s method, we compute the followingvalues. In the latter case we print only selected values.

Euler’s Method

t y h =( )0.1 y h =( )0.01 t y h =( )0.1 y h =( )0.01

0 1 1.0000 0.6 0.3777 0.4219

0.1 0.8 0.8213 0.7 0.3621 0.4039

0.2 0.65 0.6845 0.8 0.3597 0.3983

0.3 0.540 0.5819 0.9 0.3678 0.4029

0.4 0.4620 0.5071 1 0.3842 0.4158

0.5 0.4096 0.4552

(b) From Problem 36, we found the general solution of DE to be

y t ce tt( ) = + −−2 12

14

.


Using IC y 0 1( ) = yields c =54

. The solution of the IVP is

y t e tt( ) = + −−54

12

14

2 ,

and to 4 places, we have

y e1 54

12

14

041922( ) = + − =− . .

(c) The error for y 1( ) using step size h = 01. in Euler’s approximation is

ERROR = − =0 4192 0 3842 0 035. . .

Using step size h = 0 01. , Euler’s method gives

ERROR = − =0 4192 0 4158 0 0035. . . ,

which is much smaller.

(d) The accuracy of Euler’s method can be greatly improved by using a smaller step size.

43. Sample analysis: ′ − =y y e t3 , y 0 1( ) = , y 1( ) .

Exact solution is y e et t= +05 05 3. . , so y 1 114019090461656( ) = . to thirteen decimal places.

(a) y 1 95944( ) ≈ . for step size h = 01. ,

y 1 11401909375( ) ≈ . by Runge-Kutta (correct to six decimal places).

(b) From Problem 24, we found the general solution of the DE to be y ce et t= +12

3 .

(c) The accuracy of Euler’s method can be greatly improved by using a smaller step size; butit still is not correct to even one decimal place for step size 0.01.

y 1 112020610085( ) ≈ . for step size h = 001.

(d) MORAL: Euler’s method converges ever so slowly to the exact answer—clearly a farsmaller step would be necessary to approach the accuracy of the Runge-Kutta method.


44. ′ + =y ty2 1 , y 0 1( ) =

(a) Using step size h = 01. and h = 0 01. and Euler’s method, we compute the followingvalues.

Euler’s Method

t y h =( )0.1 y h =( )0.01 t y h =( )0.1 y h =( )0.01

0 1 1.0000 0.6 1.2308 1.1780

0.1 1.1 1.0905 0.7 1.1831 1.1288

0.2 1.178 1.1578 0.8 1.1175 1.0648

0.3 1.2309 1.1999 0.9 1.0387 0.9905

0.4 1.2570 1.2165 1 0.9517 0.9102

0.5 1.2564 1.2084

y 1 0 905958b g ≈ . by Runge-Kutta method using step size h = 01. (correct to six decimal

places).

(b) From Problem 41, we found the general solution of DE to be y t ce e e dtt t tb g = +− − z2 2 2.

Using IC y 0 1( ) = , yields c = 1. The solution of the IVP is

y t e e dtt tt( ) = +− z2 2

10

( )' '

and so to 10 places, we have y 1 0 9059589485b g = .

(c) The error for y 1( ) using step size h = 01. in Euler’s approximation is

ERROR = − =0 9517 0 9059 0 0458. . . .

Using step size h = 001. , Euler’s method gives

ERROR = − =0 9102 0 9060 0 0043. . . ,

which is much smaller. Using step size h = 01. in Runge-Kutta method gives ERRORless then 0.000001.

(d) The accuracy of Euler’s method can be greatly improved by using a smaller step size, butthe Runge-Kutta method has much better performance because of higher degree ofaccuracy.

!!!! Direction Fields of Homogeneous Equations

45. (a) (A) is linear homogeneous, (B) is linear nonhomogeneous, (C) is nonlinear.


(b) If y1 and y2 are solutions of a linear homogeneous equation,

′ + ( ) =y p t y 0 ,

then ′ + ( ) =y p t y1 1 0 , and ′ + ( ) =y p t y2 2 0 . We can add these equations, to get

′ + ( ) + ′ + ( ) =y p t y y p t y1 1 2 2 0a f a f .

Because this equation can be written in the equivalent form

y y p t y y1 2 1 2 0+ ′ + ( ) + =a f a f ,

then y y1 2+ is also a solution of the given equation.

(c) The sum of any two solutions follows the direction field only in (A). For the linearhomogeneous equation (A) you plot any two solutions y1 and y2 by simply followingcurves in the direction field, and then add these curves, you will see that the sum y y1 2+

also follows the direction field.

However, in equation (B) you can observe a straight line solution, which is

y t112

14

= − .

If you add this to itself you get y y y t1 2 12 12

+ = = − , which clearly does not follow the

direction field and hence is not a solution. In equation (C) y1 1= is a solution but if youadd it to itself you can see from the direction field that y y1 2 2+ = is not a solution.

!!!! Recognizing Linear Homogeneous DEs from Direction Fields

46. For (A) and (D): The direction fields appear to represent linear homogeneous DEs because thesum of any two solutions is a solution and a constant times a solution is also a solution. (Justfollow the direction elements.)

For (B) and (C): These direction fields cannot represent linear homogeneous DEsbecause the zero function is a solution of linear homogeneous equations, and these directionfields do not indicate that the zero function is a solution.

For (E): This cannot be linear because the zero function is not a solution. Also, it appearsthat y t( ) ≡ −1 is a solution, which cannot be for a linear homogeneous equation.


47. Student Project

SECTION 2.3 Growth and Decay Phenomena 97

2.3 Growth and Decay Phenomena

!!!! Half-Life

1. (a) The half-life th is the time required for the solution to reach 12 0y . Hence, y e ykt

0 012

= .

Solving for th , yields kth = − ln2 , or

tkh = −1 2ln .

(b) The solution to ′ =y ky is y t y ekt( ) = 0 so at time t t= 1, we have

y t y e Bkt1 0 1a f = = .

Then at t t th= +1 we have

y t t y e y e e y e e Be Be Bhk t t kt kt kt k kh h1 0 0 0

2 2 1 21 1 112

+ = = = = = =+ − ( ) −a f a f a fln ln ln .

!!!! Doubling Time

2. For doubling time td , we solve y e yktd0 02= , which yields

tkd =1 2ln .

!!!! Interpretation of 1k

3. If we examine the value of the decay curve

y t y ekt( ) = 0

we find

yk

y e y e y

y

k k= FH IK = = = ( )

≈

− −1 0 3678794

3

01

01

0

0

a f .

.

…

Hence,

1k

0

y

t4

0.2

0.4

0.6

0.8

1

1 2

y y ekt= 0

k = –1

1/e


is a crude approximation of the third-life of a decay curve. In other words, if a substance decaysand has a decay constant k = −0 02. , and time is measured in years, then the third-life of the

substance is roughly 1002

50.

= years. That is, every 50 years the substance decays by 23

. Note

that the curve in the figure falls to 13

of its value in approximately t = 1 unit of time.

!!!! Decay of Materials

4. dQdt

kQ= has the general solution

Q t cekt( ) = .

Initial condition Q 0 100( ) = gives Q t ekt( ) = 100 where Q is measured in grams. We also have theinitial condition Q 50 75( ) = , from which we find

k = ≈ −1

5034

0 0058ln . .

The solution is

Q t e t( ) ≈ −100 0 0058.

where t is measured in years. The half-life is

th = ≈ln.

20 0058

120 years.

!!!! Determining Decay from Half-Life

5. dQdt

kQ= has the general solution Q t cekt( ) = . With half-life th = 5 hours, the decay constant has

the value k = − ≈ −15

2 014ln . . Hence,

Q t Q e t( ) = −0

014. .

Calling tt the time it takes to decay to 110

the original amount, we have

110 0 0

014Q Q e tt= − . ,

which we can solve for tt getting

tt = ≈5 10

216 6ln

ln. hours.


!!!! Thorium-234

6. (a) The general decay curve is Q t cekt( ) = . With the initial condition Q 0 1( ) = , we haveQ t ekt( ) = . We also are given Q 1 08( ) = . so ek = 08. , or k = ( ) ≈ −ln . .08 022. Hence, we

have

Q t e t( ) = −0 22.

where Q is measured in grams and t is measured in weeks.

(b) tkh = − = ≈

ln ln.

.2 2022

31 weeks (c) Q e10 01070 22 10( ) ≈− ( ). . grams

!!!! Dating Sneferu’s Tomb

7. The half-life for Carbon-14 is th = 5600 years, so

kth

− = − ≈ −1 2 2

56000000124ln ln . .

Let tc be the time the wood has been aging, and y0 be the original amount of carbon. Fifty-fivepercent of the original amount is 055 0. y . The length of time the wood has aged satisfies the

equation

y e yte00 000124

0055− =. . .

Solving for tc gives

tc = − ≈5600 0 55

24830ln .

ln years.

!!!! Newspaper Announcement

8. For Carbon-14, kth

= − ≈ −ln .2 0000124 . If y0 is the initial amount, then the final amount of

Carbon-14 present after 5000 years will be

y e y05000 0 000124

0054−( ) =. . .

In other words, 54% of the original carbon was still present.

!!!! Radium Decay

9. 6400 years is 4 half-lives, therefore 12

625%4

≈ . .


!!!! General Half-Life Equation

10. We are given the two equations

Q Q eQ Q e

kt

kt1 0

2 0

1

2

=

= .

If we divide, we get

QQ

ek t t1

21 2= −a f

or

k t t QQ1 2

1

2− =a f ln

or

kt t

QQ=

−

ln 12

1 2.

Substituting in tkh = −1 2ln yields the general half-life of

tt t t t

h QQ

QQ

= −−

=−1 2 2 12 2

1

2

1

2

b g b glnln

lnln

.

!!!! Nuclear Waste

11. We have

kth

= − = − ≈ −ln ln .2 2

2580 00268

and solve for t in y e yt0

0 002680005− =. . . Thus

t = ≈258 20

21115ln

ln, years.

!!!! Bombarding Plutonium

12. We are given k = − ≈ −ln.

.2015

4 6209812 . The differential equation for the amount present is

dAdt

kA= + 0 00002. , A 0 0( ) = .

Solving this initial-value problem we get the particular solution

A t cek

kt( ) = −000002.


where ck

= ≈ −000002 0000004. . . Plugging in these values gives the total amount

A t e t( ) ≈ − −0 000004 1 4 6. .b gmeasured in micrograms.

