and CHAPTER
2
Linearity Nonlinearity
2.1
Linear Equations: The Nature of Their Solutions
! 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. ! 11. 13. 15. 16.
Classification First-order, nonlinear First-order, linear, nonhomogeneous, variable coefficients Second-order, linear, homogeneous, variable coefficients Second-order, linear, nonhomogeneous, variable coefficients Third-order, linear, homogeneous, constant coefficients Third-order, linear, nonhomogeneous, constant coefficients Second-order, linear, nonhomogeneous, variable coefficients Second-order, nonlinear Second-order, linear, homogeneous, variable coefficients Second-order, nonlinear Pop Quiz y + y = 2 y(t ) = ce-t + 2 y - 3 y = 5 y(t ) = ce3t - 5 3 12. 14. y + 2 y = 1 y(t ) = ce-2 t + 1 2
y - 0.08 y = 100 y(t ) = ce0.08t - 1250
y + 2 y = 4 , y(0) = 1 y(t ) = ce-2t + 2 , y(0) = 1 c = -1. Hence, y(t ) = 2 - e-2t . y + 5y = 1, y(1) = 0 y(t ) = ce-5t +
1 1 1 , y(1) = 0 c = - e5 . Hence, y(t ) = 1 - e-5(t -1) . 5 5 5
b
g
69
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! 17.
Nonhomogeneous Principle I The homogeneous equation is 2 x - y = 0 , which has solutions
Xh = ( , 2 ) arbitrary
homogeneous solution 2 Xh 5 y = 2x
k
p
2 4 X = Xh + X p Xp general solution
and describes the line y = 2 x in the xy-plane. A particular solution of the nonhomogeneous equation 2 x - y = 5 is X p = (0, - 5) . Hence the solution of 2 x - y = 5 consists of
X = {(0, - 5) + (1, 2)}
y = 2x - 5
where is any real number; this solution is the parametric form of the line y = 2 x - 5 in the xyplane. ! 18. Superposition Principle If y1 and y2 are solutions of y + p(t ) y = 0, then y1 + p(t ) y1 = 0 y2 + p(t ) y2 = 0. Adding these equations gives y1 + y2 + p(t ) y1 + p(t ) y2 = 0 or
a y + y f + p(t )a y + y f = 0 ,
1 2 1 2
which shows that y1 + y2 is also a solution of the given equation. If y1 is a solution, we have y1 + p(t ) y1 = 0 and multiplying by c we get
acy f + p(t )acy f = 0 ,
1 1
c y1 + p(t ) y1 = 0 cy1 + cp(t ) y1 = 0
a
f
which shows that cy1 is also a solution of the equation.
SECTION 2.1
Linear Equations: The Nature of Their Solutions
71
! 19.
Second-Order Superposition Principle If y1 and y2 are solutions of y + p(t ) y + q(t ) y = 0, we have y1 p(t ) y1 + q(t ) y1 = 0 + y2 + p(t ) y2 + q(t ) y2 = 0. Multiplying these equations by c1 and c2 respectively, then adding and using properties of the derivative, we arrive at
ac y + c y f + p(t )ac y + c y f + q(t)ac y + c y f = 0 ,
1 1 2 2 1 1 2 2 1 1 2 2
which shows that c1 y1 + c2 y2 is also a solution. ! 20. Linear and Nonlinear Operations L( y) = y + 2 y L(cy) = (cy) + 2(cy) = cy + 2cy = c( y + 2 y) = cL( y) Hence, L is a linear operator. 21. L( y) = y + y 2 L(cy) = (cy) + (cy)2 = cy + c2 y 2 c y + y2 = cL( y) Hence, L is not a linear operator. 22. L( y) = y + 2ty L y1 + y2 = y1 + y2 + 2t y1 + y2
1 1 2 1 2
b
g
a
f a f a f = a y + 2ty f + a y + 2ty f = La y f + La y f
2
L(cy) = (cy) + 2t (cy) = c( y + 2ty) = cL( y ) Hence, L is a linear operator. This problem illustrates the fact that the coefficients of a DE can be functions of t and the operator will still be linear. 23. L( y) = y - et y L y1 + y2 = y1 + y2 - et y1 + y2
1 t 1 2 t 1 2
a
f a f a f = b y - e y g + b y - e y g = La y f + La y f L(cy) = (cy) - e (cy) = cb y - e yg
2 t t
= cL( y )
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Hence, L is a linear operator. This problem illustrates the fact that a linear operator need not have coefficients that are linear functions of t. 24. L( y) = y + (sin t ) y L y1 + y2 = y1 + y2 + (sin t ) y1 + y2 = y1 (sin t ) y1 + y2 + (sin t ) y2 +
1 2
a
f a k
a p k = La y f + La y f
= cL( y )
f
f
p
L(cy) = (cy) + (sin t )(cy) = c{ y + (sin t ) y}
Hence, L is a linear operator. This problem illustrates the fact that a linear operator need not have coefficients that are linear functions of t. 25.