!!!! Blood Alcohol Levels

13. (a) First, because the initial blood-alcohol level is 0.2%, we have P 0 02( ) = . . After one hour,

the level reduces by 10%, therefore, we have

P P1 0 9 0 09 02( ) = ( ) = ( ). . . .

From the decay equation we have P ek1 02( ) = . , hence we have the equation

0 2 09 02. . .ek = ( )

from which we find k = ≈ −ln . .0 9 0105. Thus our decay equation is

P t e et t( ) = ≈( ) −0 2 020 9 0 105. .ln . . .

(b) The person can legally drive as soon as P t( ) < 01. . Setting

P t e t( ) = =−0 2 010105. ..

and solving for t, yields

t = −−

≈ln.

.20105

66 hours.

!!!! Exxon Valdez Problem

14. The measured blood-alcohol level was 0.06%, which had been dropping at a rate of 0.015percentage points per hour for nine hours. This being the case, the captain’s initial blood-alcohollevel was

0 06 9 0 015 0195%. . .+ ( ) = .

The captain definitely could be liable.

!!!! Sodium Pentathal Elimination

15. The half-life is 10 hours. The decay constant is

k = − ≈ −ln .210

0069 .

Ed needs

50 100 kg 5000 mgmg kga f( ) =


of pentathal to be anesthetized. This is the minimal amount that can be presented in hisbloodstream after three hours. Hence,

A A e A3 0813 500000 069 3

0( ) = ≈ =− ( ). . .

Solving for A0 yields A0 61557= , . milligrams or an initial dose of 6.16 grams.

!!!! Moonlight at High Noon

16. Let the initial brightness be I0. At a depth of d = 25 feet, we have that 15% of the light is lost,and so we have I I25 085 0( ) = . . Assuming exponential decay, I d I ekd( ) = 0 , we have the equation

I I e Ik25 085025

0( ) = = .

from which we can find

k = ≈ −ln . .0 85

250 0065 .

To find d, we use the equation

I e Id0

0 00650

1300 000

− =.,

,

from which we determine the depth to be

d = ( ) ≈1

0 0065300 000 1940

.ln , feet.

!!!! Tripling Time

17. Here k =ln210

. We can find the tripling time by solving for t in the equation

y e yt0

2 1003ln( ) = ,

giving ln ln210

3FHIK =t or

t = ≈10 3

21585ln

ln. hours.

!!!! Extrapolating the Past

18. If P0 is the initial number of bacteria present (in millions), then we are given P e k0

6 5= and

P e k0

9 8= . Dividing one equation by the other we obtain e k3 85

= , from which we find

k =ln 8

53

.


Plugging this value into the first equation gives P e02 8 5 5lna f = , in which we can solve for

P e02 8 55 195= ≈− ln .a f million bacteria.

!!!! Unrestricted Yeast Growth

19. From Problem 2, we are given

k = =ln ln21

2

with the initial population of P0 5= million. The population at time t will be 5 2e tln( ) million, so

at t = 4 hours, the population will be

5 5 16 804 2

eln

= ⋅ = million.

!!!! Unrestricted Bacterial Growth

20. From Problem 2, we are given k =ln212

so the population equation is

P t P et( ) = ( )0

2 12ln .

In order to have five times the starting value, we require P e Pt0

2 1205ln( ) = , from which we can

find

t = ≈12 52

27 9lnln

. hours.

!!!! Growth of Tuberculosis Bacterial

21. We are given the initial number of cells present is P0 100= , and that P 1 150( ) = (1.5 times

larger), then 100 ek 1 150( ) = , which yields k = ln 32

. Therefore, the population P t( ) at any time t

is

P t e et t( ) = ≈100 1003 2 0 405ln .a f cells.

!!!! Cat and Mouse Problem

22. (a) For the first 10 years, the mouse population simply had exponential growth

M t M ekt( ) = 0 .

Because the mouse population doubled to 50,000 in 10 years, the initial population must

have been 25,000, hence k =ln210

. For the first 10 years, the mouse population (in

thousands) was

M t et( ) = ( )25 2 10ln .


Over the next 10 years, the differential equation was dMdt

kM= − 6 , where M 0 50( ) = ; t

now measures the number of years after the arrival of the cats. Solving this differentialequation yields

M t cek

kt( ) = +6 .

Using the initial condition M 0 50( ) = , we find ck

= −50 6 . The number of mice (in

thousands) t years after the arrival of the cats is

M tk

ek

kt( ) = −FH IK +50 6 6

where the constant k is given by k = ≈ln .210

0069 . Plugging this in we get

M t e t( ) = − +37 870 069. .

(b) M e10 87 37 1320 069 10( ) = − ≈( ). . thousand mice.

(c) Starting with M0 50= thousand mice a 10% harvest would give M t M ekt( ) = 0 where weknow M M1 09 0( ) = . . Hence, we have M e Mk

01

00 9( ) = . , which gives k = ( ) ≈ −ln . .09 0105.

Thus the mouse population (in thousands) would decline according to

M t M e et t( ) = =− −0

0105 010550. .

where t is the number of years after the arrival of the cats. After 10 years, the mousepopulation would have fallen to

M e10 50 17 50105 10( ) = ≈− ( ). . thousand mice.

!!!! Banker’s View of e

23. The amount of money in a bank account that collects compound interest with continuouscompounding is given by

A t A ert( ) = 0

where A0 is the initial amount and r is an annual interest rate. If A0 = $1 is initially deposited,

and if the annual interest rate is r = 010. (10%), then after 10 years the account value will be

A e10 720 10 10( ) = ⋅ ≈( )$1 $2.. .


!!!! Rule of 70

24. The doubling time is given in Problem 2 by

tr r rd = ≈ =

ln .2 070 70100

where 100r is an annual interest rate (expressed as a percentage). The rule of 70 makes sense.

!!!! Power of Continuous Compounding

25. The future value of the account will be

A t A ert( ) = 0 ,

If A0 50= $0. , r = 0 06. and t = 160, then the value of the account after 160 years will be

A e160 05 390 06 160( ) = ≈( )( ). $7382.. .

!!!! Credit Card Debt

26. If Meena borrows A0 = $5000 at an annual interest rate of r = 01995. (i.e., 19.95%), compounded

continuously, then the total amount she owes (initial principle plus interest) after t years is

A t A e ert t( ) = =001995$5000 . .

After t = 4 years, she owes

A e4 1054701995 4( ) = ≈( )$5000 $11, .. .

Hence, she pays $11, . $5000 $6, .10547 10574− = interest for borrowing this money.

!!!! Compound Interest Thwarts Hollywood Stunt

27. The growth rate is A t A ert( ) = 0 . In this case A0 3= , r = 0 08. , and t = 320. Thus, the total bottles

of whiskey will be

A e320 3 393 600 000 0000 08 320( ) = ≈( )( ). , , , .

That’s 393.6 billion bottles of whiskey!

!!!! It Ain’t Like It Use to Be

28. The growth rate is A t A ert( ) = 0 , where A0 1= , t = 50, and A 50 18( ) = (using thousands of dol-

lars). Hence, we have 18 50= e r , from which we can find r = ≈ln .1850

0 0578 , or 5.78%.


!!!! How to Become a Millionaire

29. (a) From Equation (11) we see that the equation for the amount of money is

A t A e dr

ert rt( ) = + −0 1b g .

In this case, A0 0= , r = 0 08. , and d = $1000. The solution becomes

A t e t( ) = −1000008

10 08.

.b g .

(b) A d e40 000 0000 08

10 08 40( ) = = −( )$1, ,.

.b g . Solving for d, we get that the surviving annual

deposit d = $3399.55.

(c) Ar

e r40 000 000 2500 140( ) = = −$1, , b g . To solve this equation for t we require a computer.

Using Maple, we find the interest rates r = 0 090374. ≈( )9 04%. . You can confirm this

result using direct substitution.

!!!! Living Off Your Money

30. A t A e dr

ert rt( ) = − −0 1b g . Setting A t( ) = 0 and solving for t gives

tr

dd rA

=−FHG

IKJ

1

0ln

Notice that when d rA= 0 this equation is undefined, as we have division by 0; if d rA< 0 , this

equation is undefined because we have a negative logarithm. For the physical translation of thesefacts, you must return to the equation for A t( ) . If d rA= 0 , you are only withdrawing the interest,and the amount of money in the bank remains constant. If d rA< 0 , then you aren’t evenwithdrawing the interest, and the amount in the bank increases and A t( ) never equals zero.

!!!! How Sweet It Is

31. From the equation of Problem 30, we have

A t e et t( ) = − −$1, , ,.

. .000 000 100 000008

10 08 0 08b g .

Setting A t( ) = 0, and solving for t, we have

t = ≈ln.

.50 08

201 years,

the time that the money will last.


!!!! The Real Value of the Lottery

32. Following the hint, we let

′ = −A A010 50 000. , .

Solving this equation with initial condition A A0 0( ) = yields

A t A e t( ) = − +0010500 000 500 000, ,.a f .

Setting A 20 0( ) = and solving for A0 we get

Ae

e0

2

2500 000 1

332=−

≈,

$432,b g

.

!!!! Continuous Compounding

33. (a) After one year compounded continuously the value of the account will be

S S er1 0( ) = .

With r = 0 08. (8%) interest rate, we have the value

S S e S1 083300 08

0( ) = ≈. $1. .

This is equivalent to an annual compound of 8.33%.

(b) If we set the annual yield from a single compounding with interest reff , S r0 1+ effa f equalto the annual yield from continuous compounding with interest r, S er

0 , we have

S r S er0 01+ =effa f .

Solving for reff yields r ereff = −1.

(c) rdaily 0.083277539= +FHGIKJ − =1 0 008

3651

365. (i.e., 8.328%) effective annual interest rate

!!!! Good Test Equation for Computer or Calculator

34. Student Project.

!!!! Your Financial Future

35. We can write annuity equation as

A A' . * ( )= +0 08 5000 , A( )0 0= .

The exact solution is

A e t= −5000 10 08( ).


and this formula would represent the value of the annuity for the first 20 years (we plot thesolution for t = 0 , 1, 2, … 20). We have also rounded the solution to thousands.

Annuity Problem

t A(t) t A(t)

1 $416,435 11 $7054,498

2 $867,554 12 $8058,482

3 $1356,246 13 $9146,085

4 $1885,639 14 $10324,271

5 $2459,124 15 $11600,585

6 $3080,372 16 $12983,199

7 $3753,362 17 $14480,966

8 $4482,404 18 $16103,479

9 $5272,1661 19 $17861,126

10 $6127,705 20 $19765,162

After 20 years at 8% the account has grown to $19765,162. Experiment will show that theinterest rate is more important than the annual deposit.