L( y) = y + 1 - y 2 y + y
b
g
L(cy) = (cy) + 1 - (cy)2 y + (cy)
2
n s cl y + b1 - y g y + yq
= cL( y)
Hence, L( y ) is not a linear operator. ! 26. Many from One Because y(t ) = t 2 is a solution of a linear homogeneous equation, we know by parquetry (A) that ct 2 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 firstorder equations given for Problems 2731 the general solutions come in two parts: solutions to the associated homogeneous equation (which could be found by separation of variables) plus a particular solution of the nonhomogeneous equation. For second-order linear equations as Problems 3235 we can also sometimes find solutions by inspection. 27. 29. 31. 32. 1 y + y = et y(t ) = ce-t + et 2 y - y = et y(t ) = cet + tet y + y 3 c t4 = t y(t ) = + t t 5 28. 30. y + y = e-t y(t ) = ce- t + te- t
y - ty = 0 y(t ) = cet
2
2
y - a 2 y = 0 y(t ) = c1eat + c2e-at . An alternative form is y(t ) = c1 sinh(at ) + c2 cosh(at ) .
SECTION 2.1
Linear Equations: The Nature of Their Solutions
73
33. 35. ! 36.
y + a 2 y = 0 y(t ) = c1 sin at + c2 cos at y - y = 0 y(t ) = c1 + c2et Linear Algebraic Equations The homogeneous system corresponding to
34.
y + y = 0 y(t ) = c1 + c2e-t
x+z=2 y-z=0 is x+z=0 y-z=0 The solution of this system is
Xh = (- , , ) arbitrary ,
k
p
which is a line in a three-dimensional space. A particular solution of the nonhomogeneous system is
X p = ( x, y, z) = (2, 0, 0) .
Hence, the general solution of the nonhomogeneous solution is = (2 - , , ) arbitrary . ! Nonhomogeneous Principle II
X = Xh + X p = (2, 0, 0) + (- , , ) arbitrary
k
k
p
p
In these problems, the verification of y p is a straightforward substitution. To find the rest of the solution we simply add to y p all the homogeneous solutions yh , which we find by inspection or separation of variables. 37. 38. 39. 40. ! 41. y - y = 3et has general solution y(t ) = yh + y p = cet + 3tet
y + 2 y = 10sin t has general solution y(t ) = yh + y p = ce-2t + 4 sin t - 2 cos t
y - y +
2 y = y 2 has general solution y(t ) = yh + y p = ct 2 + t 3 t c t 2 + 2t 1 + y = 2 has general solution y(t ) = yh + y p = t +1 t +1 t +1
Suggested Journal Entry Student Project
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CHAPTER 2
Linearity and Nonlinearity
2.2
Solving the First-Order Linear Differential Equation
!
General Solutions
The solutions for Problems 115 can be found using either the Euler-Lagrange method or the integrating factor method. For problems where we find a particular solution by inspection (Problems 2, 6, 7) we use the Euler-Lagrange method. For the other problems we find it more convenient to use the integrating factor method, which gives both the homogeneous solutions and a particular solution in one swoop. You can use the Euler-Lagrange method to get the same results. 1.
y + 2 y = 0
By inspection we have y(t ) = ce-2t . 2. y + 2 y = 3et We find the homogeneous solution by inspection as yh = ce-2t . A particular solution on the nonhomogeneous equation can also be found by inspection, and we see y p = et . Hence the general solution is y(t ) = ce-2t + et . 3. y - y = 3et We multiply each side of the equation by the integrating factor
t =e
giving
b g z bg
p t dt
=e
zb
-1 dt
g
= e- t
e- t { y - y} = 3 or simply d ye-t = 3. dt Integrating, we find ye-t = 3t + c or y(t ) = cet + 3tet .
l q
SECTION 2.2
Solving the First-Order Linear Differential Equation
75
4.
y + y = sin t
We multiply each side of the equation by the integrating factor (t ) = et , giving et { y + y} = et sin t or d yet = et sin t . dt Integrating, we get yet = Solving for y, we find 1 1 y(t ) = ce -t + sin t - cos t . 2 2 5. y + y = 1 1 + et 1 t e (sin t - cos t ) + c . 2
l q
We multiply each side of the equation by the integrating factor (t ) = et , giving et { y + y} = et 1 + et d et . yet = 1 + et dt
l q b
Integrating, we get
yet = ln 1 + et + c .
g
Hence, y t = ce - t + e - t ln 1 + et . 6.
y + 2ty = t
bg
d
i
In this problem we see that y p (t ) =
1 is a solution of the nonhomogeneous equation (there are 2
other single solutions, but this is the easiest to find). Hence, to find the general solution we solve the corresponding homogeneous equation
y + 2ty = 0
by separation of variables getting dy = -2tdt , y
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Linearity and Nonlinearity
which has the general solution
y = ce-t
2
where c is any constant. Adding the solutions of the homogeneous equation to the particular 1 solution y p = we get the general solution of the nonhomogeneous equation: 2 y(t ) = ce-t +
2
1 . 2
7.
y + 3t 2 y = t 2 In this problem we see that y p (t ) = 1 is a solution of the nonhomogeneous equation (there are 3
other single solutions, but this is the easiest to find). Hence, to find the general solution, we solve the corresponding homogeneous equation y + 3t 2 y = 0 by separation of variables, getting dy = -3t 2dt , y which has the general solution
y(t ) = ce-t
3
where c is any constant. Adding the solutions of the homogeneous equation to the particular 1 solution y p = , we get the general solution of the nonhomogeneous equation 3 y(t ) = ce -t +
3
1 . 3
8.
1 1 y + y = 2 , (t 0) t t We multiply each side of the equation by the integrating factor
(t ) = e
giving
z
dt t
= eln t = t
1 1 t y + y = t t d 1 {ty} = . dt t
R S T
U V W
SECTION 2.2
Solving the First-Order Linear Differential Equation
77
Integrating, we find
ty = ln t + c .
Solving for y, we get y(t ) = 9.
ty + y = 2t
1 c ln t . + t t
FI HK
We rewrite the equation as 1 y + y = 2 t and multiply each side of the equation by the integrating factor
(t ) = e
giving
z
dt t
= eln t = t ,
1 t y + y = 2t t d {ty} = 2t . dt Integrating, we find ty = t 2 + c . Solving for y, we get y(t ) = 10. c +t . t
R S T
U V W
bcos t g y + sin ty = 1
We rewrite the equation as y + tan t y = sec t and multiply each side of the equation by the integrating factor
b g
(t ) = e z tan tdt = e- ln(cos t ) = eln(cos t ) = sec t ,
-1
giving sec t y + tan t y = sec2 t
2
m b gr d mbsec t g yr = sec t . dt
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Integrating, we find
bsec t g y = tan t + c .
Solving for y, we get y(t ) = c cos t + sin t . 11. y - 2 y = t 2 cos t , (t 0) t
We multiply each side of the equation by the integrating factor
(t ) = e- z a2 t fdt = e-2 ln t = eln t = t -2 ,
-2
giving
t -2 y -
R S T
U V W d lt yq = cos t . dt
2 y = cos t t
-2
Integrating, we find t -2 y = sin t + c . Solving for y, we get y(t ) = ct 2 + t 2 sin t . 12. y + sin t 3 y = 3 , (t 0) t t
We multiply each side of the equation by the integrating factor
(t ) = e
giving
za
3 t dt
f
= e3 ln t = elnbt g = t 3 ,
3
t 3 y +
R S T
U V W d lt yq = sin t . dt
3 y = sin t t
3
Integrating, we find t 3 y = - cos t + c . Solving for y, we get y(t ) = c 1 - cos t . t3 t3
SECTION 2.2
Solving the First-Order Linear Differential Equation
79
13.
b1 + e g y + e y = 0
t t
We rewrite the equation as y +
FG e IJ y = 0 H1+ e K
t t
and then multiply each side of the equation by the integrating factor
(t ) = e
giving
z
e t 1+ e t dt
b g
= elnb1+ e g = 1 + et ,
t
d 1 + et y = 0 . dt Integrating, we find
od
it
b1 + e gy = c .
t
Solving for y, we have y(t ) = 14. c . 1 + et
bt + 9gy + ty = 0
2
We rewrite the equation as y +
F Ht
2
t y=0 +9
I K
and then multiply each side of the equation by the integrating factor
(t ) = e
giving
z
t t 2 + 9 dt
b
g
= ea1 2f lnbt
2 +9
g=
t2 + 9 ,
d dt Integrating, we find
{
t2 + 9 y = 0 .
}
t2 + 9 y = c. Solving for y, we find y(t ) = c t2 +9 .
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15.
y +
F 2t + 1I y = 2t , (t 0) H t K
(t ) = e
We multiply each side of the equation by the integrating factor
z
(2t +1) tdt
= te2t ,
giving d te2t y = 2t 2e2t . dt Integrating, we find 1 te2t y = t 2e2t - te2t + e2t + c . 2 Solving for y, we have y(t ) = c ! 16. Initial-Value Problems
y - y = 1, y(0) = 1
l q
FG e IJ + 1 + t - 1 . H t K 2t
-2t
By inspection, the homogeneous solutions are yh = cet . A particular solution of the nonhomogeneous can also be found by inspection to be y p = -1. Hence, the general solution is y = yh + y p = cet - 1. Plugging in y(0) = 1 gives c - 1 = 1 or c = 2 . Hence, the solution of the IVP is y(t ) = 2et - 1. 17. y + 2ty = t 3 , y(1) = 1 We can solve the differential equation using either the Euler-Lagrange method or the integrating factor method to get y(t ) = 1 2 1 2 t - + ce-t . 2 2
Plugging in y(1) = 1 we find ce-1 = 1 or c = e. Hence, the solution of the IVP is y(t ) = 1 2 1 1-t 2 t - +e . 2 2
SECTION 2.2
Solving the First-Order Linear Differential Equation
81
18.