36. Student Project

SECTION 2.4 Linear Models: Mixing and Cooling 109

2.4 Linear Models: Mixing and Cooling

!!!! Mixing Details

1. Separating variables, we find

dxx t

dt=−

2100

from which we get

ln lnx t c= − +2 100 .

We can solve for x t( ) using properties of the logarithm, getting

x e e Ce C tc t t= = = −( )− −( )2 100 100 22100ln ln

where C ec= > 0 is an arbitrary positive constant. Hence, the final solution is

x t C t c t( ) = ± −( ) = −( )100 10021

2

where c1 is an arbitrary constant.

!!!! English Brine

2. (a) Salt inflow is

2 3 6 lbs lbs gal gal min mina fa f = .

Salt outflow is

Q lbs gal gal min Q lbs min300

3100

FH

IK =a f .

The differential equation for Q t( ) , the amount of salt in the tank, is

dQdt

Q= −6 0 01. .

Solving this equation with initial condition Q 0 50( ) = yields

Q t e t( ) = − −600 550 0 01. .

(b) The concentration conc t( ) of salt is simply the amount Q t( ) divided by the volume

(which is constant at 300). Hence the concentration at time t is given by the expression

conc t Q t e t( ) =( )

= − −

3002 11

60 01. .

(c) As t → ∞ , e t− →0 01 0. . Hence Q t( ) → 600 lbs of salt in the tank.


(d) Either take the limiting amount and divide by 300, or take the limit as t → ∞ of conc t( ).The answer is 2 lbs gal in either case.

(e) Note that the graphs of Q t( ) and of conc t( ) differ only in the scales on the vertical axis,

because the volume is constant.

0 t400

200

300

400

500

600

100 200

100

300

Q t( )

Number of lbs of salt in the tank

conc t( )

0 t400

1

2

100 200 300

Concentration of salt in the tank

!!!! Metric Brine

3. (a) The salt inflow is

01 4 04. . kg liter liters min kg mina fa f = .

The outflow is 4100

Q kg min . Thus, the differential equation for the amount of salt is

dQdt

Q= −04 004. . .

Solving this equation with the given initial condition Q 0 50( ) = gives

Q t e t( ) = + −10 40 0 04. .

(b) The concentration conc t( ) of salt is simply the amount Q t( ) divided by the volume

(which is constant at 100). Hence the concentration at time t is given by

conc tQ t

e tb g b g= = − −

10001 0 4 0.04. . .

(c) As t → ∞ , e t− →0 04 0. . Hence Q t( ) → 10 kg of salt in the tank.

(d) Either take the limiting amount and divide by 100 or take the limit as t → ∞ of conc t( ).The answer is 01. kg liter in either case.


!!!! Salty Goal

4. The salt inflow is given by

2 3 6 lb gal gal min lbs mina fa f = .

The outflow is 320

Q . Thus,

dQdt

Q= −6 320

.

Solving this equation with the given initial condition Q 0 5( ) = yields the amount

Q t e t( ) = − −40 35 3 20.

To determine how long this process should continue in order to raise the amount of salt in thetank to 25 lbs, we set Q t( ) = 25 and solve for t to get

t = ≈203

73

56ln . minutes.

!!!! Mysterious Brine

5. Input in lbs min is 2x (where x is the unknown concentration of the brine). Output is

2100

Q lbs min .

The differential equation is given by

dQdt

x Q= −2 001. ,


Q t x ce t( ) = + −200 0 01. .

Because the tank had no salt initially, Q 0 0( ) = , which yields c x= −200 . Hence, the amount of

salt in the tank at time t is

Q t x e t( ) = − −200 1 0 01.b g.We are given that

Q 120 14 200 280( ) = ( )( ) =. ,

from which we solve for x, to get x ≈ 2 0 . lb gal.


!!!! Correcting a Goof

6. Input in lbs min is 0 (she’s not adding any salt). Output is 0 03. Q lbs min . The differential equa-

tion is

dQdt

Q= −0 03. ,


Q t ce t( ) = −0 03. .

Using the initial condition Q 0 20( ) = , we get the particular solution

Q t e t( ) = −20 0 03. .

Because she wants to reduce the amount of salt in the tank to 10 lbs, we set

Q t e t( ) = = −10 20 0 03. .

Solving for t, we get

t = ≈100

32 23ln minutes.

!!!! Cleaning Up Lake Erie

7. (a) The inflow of pollutant is

40 0 01% 0 004mi yr mi yr3 3b g . .( ) = ,

and the outflow is

40 04mi yr mi100 mi

mi yr33

33b gV t V t( )

= ( ). .

Thus, the DE is

dVdt

V= −0004 04. .

with the initial condition

V 0 0 05% 100 mi 0 05( ) = ( ) =. .3 3 mib g .

(b) Solving the IVP in part (a) we get the expression

V t e t( ) = + −001 004 0 4. . .

where V is the volume of the pollutant in cubic miles.


(c) A pollutant concentration of 0.02% corresponds to

0 02% 100 mi 0 02. .3 3 mib g =of pollutant. Finally, setting V t( ) = 0 02. gives the equation

0 02 0 01 0 04 0.4. . .= + −e t ,

which yields

t = ( ) ≈2 5 4 35. ln . years.

!!!! Cascading Tanks

8. (a) The inflow of salt into tank A is zero because fresh water is added. The outflow of salt is

Q QAA100

150

lbs gal 2 gal min lb minFH IK =a f .

Tank A initially has a total of 05 100 50. ( ) = pounds of salt, so the initial-value problem is

dQdt

QA A= −50

, QA 0 50( ) = lbs.

(b) Solving for QA gives

Q t ceAt( ) = − 50

and with the initial condition QA 0 50( ) = gives

Q t eAt( ) = −50 50.

(c) The input to the second tank is

Q Q eAA

t100

2 150

50 lb gal gal min lb min lb minFH IK = = −a f .

The output from tank B is

Q QBB100

2 150

lb gal gal min lbs minFH IK =a f .

Thus the differential equation for tank B is

dQdt

e QB tB= −− 50 1

50

with initial condition QB 0 0( ) = .

(d) Solving the initial-value problem in (c), we get

Q t teBt( ) = − 50 pounds.


!!!! More Cascading Tanks

9. (a) Input to the first tank is

0 gal 1 0 gal alch gal gal min alch mina fa f = .

Output is

12 0x gal alch min .

The tank initially contains 1 gallon of alcohol, or x0 0 1( ) = . Thus, the differential

equation is given by

dxdt

x00

12

= − .

Solving, we get x t ce t0

2( ) = − . Plugging in x0 0( ) , we get c = 1, so the first tank’s alcohol

content is

x t e t0

2( ) = − .

(b) The first step of a proof by induction is to check the initial case. In our case we checkn = 0 . For n = 0 , t0 1= , 0 1! = , 2 10 = , and hence the given equation yields x t e t

02( ) = − .

This result was found in part (a). The second part of an induction proof is to show thatthe statement holds for case n, and then prove the statement holds for case n + 1. Hence,we assume

x t t enn

n t

n( ) =− 2

2!,

which means the concentration flowing into the next tank will be xn2

(because the

volume is 2 gallons). The input of the next tank is xn12

and the output 12 1x tn+ ( ) . The

differential equation for the n +( )1 tank will be

dxdt

x t en

nn

n t

n+

+

−

++ =11

2

112 2!

, xn+ ( ) =1 0 0.

Solving this IVP, we find

x t t enn

n t

n+

+ −

+( ) =

+( )1

1 2

11 2!

which is what we needed to verify. The induction step is complete.


(c) To find the maximum of x tn( ) , we take its derivative, getting

′ =−

−− − −

+x t en

t enn

n t

n

n t

n

1 2 2

11 2 2b g! !.

Setting this value to zero, the equation reduces to 2 01nt tn n− − = , and thus has rootst = 0, 2n. When t = 0 the function is a minimum, but when t n= 2 , we apply the firstderivative test and conclude it is a local maximum point. Plugging this into x tn( ) yields

the maximum value

x n M n en

n enn n

n n

n

n n2 2

2( ) ≡ =

( )=

− −

! !.

We can also see that x tn( ) approaches 0 as t → ∞ and so we can be sure this point is aglobal maximum of x tn( ) .

(d) Direct substitution. Stirling’s approximation for n! into the formula for Mn in part (c)

gives M nn ≈ −2 1 2πb g / .

!!!! Drug Metabolism

10. The drug concentration C t( ) satisfies

dCdt

a bC= −

where a and b are constants, with C 0 0( ) = . Solving this IVP gives

C t ab

e bt( ) = − −1b g .

As t → ∞ , we have e bt− → 0 (as long as b is positive), so the limiting concentration of C t( ) isab

. Notice that b must be positive or for large t we would have C t( ) < 0, which makes no sense,

because C t( ) is the amount in the body. To reach one-half of the limiting amount of ab

we set

ab

ab

e bt2

1= − −b g

and solve for t, getting tb

=ln2 .

!!!! Another Solution Method

11. Separating variables, we get

dTT M

kdt−

= − .


Solving this equation yields

ln T M kt c− = − +

or

T M e ec kt− = − .

Eliminating the absolute value, we can write

T M e e Cec kt kt− = ± =− −

where C is an arbitrary constant. Hence, we have

T t M Ce kt( ) = + − .

Finally, using the condition T T0 0( ) = gives

T t M T M e kt( ) = + − −0a f .

!!!! Still Another Approach

12. If y t T t M( ) = ( ) − , then dydt

dTdt

= , and

T t y t M( ) = ( ) + .

Hence the equation becomes

dydt

k y M M= − + −( )

or

dydt

ky= − ,

a decay equation.

!!!! Using the Time Constant

13. (a) T t T e M ekt kt( ) = + −− −0 1b g, from Equation (8). In this case, M = 95, T0 75= , and k =

14

,

yielding the expression

T t e et t( ) = + −− −75 95 14 4b gwhere t is time measured in hours. Plugging in t = 2 in this case (2 hours after noon),yields T 2 82 9( ) ≈ . °F.


(b) Setting T t( ) = 80 and simplifying for T t( ) yields

t = − ≈4 34

115ln . hours,

which translates to 1:09 P.M.

!!!! A Chilling Thought

14. (a) T t T e M ekt kt( ) = + −− −0 1b g, from Equation (8). In this problem, T0 75= , M = 10 , and

T 12

50FHIK = (taking time to be in hours). Thus, we have the equation 50 10 60 2= + −e k ,

from which we can find the rate constant

k = − ≈2 23

081ln . .