y -
F 3I y = t HtK
3,
y(1) = 4
We find the integrating factor to be
(t ) = e-
Multiplying the DE by this, we get
za
3 t dt
f
= e-3 ln t = eln t
-3
= t -3 .
d -3 t y = 1. dt Hence, t -3 y = t + c or y = ct 3 + t 4 . Plugging in y 1 = 4 gives c + 1 = 4 or c = 3 . Hence, the solution of the IVP is y(t ) = 3t 3 + t 4 . 19.
y + 2ty = t , y(0) = 1
l q
bg
We solved this differential equation in Problem 6 using the integrating factor method and found y(t ) = ce-t +
2
1 . 2
Plugging this in y(0) = 1 gives c +
1 1 = 1 or c = . Hence, the solution of the IVP is 2 2 y(t ) = 1 -t 2 1 e + . 2 2
20.
b1 + e g y + e y = 0 , y(0) = 1
t t
We solved this DE in Problem 13 and found y(t ) = Plugging in y(0) = 1 gives c . 1 + et
c = 1 or c = 2 . Hence, the solution of the IVP is 2 y(t ) = 2 . 1 + et
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! 21.
Synthesizing Facts (a) (b) (c) y(t ) = t 2 + 2t , (t > -1) t +1
3 y
y(t ) = t + 1, (t > -1)
The algebraic solution given in Example 1 for k = 1 is y(t ) = t 2 + 2t + 1 (t + 1)2 = . t +1 t +1
3
3
t
3
Hence, when t -1 we have y = t + 1. (d) The solution passing through the origin (0, 0) asymptotically approaches the line
y = t + 1 as t , which is the solution passing through y(0) = 1. The entire line y = t + 1 is not a solution of the DE, as the slope is not defined when t = -1. The segment
of 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 solution would be the segment of the line y = t + 1 for t less than 1. Notice in the direction field the 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 + 2 y = 0
Here p(t ) = 2 , therefore
(t ) = e
z
p(t )dt
=e
z
2dt
= e2 t .
Multiplying each side of the equation y + 2 y = 0 by e2t yields d ye2t = 0 . dt Integrating gives ye2t = c . Solving for y gives y(t ) = ce-2t .
l q
SECTION 2.2
Solving the First-Order Linear Differential Equation
83
23.
y + 2 y = 3et Here p(t ) = 2 , therefore
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
z
2dt
= e2 t .
y + 2 y = 3et by e2t yields d ye2t = 3e3t . dt Integrating gives ye2t = e3t + c . Solving for y gives y(t ) = ce-2t + et . 24. y - y = e3t Here p(t ) = -1, therefore
l q
(t ) = e
Multiplying each side of the equation
z
p(t )dt
= e-
z
dt
= e-t .
y - y = e3t by e -t yields d ye-t = e2t . dt Integrating gives ye -t = Solving for y gives 1 y(t ) = cet + e3t . 2 25.
y + y = sin t
l q
1 2t e +c. 2
Here p(t ) = 1 therefore the integrating factor is
(t ) = e
z
p(t )dt
=e
z
dt
= et .
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Multiplying each side of the equation
y + y = sin t
by et gives the differential d yet = et sin t . dt Integrating gives yet = Solving for y gives 1 y(t ) = {sin t - cos t} + ce-t . 2 26. y + y = 1 1 + et 1 t e sin t - cos t + c . 2
l q l
q
Here p(t ) = 1 therefore
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
z
dt
= et .
y + y = by et yields
1 1 + et
d et . yet = 1 + et dt
l q b
Integrating gives
yet = ln 1 + et + c .
g
Solving for y gives
y(t ) = e-t ln 1 + et + ce-t .
b
g
27.
y + 2ty = t
Here p(t ) = 2t , therefore
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
z
2tdt
= et .
2
y + 2ty = t
SECTION 2.2
Solving the First-Order Linear Differential Equation
85
by et yields d 2 2 yet = tet . dt Integrating gives yet = Solving for y gives y(t ) = ce-t +
2 2
2
n s
1 t2 e +c. 2
1 . 2
28.
y + 3t 2 y = t 2 Here p(t ) = 3t 2 , therefore
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
z
3t 2 dt
= et .
3
y + 3t 2 y = t 2 by et yields d 3 3 yet = t 2et . dt Integrating gives 1 3 3 yet = et + c . 3 Solving for y gives y(t ) = ce -t +
3 3
n s
1 . 3
29.