After one hour, the temperature will have fallen to

T e1 10 60 10 60 49

36 23

3672 2 3( ) ≈ + = + = ≈ln .a f ° F.

(b) Setting T t( ) = 15 gives the equation

15 10 60 2 2 3= + e t lna f.

Solving for t gives

t = − ≈lnln

.122

30623b g

hours (3 hrs, 3.6 min).

!!!! Temperature in the Fridge

15. T t T e M ekt kt( ) = + −− −0 1b g. Taking t in hours, we measure T 1

2FHIK and T 1( ). We then subtract M

from each value and divide, getting the equation

T MT M

e k112

2b gb g

−−

= −

from which we can find

kT MT M

= −−−

2112

ln b gb g .

Once we have k, we solve for T0 in the equation for T 1( ) or T 12FH IK , getting the value

T T M e Mk0 1= ( ) −( ) + .


Finally setting M = 70, T 12

45FH IK = , and T 1 55( ) = , we get k = 1022. . Hence T0 283≈ °. F (below

freezing!).

!!!! Warm or Cold Beer?

16. Again, we use

T t M T M e kt( ) = + − −0a f .

In this case, M = 70, T0 35= . If we measure t in minutes, we have T 10 40( ) = , giving

40 70 35 10= − −e k .

Solving for the decay constant k, we find

k = − ≈ln

.67

1000154

b g.

Thus, the equation for the temperature after t minutes is

T t e t( ) ≈ − −70 35 0 0154. .

Substituting t = 20 gives T 20 44 3( ) ≈ . ° F.

!!!! Case of the Cooling Corpse

17. T t T e M ekt kt( ) = + −− −0 1b g. We know that M = 50 and T0 986= . °F. The first measurement takes

place at unknown time t1 so

T t e kt1 70 50 486 1a f = = + −.

or 48 6 201. e kt− = . The second measurement is taken two hours later at t1 2+ , yielding

60 50 48 6 1 2= + − +. e k ta f

or 48 6 101 2. e k t− + =a f . Dividing the second equation by the first equation gives the relationship

e k− =2 12

from which k =ln22

. Using this value for k the equation for T t1a f gives

70 50 48 6 2 2= + −. lne t

from which we find t1 2 6≈ . hours. Thus, the person was killed approximately 2 hours and 36

minutes before 8 P.M., or at 5:24 P.M.

!!!! Professor Farlow’s Coffee

18. T t T e M ekt kt( ) = + −− −0 1b g. For this problem, M = 70 and T0 200= °F. The equation for the

coffee temperature is

T t e kt( ) = + −70 130 .


Measuring t in hours, we are given

T e k14

120 70 130 4FH IK = = + − ;

so the rate constant is

k = − ≈4 513

38ln . .

Hence

T t e t( ) = + −70 130 38. .

Finally, setting T t( ) = 90 yields

90 70 130 38= + −e t. ,

from which we find t ≈ 0 49. hours, or 29 minutes and 24 seconds.

!!!! The Coffee and Cream Problem

19. The basic law of heat transfer states that if two substances at different temperatures are mixedtogether, then the heat (calories) lost by the hotter substance is equal to the heat gained by thecooler substance. The equation expressing this law is

M S t M S t1 1 1 2 2 2∆ ∆=

where M1 and M2 are the masses of the substances; S1 and S2 are the specific heats, and ∆ t1and ∆ t2 are the changes in temperatures of the two substances, respectively.

In this problem we assume the specific heat of coffee (the ability of the substance to hold heat) isthe same as the specific heat of cream. Defining

CRT

0( ) == ( )=

initial temperature of the coffeeroom temperature temp of the creamtemperature of the coffee after the cream is added

we have

M C T M T R1 20( ) −( ) = −( ).

If we assume the mass of the cream is 110

the mass Mg of the coffee (the exact fraction does not

effect the answer), we have

10 0C T T R( ) −( ) = − .


The temperature of the coffee after John initially adds the cream is

T C R=

( ) +10 011

.

After that John and Maria’s coffee cools according to the basic law of cooling, or

John: 10 011

C R ekt( ) +FH IK , Maria: C ekt0( )

where we measure t in minutes. At time t = 10 the two coffees will have temperature

John: 10 011

10C R e k( ) +FH IK , Maria: C e k0 10( ) .

Maria then adds the same amount of cream to her coffee, which means John and Maria’s coffeesnow have temperature

John: 10 011

10C R e k( ) +FH IK , Maria: 10 011

10C e Rk( ) +FHG

IKJ .

Multiplying each of these temperatures by 11, subtracting 10 0 10C e k( ) and using the fact that

Re10k R> , we conclude that John drinks the hotter coffee.

!!!! A Real Mystery

20. T t T e M ekt kt( ) = + −− −0 1b g

While the can is in the refrigerator T0 70= and M = 40, yielding the equation

T t e kt( ) = + −40 30 .

Measuring time in minutes, we have

T e k15 40 30 6015( ) = + =− ,

which gives k = − ≈ln .23

150027. Letting t1 denote the time the can was removed from the

refrigerator, we know that the temperature at that time is given by

T t e kt1 40 30 1a f = + − ,

which would be the W0 for the warming equation W t( ) , the temperature after the can is removed

from the refrigerator

W t W e ktb g b g= + − −70 700

(the k of the can doesn’t change). Plugging in W0 where t t= 1 and simplifying, we have

W t e ekt kt( ) = + −− −70 30 11b g .


The initial time for this equation is t1 (the time the can was taken out of the refrigerator), so thetime at 2 P.M. will be 60 1− t minutes yielding the equation in t1 :

W t60 1− =a f 60 70 30 11 160= + −− − −e ekt k tb g a f .

This simplifies to

13

60 60 1= −− − −e ek k ta f ,

which is relatively easy to solve for t1 (knowing that k ≈ 0027. ). The solution is t1 37≈ ; hence

the can was removed from the refrigerator at 1:37 P.M.

!!!! Computer Mixing

21. ′ ++

=yt

y11

2, y 0 0( ) =

When the inflow is greater than the out-flow, the amount of dissolved substancekeeps growing without end.

y

0

4

t30

22. ′ +−

=yt

y11

2, y 0 0( ) =

When the inflow is less than the outflow,we note that the amount of salt y t( ) in the

tank becomes zero when t = 1, which isalso when the tank is emptied.

0

1y

10t


23. Student Project


2.5 Nonlinear Models: Logistic Equation

!!!! Equilibrium Points

Note; Problems 1–6 are all autonomous equations, so lines of constant slope (isoclines) are horizontallines.

1. ′ = +y ay by2 , a b> >( )0 0,

We find the equilibrium points by solving

′ = + =y ay by2 0,

getting y = 0, −ab

. By inspecting

′ = +( )y y a by ,

we see that solutions have positive slope ′ >( )y 0 when y > 0 or y ab

< − and negative slope

′ <( )y 0 for − < <ab

y 0 . Hence, the equilibrium solution y t( ) ≡ 0 is unstable, and the equilibrium

solution y t ab

( ) ≡ − is stable.

y

y = 0

y = –a/bstable equilibrium

unstable equilibriumt

2. ′ = −y ay by2 , a b> >( )0 0,


′ = − =y ay by2 0,

getting y = 0, ab

. By inspecting

′ = −( )y y a by ,

SECTION 2.5 Nonlinear Models: Logistic Equation 123

we see that solutions have negative slope ′ <( )y 0 when y < 0 or y ab

> and positive slope

′ >( )y 0 for 0 < <y ab

. Hence, the equilibrium solution y t( ) ≡ 0 is unstable, and the equilibrium

solution y t ab

( ) ≡ is stable.

y

y = 0

y = a/bstable equilibrium

unstable equilibriumt

3. ′ = − +y ay by2 , a b> >( )0 0,


′ = − + =y ay by2 0,

getting y = 0, ab

. By inspecting

′ = − +( )y y a by ,

we see that solutions have positive slope when y < 0 or y ab

> and negative slope for 0 < <y ab

.

Hence, the equilibrium solution y t( ) ≡ 0 is stable, and the equilibrium solution y t ab

( ) ≡ is

unstable.y

y = 0

y = a/b unstable equilibrium

stable equilibriumt


4. ′ = − −y ay by2 , a b> >( )0 0,


′ = − − =y ay by2 0,

getting y = 0, −ab

. By inspecting

′ = − +( )y y a by ,

y

ty = 0

y = –a/b

stable equilibrium

unstable equilibrium

we see that solutions have negative slope when y > 0 or y ab

< − and positive slope for

− < <ab

y 0 . Hence, the equilibrium solution y t( ) ≡ 0 is stable, and the equilibrium solution

y t ab

( ) ≡ − is unstable.

5. ′ = −y ey 1

Solving for y in the equation

′ = − =y ey 1 0,

we get y = 0, hence we have one equilibrium(constant) solution y t( ) ≡ 0. Also ′ >y 0 for ypositive, and ′ <y 0 for y negative. This says thaty t( ) ≡ 0 is an unstable equilibrium point.

y

y = 0 unstable equilibriumt

6. ′ = −y y y

Setting ′ =y 0 we find equilibrium points at

y = 0 and 1.

The equilibrium at y = 0 is stable; that at y = 1 is

unstable. Note also that the DE is only definedwhen y ≥ 0.

stable equilibriumy = 0

y = 1unstable equilibrium

DE not defined

–2

2y

2–2t


!!!! Nonautonomous Sketching

For nonautonomous equations, the lines of constant slope are not horizontal lines as they were in theautonomous equations in Problems 1–6.

7. ′ = −( )y y y t

In this equation we observe that ′ =y 0 when y = 0 , and when y t= ; y ≡ 0 is equilibrium, but

y t= is just an isocline of horizontal slopes. We can draw these lines in the ty-plane with

horizontal elements passing through them.

We then observe from the DE that when

y > 0 and y t> the slope is positive

y > 0 and y t< the slope is negative

y < 0 and y t> the slope is negative

y < 0 and y t< the slope is positive.

From the preceding facts, we surmise that the solutions behave according to our simple analysisof the sign ′y . As can be seen from this figure, the equilibrium y ≡ 0 is stable at t > 0 and

unstable at t < 0 .

t

–4

4y

4–4y = 0

Isocline of slope 0.

Isocline. of slope 0.

y = t


8. ′ = −( )y y t 2

In this equation we observe that ′ =y 0 when y t= . We can draw isoclines y t c− = and elementswith slope ′ =y c2 passing through them. Note that the solutions for c = ±1 are also solutions to

the DE. Note also that for this DE the slopes are all positive.