1 1 y + y = 2 t t 1 Here p(t ) = , therefore t
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
za
1 t dt
f
= eln t = t .
1 1 y + y = 2 t t
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by t yields d 1 {ty} = . dt t Integrating gives
ty = ln t + c .
Solving for y gives 1 1 y(t ) = c + ln t . t t 30.
ty + y = 2t
1 Here p(t ) = , therefore t
(t ) = e
Multiplying each side of the equation
z
p(t )dt
=e
za
1 t dt
f
= eln t = t .
y + by t yields
y =2 t
d {ty} = 2t . dt Integrating gives ty = t 2 + c . Solving for y gives 1 y( t ) = c + t . t ! 31. Switch for Linearity dy 1 = , y(-1) = 0 dt t + y dt dt = t + y , or -t = y. dy dy
Flipping both sides of the equation, yields the equivalent linear form Solving this equation we get t ( y) = ce y - y - 1.
Using the condition y(-1) = 0, we find -1 = ce0 - 1, and so c = 0 . Thus, we have t = - y - 1 and solving for y gives y(t ) = -t - 1.
SECTION 2.2
Solving the First-Order Linear Differential Equation
87
! 32.
The Tough Made Easy
dy y2 = y dt e - 2ty
We flip both sides of the equation, getting dt e y - 2ty = dy y2 or dt 2 ey + t= 2. dy y y We solve this linear DE for t ( y) getting t ( y) = ey + c . y2
! 33.
A Useful Transformation (a) Letting z = ln y , we have y = ez and dy dz = ez . Now the equation dt dt dy + ay = by ln y dt can be rewritten as ez dz + ae z = bze z . dt
Dividing by e z gives the simple linear equation dz - bz = - a . dt Solving yields z = cebt + and using z = ln y , the solution becomes y(t ) = eaa b f+ ce .
bt
a b
(b)
If a = b = 1, we have y(t ) = eb1+ ce g .
t
Note that when c = 0 we have the constant solution
y = e.
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! 34.
Bernoulli Equation y + p(t ) y = q(t ) y , 0 , 1 (a) Let y = 1 (1- ) , with y = (1- ) . Hence, dy 1 (1- ) d = . dt 1 - dt Plugging everything in the original equation yields
(1- ) d + p(t )1 (1- ) = q(t ) (1- ) . 1 - dt
We can now rewrite d + (1 - ) p(t ) = (1 - )q(t ) , dt which is a linear DE in (t ) . (b)
= 3, p(t ) = -1, and q(t ) = 1; hence
d + 2 = -2 , which has the general solution dt
(t ) = -1 + ce-2t .
Because = 1 , this yields y2
y(t ) = -1 + ce-2t
b
g
-1 2
.
Note, too, that y = 0 satisfies the given equation. (c) When = 0 the Bernoulli equation is dy + p(t ) y = q(t ) , dt which is the general first-order linear equation we solved by the integrating factor method and the Euler-Lagrange method. When = 1 the Bernoulli equation is dy + p(t ) y = q(t ) y dt or dy + ( p(t ) - q(t )) y = 0, dt which can be solved by separation of variables.
SECTION 2.2
Solving the First-Order Linear Differential Equation
89
! 35.
Ricatti Equation y = p(t ) + q(t ) y + r(t ) y 2 (a) Suppose y1 satisfies the DE so that dy1 2 = p(t ) + q(t ) y1 + r(t ) y1 . dt If we define a new variable y = y1 + 1
, then
dy dy1 1 d . = - dt dt 2 dt Substituting for dy1 yields dt
F I H K
dy 1 d 2 . = p(t ) + q(t ) y1 + r(t ) y1 - 2 dt dt Now, if we require, as suggested, that satisfies the linear equation d = - q(t ) + 2r(t ) y1 - r(t ) , dt
F I H K
a
f
then substituting in the previous equation gives dy q(t ) 2r(t ) y1 r (t ) 2 = p(t ) + q(t ) y1 + r (t ) y1 + + + 2 , dt which simplifies to dy 1 y 1 = p t + q t y1 + + r t y12 + 2 1 + 2 = p t + q t y + r t y 2 . dt Hence, y = y1 + equation. (b) y = -1 + 2 y - y 2 Let y1 = 1 so y1 = 0, and substitution in the DE gives
2 0 = -1 + 2 y1 - y1 = -1 + 2 - 1 = 0 .
b g b gFGH
1
IJ b gFG K H
FG IJ H K
IJ b g b g b g K
satisfies the Ricatti equation as well, as long as satisfies its given
Hence, y1 satisfies the given equation. To find and then y, note that p(t ) = -1, q(t ) = 2 , r (t ) = -1. Now find from the assumed requirement that d = -(2 + 2(-1)(1)) - (-1) , dt
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Linearity and Nonlinearity
which reduces to
d = 1. This gives (t ) = t + c , hence dt y(t ) = y1 + 1
= 1+
1 . t +c
! 36.