–2

2y

2–2t

y t= + 1

unstable solution

stable solution

9. ′ = ( )y ytsin

Isoclines of horizontal slopes (dashed) are hyper-bolas yt n= ± π for n = 0 1 2,, , … . On the com-

puter drawn graph you can sketch the hyperbolasfor isoclines and verify the alternating occur-rence of positive and negative slopes betweenthem as specified by the DE.

y = 0

–5

5y

stable solution π π

t–

Only y = 0 is an equilibrium (unstable for t < 0 , stable for t > 0).

!!!! Inflection Points

10. ′ = −FHGIKJy r y

Ly1

We differentiate with respect to t (using the chain rule), and then substitute for dydt

from the DE.

This gives

d ydt

ddt

dydt

ryL

dydt

r yL

dydt

ryL

r ry yL

2

21 1 2 1= FHG

IKJ = −FHG

IKJ + −FHG

IKJ = − +FHG

IKJ −FHG

IKJ .

Setting d ydt

2

2 0= and solving for y yields y = 0, L, L2

. Values y = 0 and y L= are equilibrium

points; y L=

2 is an inflection point. See text Figure 2.5.8.


11. ′ = − −FHGIKJy r y

Ty1

We differentiate with respect to t (using the chain rule), and then substitute for dydt

from the DE.

This gives

d ydt

ddt

dydt

ryT

dydt

r yT

dydt

ryT

r r yT

y2

21 1 2 1= FHG

IKJ = − −FHG

IKJ − −FHG

IKJ = − −F

HGIKJ −FHG

IKJ .

Setting d ydt

2

2 0= and solving for y yields y = 0, T, T2

. Values y = 0 and y T= are equilibrium

points; y T=

2 is an inflection point. See text Figure 2.5.9.

12. ′ = −( )y y tcos

We differentiate ′y with respect to t (using the chain rule), and then substitute for dydt

from the

DE. This gives

d ydt

ddt

dydt

y t dydt

y t y t2

2 = FHGIKJ = − − = − − −sin sin cosb g b g b g .

Setting d ydt

2

2 0= and solving for y yields y t n− = π , y t n n− = +π π2

for n = ± ±0 1 2,, , … .

Note the inflection points change with t in this nonautonomous case. See text Figure 3.5.3, graphfor (2) this section to see that the inflection points occur only when y = −1, so they lie along thelines y t m= + π where m is an odd integer.

!!!! Logistic Equation ′ = −FH IKy r yL

y1

13. (a) We rewrite the logistic DE by separation of variables and partial fractions to obtain

11

1

ydy rdtL

yL

+−

FHG

IKJ = .

Integrating gives

ln lny yL

rt c− − = +1 .


If y L0 > , we know by qualitative analysis (see text Figure 2.5.8) that y L> for all

future time. Thus ln ln1 1− = −FH IKyL

yL

in this case, and the implicit solution (8) becomes

y CeyL

rt

−=

1, with C y

yL

=−0

0 1.

Substitution of this new value of C and solving for y gives y L0 > gives

y t LeL

yrt

( ) =+ − −1 1

0e jwhich turns out (surprise!) to match (10) for y L0 < . You must show the algebraic steps

to confirm this fact.

(b) The derivation of formula (10) required by 1−yL

, which is undefined if y L= . Thus,

although formula (10) happens to evaluate also to y L≡ if y L≡ , our derivation of the

formula is not legal in that case, so it is not legitimate to use (10).

However the original logistic DE ′ = −FH IKy r yL

y1 is fine if y L≡ and reduces in

that case to dydt

= 0 , so y = a constant (which must be L if y L0( ) = ).

(c) The solution formula (10) states

y t LeL

yrt

( ) =+ − −1 1

0e j.

If 0 0< <y L , the denominator is greater than 1 and as t increases, y t( ) approaches L

from below.

If y L0 > , the denominator is less than 1 and as t increases, y t( ) approaches L

from above.

If y L0 = , y t L( ) = . These implovations of the formula are confirmed by the

graph of the Figure 2.5.8.

(d) By straightforward though lengthy computations, taking the second derivative of

y t LeL

yrt

( ) =+ − −1 1

0e j

(10)


gives (with the help of Maple)

′′ =− − − + −

+ −

− − −

−y

L r e e e

e

Ly

rt Ly

rt Ly

rt

Ly

rt0 0 0

0

1 2 1 1 1

1 1

2

3e j e j e jo t

e j.

Setting ′′ =y 0 , we get

2 1 1 1 00 0

Ly

e Ly

ert rt−FHGIKJ − + −FHG

IKJ

LNM

OQP =

− −

or

Ly

e rt

01 1−FHGIKJ =− .

Solving for t, we get tr

Ly

* ln= −FHGIKJ

1 10

. Plugging this value into the analytical solution

for the logistic equation, we get y t L*( ) =2

.

At t* the rate ′y is

rL

L r L rLL

12 2 2 4

2−FHGIKJFHIK = FH

IK = .

!!!! Fitting the Logistic Law

14. The logistic equation is

′ = −FH IKy ry yL

1 .

If initially the population doubles every hour, we have

trd = =

ln2 1

which gives the growth rate r = ≈12

14ln

. . We are also given L = ×5 109 . The logistic curve after

4 hrs is calculated from the analytic solution formula,

y t Le e eL

yrt

( ) =+ −

=×

+ −=

×+

≈ ×− × − ( ) −1 1

5 101 1

5 101 4

4 9 100

99

9

5 1010

14 4

9

5 69

e j d i . . . .


!!!! Logistic Model with Harvesting

15. ′ = −FHGIKJ −y ry y

Lh t1 b g

(a) Graphs of ′y versus y for different val-

ues of harvesting h are shown. Feasibleharvests are those values of h that keepthe slope ′y positive for some

0 < <y L .

Because the curve ′y versus y is always

a maximum at

y L=

2,

y0.8

–0.2

0.2

0.2 0.4 0.6 1

y'rL

==11

h = 0

′ = −( ) −y y y1 h

h = 0.1

h = 0.2h = 0.25

we find the value of h that gives ′FHIK =y L

20 ; this will be the maximum sustainable value

hmax harvesting. By setting ′FHIK =y L

20 , we find h rL

=4

.

(b) As a preliminary to graphing, we find the equilibrium solutions under harvesting bysetting ′ =y 0 in the equation

′ = −FHIK −y r y

Ly h1

getting

r yL

y h1 0−FHIK − = or y rLy hL2 0− + = .


y yL L hL

r1 2

2 2 2

2, =

± − b g

where both roots are positive. The smaller root represents the smaller equilibrium(constant) solution, which is unstable, and the larger root is the larger stable equilibriumsolution. As we say in part (a), harvesting (fish/day or some similar unit) must satisfy

h rL<

2.


y

20ty = 0

y = 1stable equilibrium


Straight logistic

′ = −( )y y y1

0

2y

20t

semistable equilibrium y = 0.5

Logistic with maximum 1 sustainableharvesting

′ = −( ) −y y y1 0 25.

Note that the equilibrium value with harvesting h = 025. is lower than the equilibriumvalue without harvesting. Note further that maximum harvesting has changed the phaseline and the direction of solutions below equilibrium. The harvesting graph implies thatfishing is fine when the population is above equilibrium, but wipes out the populationwhen it is below equilibrium.

!!!! Semistable Equilibrium

16. ′ = −( )y y1 2

We draw upward arrows on the y-axis for

y ≠ 1

to indicate the solution is increasing. When

y = 1

we have a constant solution.–1

2y

6t

y = 1semistable equilibrium

Because the slope lines have positive slope both above and below the constant solutiony t( ) ≡ 1, we say that the solution y t( ) ≡ 1, or the point 1, is semistable (stable from below,

unstable from above). In other words, if a solution is perturbed from the value of 1 to a valuebelow 1, the solution will move back towards 1, but if the constant solution y t( ) ≡ 1 is perturbed

to a value larger than 1, it will not move back towards 1. Semistable equilibria are customarilynoted with half-filled circles.


!!!! Gompertz Equation dydt

y b y= −( )1 ln

17. (a) Letting z y= ln and using the chain rule we get

dzdt

dzdy

dydt y

dydt

= FHGIKJFHIK =FHGIKJFHIK

1 .

Hence, the Gompertz equation becomes

dzdt

a bz= − .

(b) Solving this DE for z we find

z t Ce ab

bt( ) = +− .

Substituting back z y= ln gives

y t e ea b Ce bt( ) = − .

Using the initial condition y y0 0( ) = , we finally get C y ab

= −ln 0 .

(c) From the solution in part (b), limt

a by t e→∞

( ) = when b > 0 , y t( ) → ∞ when b < 0 .

!!!! Fitting the Gompertz Law

18. (a) From Problem 17,

y t e ea b ce bt( ) = −

where c y ab

= −ln 0 . In this case y 0 1( ) = , y 2 2( ) = . We note y y24 28 10( ) ≈ ( ) ≈ means

the limiting value ea b has been reached. Thus

ea b = 10 ,

so

ab

= ≈ln .10 2 3.

The constant c ab

= − = − = −ln . .1 0 2 3 2 3 . Hence,

y t e e bt( ) = − −10 2 3.

and

y e e b2 10 22 3 2( ) = =− −. .


Solving for b:

− = ≈ −

≈ − ≈

− = ( ) ≈ −≈

−

−

2 3 210

1609

16092 3

0 6998

2 0 6998 0 35701785

2

2

. ln .

..

.

ln . ..

e

e

bb

b

b

and a b= 2 3. gives a ≈ 0 4105. .

(b) The logistic equation

′ = −FHIKy ry y

L1

has solution

y t LLy

e rt( ) =

+ −FHGIKJ

−1 10

.

0

y

t12

10

24

logistic

Gompertz

y 2 2( ) =

y = 10

y 0 1( ) =

We have L = 10 and y0 1= , so

y te rt( ) =

+ −10

1 9

and

ye r2 10

1 922( ) =

+=− .

Solving for r

9 102

1 4

2 49

0 8109

0 81092

0 405

2e

r

r

r− = − =

− = ≈ −

=−

−≈

ln .

. . .


!!!! Autonomous Analysis

19. (a) ′ =y y2

(b) One semistable equilibrium at y t( ) ≡ 0 is

stable from below, unstable from above.

–4

4y

semistable equilibrium

4–4t

20. (a) ′ = − −( )y y y1

(b) The equilibrium solutions are y t( ) ≡ 0, y t( ) ≡ 1. The solution y t( ) ≡ 0 is stable. Thesolution y t( ) ≡ 1 is unstable.

–4

4y

y = 1y = 0 stable equilibrium


4–4 t

21. (a) ′ = − −FH IK −FH IKy y yL

yM

1 1 , ′ = − −( ) −( )y y y y1 1 05.

(b) The equilibrium points are y = 0 , L, M. y = 0 is stable. y M= is stable if M L> andunstable if M L< . y L= is stable if M L< and unstable if M L> .

y

t

stable equilibriumunstable equilibrium

stable equilibriumy = 0y = Ly = M


22. (a) ′ = −y y y

Note that the DE is only defined for

y ≥ 0 .

(b) The constant solution y t( ) ≡ 0 is stable,the solution y t( ) ≡ 1 is unstable.

–4

4y

–2 stable equilibrium


y = 0 ty = 1

2

23. (a) ′ = −( )y k y1 2 , k > 0

(b) The constant solution y t( ) ≡ 1 is semi-stable (unstable above, stable below).

–4

4y

4–4t

semistable equilibriumy = 1

24. (a) ′ = −y y y2 24b g(b) The equilibrium solution y t( ) ≡ 2 is stable, the solution y t( ) ≡ −2 is unstable and the

solution y t( ) ≡ 0 is semistable.

–4

4y

2–2t

semistable equilibriumy = 0

y = 2

y = –2

stable equilibrium



!!!! Hubbert Peak

25. (a) From even a hand-sketched logisticcurve you can graph its slope ′y and finda roughly bell-shaped curve for ′( )y t .

Depending on the scales used, it may besteeper or flatter than the bell curveshown in Fig. 1.3.5.

0

y, y'

t

1

2

y

y'

inflection point

(b) For a pure logistic curve, the inflection point always occurs when y L=

2. However, if

we consider models different from the logistic model that still show similar solutionsbetween 0 and the nonzero equilibrium, it is possible for the inflection point to be closerto 0. When this happens oil recovery reaches the maximum production rate much earlier.

Of course the logistic model is acrude model of oil production. For exam-ple it doesn’t take into consideration thefact that when oil prices are high, manyoil wells are placed into production.

If the inflection point is lowerthan halfway on an approximately logis-tic curve, the peak on the ′y curve occurs

sooner and lower creating an asymmetriccurve for ′y .

0

y, y'

t

1

2

inflection point

y

y'

y = 0.5

Lower inflection point


(c) These differences may or may not be significant to people studying oil production; itdepends on what they are looking for. The long-term behavior, however, is the same; thepeak just occurs sooner. After the peak occurs, if the model holds, it is downhill insofaras oil production is concerned. Typical skew of peak position is presented on thesefigures:

10y

10–4t

Typical curves with differentinflection points

4y

10–4t

′y curves for corresponding ones

!!!! Stefan’s Law Again

26. ′ = −T k M T4 4b gThe equation tells us that when

0 < <T M ,

the solution T T t= ( ) increases because ′ >T 0 ,

and when M T< the solution decreases because′ <T 0 . Hence, the equilibrium point T t M( ) ≡ is

stable. We have drawn the directional field ofStefan’s equation for M = 3, k = 0 05. .

20t

y

stable equilibriumy = 3

dTdt

T= −005 34 4. b g

To > M gives solutions falling to M.

To < M gives solutions rising to M.

The one actions match intuition and experiment.


!!!! Chemical Reactions

27. x k x x= −( ) −( )100 50

The solutions for the given initial conditions areshown on the graph. Note that all behaviors areat equilibrium or flown off scale before t = 03. !

Conclusions: Any causes x t( ) to increase

without bound. On the other hand, for any

x 0 0100( ) ∈( ),

0

200y

0.50t

stable equilibrium


the solution will approach an equilibrium value of 50, which implies the tiniest amount issufficient to start the reaction.

If you are looking for a different scenario, you might consider some other modelingoptions that appear in Problem 28.

0

200y

0.50t

Phase portrait

0

200y

0.50t

Products amount

The curve for x 0 150( ) = is almost vertical.

(a) A solution starting at x 0 0( ) = increases and approaches 50.

(b) A solution starting at x 0 75( ) = decreases and approaches 50.

(c) A solution starting at x 0 150( ) = increases without bound.

!!!! General Chemical Reaction of Two Substances

28. (a), (b) We consider the four cases when the exponent is an even positive integer and an oddpositive integer. In each case, we analyze the sign of the derivative for different values ofx. For convenience we pick a = 1, b = 2 .


dxdt

k x x= −( ) −( )1 2even even .

We have drawn a graph of dxdt

versus x.

By drawing arrows to the right when

dxdt

is positive

and arrows to the left when

dxdt

is negative,

2 x1 3semistable semistable

dx dt

Both even exponents

we have a horizontal phase line for x t( ) . We also see that both equilibrium solutionsx t( ) ≡ 1 , x t( ) ≡ 2 are unstable; although both are semistable; stable from below and

unstable from above.

dxdt

k x x= −( ) −( )1 2even odd .

Here x t( ) ≡ 1 is unstable although it isstable from below. The solution x t( ) ≡ 2

is stable. 2 x

1

–1

1 3semistable

stable

dx dt

Even and odd exponents

dxdt

k x x= −( ) −( )1 2odd even .

Here x t( ) ≡ 2 is semistable, stable from

above and unstable from below. Thesolution x t( ) ≡ 1 is stable. 2 x

1

–1

1 3stable

semistable

dx dt

Odd and even exponents


dxdt

k x x= −( ) −( )1 2odd odd .

Here the smaller of the two solutions,x t( ) ≡ 1, is stable; the larger solution,x t( ) ≡ 2 , is unstable. 2 x

1

–1

1 3stable unstable

dx dt

Both odd exponents

!!!! Useful Transformation

29. ′ = −( )y ky y1

Letting

z yy

=−1

yields

dzdt

dzdy

dydt y

dydt

= FHGIKJFHIK =

−( )FHIK

11 2 .

Substituting for dydt

from the original DE yields a new equation

1 12−( ) = −( )y dzdt

ky y ,

which gives the result

dzdt

kyy

kz=−

=1

.

Solving this first-order equation for z z t= ( ), yields z t cekt( ) = and plugging this in the transfor-

mation z yy

=−1

, we get

yy

cekt1−

= .

Finally solving this for y gives

y tec

kt( ) =+ −

11 1 ,

where 1 1 10c y

= − .


!!!! Solving the Threshold Equation

30. ′ = − −FH IKy ry yT

1

Introducing backwards time τ = −t , yields

dydt

dyd

ddt

dyd

= = −τ

ττ

.

Hence, if we run the threshold equation

dydt

ry yT

= − −FHIK1

backwards, we get

− = − −FHGIKJ

dyd

ry yTτ

1 .

Equivalently it also yields the first-order equation

dyd

ry yTτ

= −FHIK1 ,

which is the logistic equation with L T= and t = τ . We know the solution of this logistic

equation to be

y TeT

yr

ττ

( ) =+ − −1 1

0e j

.

We can now find the solution of the threshold equation by replacing τ by −t , yielding

y t TeT

yrt

( ) =+ −1 1

0e j

.

!!!! Limiting Values for the Threshold Equation

31. ′ = − −FH IKy ry yT

1

(a) For 0 0< <y T as t → ∞ the denominator of

y t TeT

yrt

( ) =+ −1 1

0e j

goes to plus infinity and so y t( ) goes to zero.


(b) For y T0 > the denominator of

y t TeT

yrt

( ) =+ −1 1

0e j

will reach zero when

1 1 00

+ −FHGIKJ =

Ty

ert .

Solving for t gives the location of a vertical asymptote on the ty graph

tr

yy T

* ln=−

FHGIKJ

1 0

0.

!!!! Pitchfork Bifurcation

32. ′ = − = −y y y y yα α3 2d i(a) For α ≤ 0 the only real root of y yα − =2 0b g is y = 0 . Because ′ = − <y y yα 2 0b g for

y > 0 and

′ = − >y y yα 2 0b gfor y > 0 , the equilibrium solution y = 0 is stable.

(b) When α > 0 the equation

′ = − =y y yα 2 0b ghas roots

y = 0 , ± α .

The points y = ± α are stable,but y = 0 is unstable as illustratedby the graph of ′ = − −y y y1 2b g .

(c)

1 2–1

–1

–2

1

y

αunstable equilibriastable equilibria

Pitchfork bifurcation at 0 0, ( )

!!!! Another Bifurcation

33. ′ = + +y y by2 1

(a) We find the equilibrium points of the equation by setting ′ =y 0 and solving for y. Doing

this we get

y b b=

− ± −2 42

.


We see that for − < <2 2b there are no (real) solutions, and thus no equilibriumsolutions. For b = −2 we have the equilibrium solution +1, and for b = +2 we haveequilibrium solution –1. For each b ≥ 2 we have two equilibrium solutions.

(b) The bifurcation points are at b = −2 and b = +2 . As b passes through –2 (increasing),the number of equilibrium solutions changes from 2 to 1 to 0, and when b passes through+2, the number of equilibrium solutions changes from 0 to 1 to 2.

(c) We have drawn some solution for each of the values b = −3, –2, –1, 0, 1, 2, and 3.

b = –3

1t


stable equilibrium

–3

3y

b = –2

1t


–3

3y

1t

–3

3y b = –1

no equilibrium points

1t

–3

3y b = 0



1t

–3

3y b = 1


b = 2

1t


–3

3y

1t

–3

3y

stable equilibrium


b = 3

(d) For b = 2 and b = −2 the single equilibrium is semistable. (Solutions above are repelled;those below are attracted.) For b > 2 there are two equilibria; the larger one is unstableand the smaller one is stable. For b < 2 there are no equilibria.

(e) The bifurcation diagram shows the loca-tion of equilibrium points for y versusthe parameter value b. Solid circles rep-resent stable equilibria; open circles rep-resent unstable equilibria. 2 4–2

–2

–4

–4

2

4y

t

unstable

unstablestable

stable

semistable

semistable

b

Equilibria of ′ = + +y y by2 1 versus b


!!!! Computer Lab: Bifurcation

34. ′ = + +y ky y2 1

(a) Set ′ =y 0 to find the equilibria use an open-ended graphic DE solver shown in samplegraphs, as in Problems 33 and 35. Setting ′ =y 0 yields two equilibria,

y kke =

− ± −1 1 42

for k <14

; none for k >14

; one for k =14

. The following phase-plane graphs illustrate the

bifurcation.

(b) Sample figures illustrates solution behavior.

–2

2y

2–2t

k = 05. No equilibrium point

–2

2y

2–2t

k = −05.