Computer Visuals (a) (b)
y + 2 y = t
yh (t ) = ce-2t , y p =
1 1 t- 2 4
The general solution is 1 1 y(t ) = yh + y p = ce-2t + t - . 2 4 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, although all solutions are attracted to y p , we would not call y p a steady-state solution because it is neither constant nor periodic; y p as t .
5
y y = 05x - 0.25 . t
5
5
5
37.
(a) (b)
y - y = e3t yh (t ) = cet , y p = 1 3t e . The general solution is 2 1 y(t ) = yh + y p = cet + e3t . 2 yh
1 :
yp
SECTION 2.2
Solving the First-Order Linear Differential Equation 3 y
91
(c)
There is no steady-state solution (including both yh and y p ) 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 to the bottom-most one.
3
3 c = 2 3
t
38.
(a)
y + y = sin t
2
y
6
6
t
c=-0.002
2
(b)
yh (t ) = ce-t , y p =
1 1 sin t - cos t . 2 2
The general solution is 1 1 y(t ) = yh + y p = ce-t + sin t - cos t . 2 2 The curves in the figure in part (a) are labeled for different values of c. (c) The sinusoidal steady-state solution yp = 1 1 sin t - cos t 2 2
occurs when c = 0 . Note that the other solutions approach this solution as t .
92
CHAPTER 2
Linearity and Nonlinearity
39.
(a)
y + y = sin 2t
2
y
6
6
t
2
(b)
1 yh (t ) = ce-t , y p = (sin 2t - 2 cos 2t ) . 5 The general solution is y(t ) = ce- t +
yh
3
sin 2t - 2 cos 2t . !# "## #5 $
yp
(c)
The steady-state solution is y p , which attracts all other solutions. The transient solution is yh .
40.
(a)
y + 2ty = 0
2 c =2 c =1 2 c = 2 2 c =0 c = 1 2 t y
(b)
This equation is homogeneous. The general solution is
yh (t ) = ce-t .
2
(c)
The equation has steady-state solution y = 0 . All solutions tend towards zero as t .
SECTION 2.2
Solving the First-Order Linear Differential Equation
93
41.
(a)
y + 2ty = 1
4 y
The approximate c values corresponding to the curves in the center counted from top to bottom, are {1; 1; 2; 2}
4
4
t
and approximately 50,000 (left curve) and 50,000 (right curve) for the side curves.
4
(b)
yh (t ) = ce-t , y p = e-t
2
2
z
et dt .
2
The general solution is
y(t ) = ce- t + e- t et dt . ! "# # $ yh
2 2 2
;
z
yp
(c)
The steady-state solution is y(t ) = 0, which is not equal to y p . Both yh and y p are transient, but as t , all solutions approach 0.
! 42.
Computer Numerics
y + 2 y = t , y(0) = 1
(a)
. Using step size h = 01 and h = 0.01 and Euler's method, we compute the following
values. In the latter case we print only selected values. Euler's Method t 0 0.1 0.2 0.3 0.4 0.5 (b) y(h = 0.1) 1 0.8 0.65 0.540 0.4620 0.4096 y(h = 0.01) 1.0000 0.8213 0.6845 0.5819 0.5071 0.4552 t 0.6 0.7 0.8 0.9 1 y(h = 0.1) 0.3777 0.3621 0.3597 0.3678 0.3842 y(h = 0.01) 0.4219 0.4039 0.3983 0.4029 0.4158
From Problem 36, we found the general solution of DE to be 1 1 y(t ) = ce -2t + t - . 2 4
94
CHAPTER 2
Linearity and Nonlinearity
Using IC y(0) = 1 yields c =
5 . The solution of the IVP is 4 y(t ) = 5 -2 t 1 1 e + t- , 4 2 4
and to 4 places, we have y(1) = (c) 5 -2 1 1 e + - = 0.4192 . 4 2 4
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) 43. The accuracy of Euler's method can be greatly improved by using a smaller step size.
Sample analysis: y - y = e3t , y(0) = 1, y(1) . Exact solution is y = 05et + 05e3t , so y(1) = 114019090461656 to thirteen decimal places. . . . (a) y(1) 9.5944 for step size h = 01 , . y(1) 11401909375 by Runge-Kutta (correct to six decimal places). . (b) (c) 1 From Problem 24, we found the general solution of the DE to be y = cet + e3t . 2 The accuracy of Euler's method can be greatly improved by using a smaller step size; but it still is not correct to even one decimal place for step size 0.01. y(1) 112020610085 for step size h = 0.01 . (d) MORAL: Euler's method converges ever so slowly to the exact answer--clearly a far smaller step would be necessary to approach the accuracy of the Runge-Kutta method.
SECTION 2.2
Solving the First-Order Linear Differential Equation
95
44.
y + 2ty = 1 , y(0) = 1 (a) Using step size h = 01 and h = 0.01 and Euler's method, we compute the following . values. Euler's Method t 0 0.1 0.2 0.3 0.4 0.5
y(h = 0.1) y( h = 0.01)
t 0.6 0.7 0.8 0.9 1
y(h = 0.1)
y( h = 0.01)
1 1.1 1.178 1.2309 1.2570 1.2564
1.0000 1.0905 1.1578 1.1999 1.2165 1.2084
1.2308 1.1831 1.1175 1.0387 0.9517
1.1780 1.1288 1.0648 0.9905 0.9102
y 1 0.905958 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 - t + e - t Using IC y(0) = 1 , yields c = 1. The solution of the IVP is y(t ) = e- t (1 +
2
bg
bg
2
2
z
e t dt .
2
and so to 10 places, we have y 1 = 0.9059589485 (c) . The error for y(1) using step size h = 01 in Euler's approximation is
ERROR = 0.9517 - 0.9059 = 0.0458 .
bg
z
t
0
et dt ' )
'2
Using step size h = 0.01 , 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 ERROR . less then 0.000001. (d) The accuracy of Euler's method can be greatly improved by using a smaller step size, but the Runge-Kutta method has much better performance because of higher degree of accuracy. ! 45. Direction Fields of Homogeneous Equations (a) (A) is linear homogeneous, (B) is linear nonhomogeneous, (C) is nonlinear.
96
CHAPTER 2
Linearity and Nonlinearity
(b)
If y1 and y2 are solutions of a linear homogeneous equation,
y + p(t ) y = 0 ,
then y1 + p(t ) y1 = 0 , and y2 + p(t ) y2 = 0 . We can add these equations, to get
a y + p(t ) y f + a y + p(t ) y f = 0 .
1 1 2 2
Because this equation can be written in the equivalent form
a y + y f + p(t )a y + y f = 0 ,
1 2 1 2
then y1 + y2 is also a solution of the given equation. (c) The sum of any two solutions follows the direction field only in (A). For the linear homogeneous equation (A) you plot any two solutions y1 and y2 by simply following curves in the direction field, and then add these curves, you will see that the sum y1 + y2 also follows the direction field. However, in equation (B) you can observe a straight line solution, which is y1 = 1 1 t- . 2 4
If you add this to itself you get y1 + y2 = 2 y1 = t -
1 , which clearly does not follow the 2 direction field and hence is not a solution. In equation (C) y1 = 1 is a solution but if you
add it to itself you can see from the direction field that y1 + y2 = 2 is not a solution. ! 46. Recognizing Linear Homogeneous DEs from Direction Fields For (A) and (D): The direction fields appear to represent linear homogeneous DEs because the sum of any two solutions is a solution and a constant times a solution is also a solution. (Just follow the direction elements.) For (B) and (C): These direction fields cannot represent linear homogeneous DEs because the zero function is a solution of linear homogeneous equations, and these direction fields 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 appears that y(t ) -1 is a solution, which cannot be for a linear homogeneous equation. ! 47. Suggested Journal Entry Student Project
SECTION 2.3
Growth and Decay Phenomena
97
2.3
Growth and Decay Phenomena
! 1.
Half-Life (a) The half-life th is the time required for the solution to reach Solving for th , yields kth = - ln2 , or 1 th = - ln 2 . k (b) The solution to y = ky is y(t ) = y0ekt so at time t = t1 , we have y t1 = y0e kt1 = B . Then at t = t1 + th we have y t1 + th = y0e k at1 +th f = y0e kt1 ekth = y0e kt1 e- k (ln 2) k = Be- ln 2 = Belna1 2f = 1 1 y0 . Hence, y0e kt = y0 . 2 2
af
a
f
1 B. 2
! 2.
Doubling Time For doubling time td , we solve y0e ktd = 2 y0 , which yields td = 1 ln 2 . k
! 3.
Interpretation of
1 k
y 1 0.8 0.6 0.4 1/e 0.2 0 1 2 4 t y = y0ekt k = 1
If we examine the value of the decay curve y(t ) = y0ekt we find y=
F 1I = y e a HkK
0
k -1 k
f = y0e-1 = y0(0.3678794...)
y 0. 3 Hence, 1 k
98
CHAPTER 2
Linearity and Nonlinearity
is a crude approximation of the third-life of a decay curve. In other words, if a substance decays and has a decay constant k = -0.02 , and time is measured in years, then the third-life of the 1 2 substance is roughly = 50 years. That is, every 50 years the substance decays by . Note 0.02 3 1 that the curve in the figure falls to of its value in approximately t = 1 unit of time. 3 ! 4. Decay of Materials dQ = kQ has the general solution dt Q(t ) = cekt . Initial condition Q(0) = 100 gives Q(t ) = 100ekt where Q is measured in grams. We also have the initial condition Q(50) = 75, from which we find k= The solution is Q(t ) 100e-0.0058t where t is measured in years. The half-life is th = ! 5. Determining Decay from Half-Life dQ = kQ has the general solution Q(t ) = cekt . With half-life th = 5 hours, the decay constant has dt 1 the value k = - ln 2 -014 . Hence, . 5 Q(t ) = Q0e-0.14t . Calling tt the time it takes to decay to 1 the original amount, we have 10 1 Q0 = Q0e-0.14tt , 10 which we can solve for tt getting tt = 5 ln 10 16.6 hours. ln 2 ln 2 120 years. 0.0058 1 3 ln -0.0058 . 50 4
SECTION 2.3
Growth and Decay Phenomena
99
! 6.