Two semistable equilibrium points

–5

5y

5–5t

k = 0 25. One semistable equilibrium point


35. ′ = + +y y y k2

(a) Setting ′ =y 0 yields two equilibria,

y ke =

− ± −1 1 42

,

for k <14

; none for k >14

; one for k =14

. The following phase-plane graphs illustrate the

bifurcation.

(b)

–4

4y

4–4t

k = 1


–4

4y k = 1/4

semistable equilibrium4–4t

–4

4y

4–4t

k = –1

stable equilibrium



!!!! Computer Lab: Growth Equation

36. ′ = −FH IKy ry yL

1

We graph the direction field of this equation for L = 1, r = 05. , 1, 2, and 5. We keep L fixedbecause all it does is raise or lower the steady-state solution to y L= . We see that the larger the

parameter r, the faster the solution approaches the steady-state L.

r = 0.5

stable equilibrium


2y

3–1t

r = 1

stable equilibrium


2y

3–1t

r = 2

stable equilibrium


2y

3–1t



r = 3

stable equilibrium


2y

3–1t

Solutions of ′ = −( )y ry y1 for different values of r

37. ′ = − −FH IKy r yT

y1

The parameter r governs the steepness of the solution curves; the higher r the more steeply yleaves the threshold level T.

38. ′ = −FH IKy r yL

y1 ln

Equilibrium at y eL= ; higher r values give steeper slopes.

39. ′ = −y re ytβ

For larger β or for larger r, solution curves fallmore steeply. Unstable equilibrium r = 1 , β = 1

–100

100y

2–2t


40. Student Project

SECTION 2.6 Systems of DEs: A First Look 149

2.6 Systems of DEs: A First Look

!!!! Predicting System Behavior

1. (a) ′ =′ = −

x yy x y3

This (linear) system has one equilibrium point at the origin, x y, ,a f a f= 0 0 , as do all

linear systems. The υ- and h-nullclines are respectively, as shown in part (b).

(b)

h

–2

2y

2–2x

-nullclineυ -nullcline

(c) A few solutions along with the verticaland horizontal nullclines are drawn.

–2

2y

2–2x

(d) The equilibrium point (0, 0) is unstable.

All solutions tend quickly to y x=

3 then move gradually towards +∞ or −∞

asymptotically along that line. Whether the motion is left or right depends on the initialconditions.

2. (a) ′ = − −

′ = −

x x yy x y

12

Setting ′ =x 0 and ′ =y 0 gives

υ - nullcline- nullcline

1 002

− − =− =x y

h x y .


From the intersection of the two nullclines we find two equilibrium points shown in thefollowing figures. We can locate them graphically far more easily than algebraically!

(b)

2–2x

–2

2y

h-nullclineυ -nullcline

(c)

2–2x

–2

2y

(d) The lower equilibrium point at

14

1 5 12

1 52( ) , ( )+ − −LNM

OQP

is unstable and the upper equilibrium at

14

1 5 12

5 12( ) , ( )− −LNM

OQP

is stable.

Most trajectories spiral counterclockwise toward the first quadrant equilibriumpoint. However, if the initial condition is somewhat left or below the 4th quadrantequilibrium, they shoot down towards −∞ . We suspect a dividing line between thesebehaviors, and we will find it in Chapter 6.

3. (a) ′ = − −

′ = − −

x x yy x y

11 2 2

∆

Setting ′ =x 0 and ′ =y 0 gives


x yh x y

+ =+ =

112 2 .

From the intersection of the two nullclines we find two equilibrium points (0, 1), (1, 0).


(b)

2–2x

–2

2y

h-nullcline υ -nullcline

(c)

2–2x

–2

2y

(d) The equilibrium at (1, 0) is unstable; the equilibrium at (0, 1) is stable. Most trajectoriesseem to be attracted to the stable equilibrium, but those that approach the lower unstableequilibrium from below or form the right will turn down toward the lower right.

4. ′ = +′ = +

x x yy x y2 2

(a) This (singular) system has an entire lineof equilibrium points on the line

x y+ = 0 .

(b) The direction field and the line of unsta-ble equilibrium points are shown at theright.

(c) We superimpose on the direction field afew solutions.

–2

2y

2–2x

(d) From part (c) we see the equilibrium points on the line x y+ = 0 are all unstable.

All nonequilibrium trajectories shoot away from the equilibria along straightlines (of slope 2), towards +∞ if the IC is above the line x y+ = 0 and toward −∞ if theIC is below x y+ = 0 .

5. (a) ′ = − −

′ = − −

x x yy x y

43 2 2

Setting ′ =x 0 and ′ =y 0 gives the intersection of the nullclines:


y xh x y

= −+ =

432 2 .

We find no equilibria because the nullclines do not intersect.


(b)

–4

4y

4–4x

h-nullcline υ -nullcline

(c)

–4

4y

4–4x

(d) There are no equilibria—all solutions head down to lower right.

6. (a) ′ =′ = +

x yy x y5 3

This (linear) system has one equilibrium point at x y, ,a f a f= 0 0 as do all linear systems.The 64-dollar question is: Is it stable? The υ- and h-nullclines: y = 0 , 5 3 0x y+ = , are

shown following and indicate that the origin (0, 0) is unstable. Hence, points startingnear the origin will leave the origin. We will see later other ways for showing that (0, 0)is unstable.

(b) The direction field and a few solutionsare drawn. Note how the solutions crossthe vertical and horizontal nullclines.

–2

2y

2–2x

h-nullcline

υ -nullcline


(c)

–2

2y

2–2x

(d) We see from the preceding figure that solutions come from infinity along a line, and thenif they are not exactly on the line head off either upwards and to the left or downwardsand go to the left on another line. Whether they go up or down depends on whether theyinitially lie above or below the line. It appears that points that start exactly on the linewill go to (0, 0). We will see later in Chapter 6 when we study linear systems usingeigenvalues and eigenvectors that the solutions come from infinity on one eigenvectorand go to infinity on another eigenvector.

7. (a) ′ = − −′ = −

x x yy x y

1

Setting ′ = ′ =x y 0 and finding the intersection of the nullclines:


y xh y x

= −=1

we find one equilibrium point 12

12

, FH IK . The arrows indicate that it is a stable equilib-

rium.

(b)

2–2x

–2

2y (c)

2–2x

–2

2y

(d) The equilibrium is stable; all other solutions spiral into it.


8. (a) ′ = +′ =

x x yy x

2

This (linear) system has one equilibrium point at the origin (0, 0), as do all linearsystems. The υ- and h-nullclines: x y+ =2 0 , x = 0 , are shown following and indicate

that the origin is (0, 0) is unstable. We will see later other ways to show that the systemis unstable.

(b)

–2

2y

2–2x

h-nullcline

υ -nullcline

(c)

–2

2y

2–2x

(d) The equilibrium point (0, 0) is unstable. Other solutions come from upper left, lowerright, heading toward origin but veers off towards ±∞ in upper right or lower left.

!!!! Sharks and Sardines with Fishing

9. (a) With fishing the equilibrium point of the system

′ = − −( )

′ = − + −( )x x a by fy y c dx f

is

~

~ .

x c fd

cd

fd

y a fb

ab

fb

e

e

=+

= +

=−

= −

With fishing we increase the equilibrium of the prey xe by fd

and decrease the equilib-

rium of the predator ye by fb

.


Using the parameters from Ex-ample 3 we set

a = 2 , b = 1, c = 3 , d = 05. ;

the new equilibrium point of the fishedmodel is

~

~ .

x c fd

cd

fd

f

y a fb

ab

fb

f

e

e

=+

= + = +

=−

= − = −

6 2

2

0

4y

150x

Shark (y) and sardine (x) trajectories

The trajectories are closed curves representing periodic motion of both sharks andsardines. The trajectories look like the trajectories of the unfished case in Example 3except the equilibrium point has moved to the right (more prey) and down (fewerpredators).

(b) With the parameters in part (a) and f = 05. the equilibrium point is (7, 1.5). This

compares with the equilibrium point (6, 2) in the unfished case.

As the fishing rate f increasesfrom 0 to 2, the equilibrium point movesalong the line from the unfished equilib-rium at (6, 2) to (10, 0). Hence, the fish-ing of each population at the same ratebenefits the sardines (x) and drives thesharks (y) to extinction. This is illus-trated in the figure. 84

y

x1262 10

2 (6, 2)

(10, 0)

(sharks)

(sardines)

1

(7, 1.5)

(c) You should fish for sardines when the sardine population is increasing and sharks whenthe shark population is increasing. In both cases, more fishing tends to move thepopulations closer to equilibrium while maintaining higher populations in the low partsof the cycle.

(d) If we look at the insecticide model and assume both the good guys (predators) and badguys (prey) are harvested at the same rate, the good guys will also be diminished and thebad guys peak again. As f → 1 (try f = 08. ) the predators get decimated first, then the

prey can peak again. If you look at part (a), you see that the predator/prey model does notallow either population to go below zero, as the x- and y-axes are solutions and thesolutions move along the axes, thus it is impossible for other solutions to cross either of


these axes. You might continue this exploration with the IDE tool, Lotka-Volterra withHarvest, as in Problem 24.

!!!! Finding the Model

10. ′ = − − −′ = − +

x ax bx dxy fxy cy dxy

2 11. ′ = +′ = − +

′ = − −

x ax bxyy cy dxy eyzz fz gx hyz2

12. ′ = − − −

′ = − +′ = − +

x ax bx cxy dxzy ey fy gxyz hz kxz

2

2

!!!! Host-Parasite Models

13. (a) A suggested model is

′ = −+

′ = − +

H aH c HP

P bP dHP1

where a, b, c, and d are positive parameters. Here a species of beetle (parasite) dependson a certain species of tree (host) for survival. Note that if the beetle were so effective asto wipe out the entire population of trees, then it would die out itself, which is reflectedin our model (note the differential equation in P). On the other hand, in the absence ofthe beetle, the host tree may or may not die out depending on the size of the parameters aand c. We would probably pick a c> , so the host population would increase in theabsence of the parasite. Note too that model says that when the parasite (P) population

gets large, it will not destroy the host entirely, as HP1+

becomes small. The modeler

might want to estimate the values of parameters a, b, c, d so the solution fits observeddata. The modeler would also like to know the qualitative behavior of P H,a f in the PH

plane.

(b) Many bacteria are parasitic on external and internal body surfaces; some invading innertissue causing diseases such as typhoid fever, tuberculosis, and pneumonia. It isimportant to construct models of the dynamics of these complex organisms.


!!!! Competition

14. (a) ′ = − −′ = − −

x x x yy y x y

( )( )4 24 2

Setting ′ =x 0 and ′ =y 0 we find

υ - nullclines - nullclines

2 4 02 4 0

x y xh x y y

+ = =+ = =

,, .

Equilibrium points: (0, 0), (0, 2), (2, 0),43

43

, FHIK . The directions of the solution

curves are shown in the figure.

50x

y

(b) It can be seen from the figure, that the equilibrium points (0, 0), (0, 2) and (2, 0) are

unstable. Only the point 43

43

, FHIK is stable because all solution curves nearby point

toward it.

(c) Some solution curves are shown in the figure.

(d) Because all the solution curves eventually reach the stable equilibrium at 43

43

, FHIK , the

two species described by this model can coexist.

15. (a) ′ = − −′ = − −

x x x yy y x y

( )( )12



x y xh x y y

+ = =+ = =

1 02 0,, .

Equilibrium points: (0, 0), (0, 2), (1, 0).The directions of the solution curves areshown in the figure.

30x

y

(b) It can be seen from the figure, that the equilibrium points (0, 0) and (1, 0) are unstable;the point (0, 2) is stable because all solution curves nearby point toward it.


(d) Because all the solution curves eventually reach the stable equilibrium at (0, 2), the xspecies always die out and the two species described by this model cannot coexist.


16. (a) ′ = − −′ = − −

x x x yy y x y

( )( )4 21 2



x y xh x y y

+ = =+ = =2 4, 0

2 1 0, .

Equilibrium points: (0, 0), (0, 1), (4, 0).The directions of the solution curves areshown in the figure.

50x

y

(b) It can be seen from the figure, that the equilibrium points (0, 0) and (0, 1) are unstable;the point (4, 0) is stable because all solution curves nearby point toward it.


(d) Because all the solution curves eventually reach the stable equilibrium at (4, 0), the yspecies always die out and the two species described by this model cannot coexist.

17. (a) ′ = − −′ = − −

x x x yy y x y

( )( )2 22 2



x y xh x y y

+ = =+ = =2 2, 0

2 2, 0.

Equilibrium points: (0, 0), (0, 2), (2, 0),23

23

, FHIK . The directions of the solution

curves are shown in the figure.

30x

y

(b) It can be seen from the figure, that the equilibrium points (0, 0) and 23

23

, FHIK are

unstable; the points (0, 2) and (2, 0) are stable because all nearby arrows point towardthem.


(d) Because all the solution curves eventually reach one of the stable equilibria at (0, 2) or(2, 0), the two species described by this model cannot coexist, unless they are exactly at

the unstable equilibrium point 23

23

, FHIK . Which species dies out is determined by the

initial conditions.


!!!! Simpler Competition

18. Setting ′ =x 0 , we find the υ-nullclines are the vertical line x = 0 and the horizontal line y ab

= .

Setting ′ =y 0 , we find the h-nullclines are the horizontal y = 0 and vertical line x cd

= . The

equilibrium points are (0, 0) and cd

ab

, FHIK . By observing the signs of ′x , ′y we find

′ > ′ >x y0 0, when x cd

y ab

< <,

′ < ′ <x y0 0, when x cd

y ab

> >,

′ < ′ >x y0 0, when x cd

y ab

> <,

′ > ′ <x y0 0, when x cd

y ab

< >,

Hence, both equilibrium points are unstable. Wecan see from the following direction field (witha b c d= = = = 1) that one of two species,depending on the initial conditions, goes toinfinity and the other toward extinction.

Equation used in the drawing on the right are:x x yy y x' ( )' ( ).= −= −

11 2

4y

x4

2

One can get the initial values for these curves directly from the graph.

!!!! Nullcline Patterns

19. (a–e) When the υ-nullcline lies above the h-nullcline, there are three equilibrium points in the

first quadrant: (0, 0), 0, df

FHGIKJ and a

b, 0FHIK . The points (0, 0), 0,

df

FHGIKJ are unstable and

ab

, 0FH IK is stable. Hence, only population x survives.


(a)–(b) y

x

a/c

d/f

d/e a/b


Nullclines and equilibria

(c)–(d) y

x

a/c

d/f

d/e a/b

h-nullcline

υ -nullcline

Sample trajectories when theυ-nullcline is above the h-nullcline

(e) Population x survives.

20. (a–e) When the h-nullcline lies above the υ-nullcline, there are three equilibrium points in the

first quadrant: (0, 0), ab

, 0FHIK and 0,

df

FHGIKJ . The points (0, 0), a

b, 0FHIK are unstable and

0, df

FHGIKJ stable. Hence, only population y survives.

(a)–(b) y

xa/b d/e

d/f

a/ch-nullclineυ -nullcline

Nullclines

(c)–(d) y

x

d/f

a/c

a/b d/e


Sample trajectories when theh-nullcline is above the υ-nullcline

(e) Population y survives.

21. (a–e) When the two nullclines intersect as they do in the figure, then there are four equilibrium

points in the first quadrant: (0, 0), ab

, 0FHIK , 0,

df

FHGIKJ , and x ye e,a f , where x ye e,a f is the

intersection of the lines bx cy a+ = , ex fy d+ = . Analyzing the sign of the derivatives inthe four regions of the first quadrant, we find that x ye e,a f is stable and the others

unstable. Hence, the two populations can coexist.


(a)–(b) y

x

d/f

a/b d/e

a/c

h-nullcline

υ -nullcline


(c)–(d) y

x

a/c

d/f

a/b d/e

h-nullcline

υ -nullcline

Typical trajectories when the nullclinesintersect and the slope of the vertical

nullcline is more negative.

(e) The two populations coexist

22. (a–e) When the two nullclines intersect as they do in the figure, then there are four equilibrium

points in the first quadrant: (0, 0), ab

, 0FH IK , 0, df

FHGIKJ , and x ye e,a f , where x ye e,a f is the

intersection of the lines bx cy a+ = , ex fy d+ = . Analyzing the sign of the derivatives in

the four regions of the first quadrant, we find ab

, 0FHIK and 0,

df

FHGIKJ are stable and the

other two unstable. Hence, only one of the two populations survives, and which survivesdepends on the initial conditions.

(a)–(b) y

x

a/c

d/e a/b

d/f

h-nullcline

υ -nullcline


(c)–(d)

��

��

y

x

d/f

a/c

d/e a/b

h-nullcline

υ -nullcline

Typical trajectories when the nullclinesintersect and the slope of the h-nullcline

is more negative.

(e) Only one of the two populations survives, and which survives depends on the initialconditions, as indicated by shading. For initial conditions in the upper region y survives;for initial conditions in the lower region, x survives.


!!!! Unfair Competition

23. ′ = −( ) −′ = −

x ax bx cxyy dy exy

1

Setting ′ = ′ =x y 0 , we find three equilibrium points:

0 0,a f , 1 0b

, FH IK , and de

a e bdce

, −FH IK .

The point (0, 0) corresponds to both populations becoming extinct, the point 1 0b

, FH IK corresponds

to the second population becoming extinct, and the point de

a e bdce

, −FH IK corresponds to either a

stable coexistent point or an unstable point. If we take the special case where 1b

de

> , e.g.,

′ = −( ) −′ = −

x x x xyy y xy

105.

where a b c e= = = = 1 , d = 05. , we have the equilibrium points (0, 0), (1, 0), and (0.5, 0.5). Ifwe draw two nullclines; v-nullcline: y x= −1 , h-nullcline: x = 05. , as shown following we see

that the equilibrium point (0.5, 0.5) is unstable. Hence, the two species cannot coexist.

1.50x0

1.5y

d/e 1/b

h-nullcline

υ -nullcline

Nullclines and equilibria for 1b

de

>

1.50x0

1.5y

d/e 1/b

Sample trajectories for 1b

de

>

The reader must check separately the cases where 1b

de

= or 1b

de

< .

!!!! Computer Lab: Parameter Investigation

24. Hold three of the parameters constant and observe how the fourth parameter affects the behaviorof the two species. See if the behavior makes sense in your understanding of the model. Keep inmind that parameter aR is a measure of how well the prey grows in the absence of the predator(large aR for rabbits), aF is a measure of how fast the predator population will decline when theprey is absent (large aF if the given prey is the only source of food for the predator), cR is a


measure of how fast the prey’s population declines per number of prey and predators, and cF is a

measure of how fast the predator’s population increases per number of prey and predators. Evenif you are not a biology major, you may still ponder the relative sizes of the four parameters inthe two predator–prey systems: foxes and rabbits, and ladybugs and aphids. You can use theseexplanations to reach the same conclusions as in Problem 9.

!!!! Computer Lab: Competition Outcomes

25. (a) Using the IDE software, hold all parameters fixed except one and observe how the lastparameters affects the solution. See if the behavior of the two species makes sense inyour understanding of the model. Play a mind game and predict if there will becoexistence between the species, whether one becomes extinct, and so on, before youmake the parameter change.

Note that in the IDE tool, Competitive Exclusion, there are six parameters; K1 ,B1 , r1 , K2 , B2 , and r2 . The parameters in our text called aR , bR , cR , aF , bF , and cF

and enter the equations slightly differently. The reason for this discrepancy is due to theway the parameters in the IDE software affect the two isoclines, called the N1 and N2

isoclines in the IDE software.

By changing the parameters K1 and K2 in the IDE software, you simply movethe respective isoclines in a parallel direction. The parameters B1 , B2 change the slopesof the nullclines. And finally, the parameters r r1 2, do not affect the nullclines, but affect

the direction field or the transient part of the solution.

Your hand-sketched phase plane for the four cases should qualitatively look likethe following four pictures, with the basins of attraction colored for each equilibrium.

0

3y

50x

Case 1: Population x dies out′ = − −( )

′ = − −( )x x x yy y x y

12

0

3y

50x

Case 2: Population y dies out′ = − −( )

′ = − −( )x x x yy y x y

4 21 2


20x0

2y

Case 3: Populations coexist′ = − −( )

′ = − −( )x x x yy y x y

2 22 2

20x0

2y

Case 4: One of the populations dies out′ = − −( )

′ = − −( )x x x yy y x y

2 22 2

Of the four different scenarios to the competitive model, in only one (Case 3)can both species coexist. In the other three cases one of the two dies out. In Case 4 thespecies that dies out depends on the initial conditions, and in Case 1 and 2 one specieswill die out regardless of the initial condition. Note too that in all four cases if onepopulation initially starts at zero, it remains at zero.

(b) The basins of attraction for each stable equilibrium are shown for each of the four cases.Compare with the figures in part (a).

��

y

x

Case 1��

�

Case 2

��

y

x

Case 3

��

��

y

x

Case 4


26. Student Project

chapter 2 linearity and nonlinearity - florida gulf coast...

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