Thorium-234 (a) The general decay curve is Q(t ) = cekt . With the initial condition Q(0) = 1 , we have Q(t ) = e kt . We also are given Q(1) = 08 so e k = 08 , or k = ln(08) -0.22. Hence, we . . . have Q(t ) = e-0.22t where Q is measured in grams and t is measured in weeks. (b) th = - ln 2 ln 2 = 31 weeks . k 0.22 (c) Q(10)e-0.22(10) 0107 grams .
! 7.
Dating Sneferu's Tomb The half-life for Carbon-14 is th = 5600 years, so k- 1 ln 2 -0.000124 . ln 2 = - th 5600
Let tc be the time the wood has been aging, and y0 be the original amount of carbon. Fifty-five percent of the original amount is 055y0 . The length of time the wood has aged satisfies the . equation y0e-0.000124te = 055 y0 . . Solving for tc gives tc = - ! 8. Newspaper Announcement For Carbon-14, k = - ln 2 -0.000124 . If y0 is the initial amount, then the final amount of th y0e5000(-0.000124) = 054 y0 . . In other words, 54% of the original carbon was still present. ! 9. Radium Decay 6400 years is 4 half-lives, therefore 1 6.25% . 24 5600 ln 0.55 4830 years. ln 2
Carbon-14 present after 5000 years will be
100
CHAPTER 2
Linearity and Nonlinearity
! 10.
General Half-Life Equation We are given the two equations Q1 = Q0e kt1 Q2 = Q0e kt2 . If we divide, we get Q1 = ek at1 - t2 f Q2 or k t1 - t2 = ln or k= ln Q1
2
a
f
Q1 Q2
Q
t1 - t2
.
1 Substituting in t h = - ln 2 yields the general half-life of k th = - ! 11. Nuclear Waste We have k=- ln 2 ln 2 =- -0.00268 th 258
bt - t g ln 2 = bt - t g ln 2 .
1 2 Q ln Q1 2 2 1 Q ln Q1 2
and solve for t in y0e-0.00268t = 0.05 y0 . Thus t= ! 12. Bombarding Plutonium We are given k = - ln 2 -4.6209812 . The differential equation for the amount present is . 015 dA = kA + 0.00002 , A(0) = 0 . dt Solving this initial-value problem we get the particular solution A(t ) = cekt - 0.00002 k 258 ln 20 1115 years. , ln 2
SECTION 2.3
Growth and Decay Phenomena
101
where c =
0.00002 -0.000004 . Plugging in these values gives the total amount k
A(t ) 0.000004 1 - e-4.6t
b
g
measured in micrograms. ! 13. Blood Alcohol Levels (a) First, because the initial blood-alcohol level is 0.2%, we have P(0) = 0.2 . After one hour, the level reduces by 10%, therefore, we have P(1) = 0.9 P(0) = 0.9(0.2) . From the decay equation we have P(1) = 0.2ek , hence we have the equation 0.2ek = 0.9(0.2) from which we find k = ln 0.9 -0105. Thus our decay equation is . P(t ) = 0.2e(ln 0.9)t 0.2e-0.105t . (b) The person can legally drive as soon as P(t ) < 01 . Setting . P(t ) = 0.2e-0.105t = 01 . and solving for t, yields t=- ! 14. Exxon Valdez Problem The measured blood-alcohol level was 0.06%, which had been dropping at a rate of 0.015 percentage points per hour for nine hours. This being the case, the captain's initial blood-alcohol level was 0.06 + 9(0.015) = 0195% . . The captain definitely could be liable. ! 15. Sodium Pentathal Elimination The half-life is 10 hours. The decay constant is k=- Ed needs ln 2 - 0.069 . 10 ln 2 6.6 hours. -0105 .
a50 mg kgf(100 kg) = 5000 mg
102
CHAPTER 2
Linearity and Nonlinearity
of pentathal to be anesthetized. This is the minimal amount that can be presented in his bloodstream after three hours. Hence, A(3) = A0e-0.069(3) 0813 A0 = 5000 . . Solving for A0 yields A0 = 6,1557 milligrams or an initial dose of 6.16 grams. . ! 16. Moonlight at High Noon 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 (25) = 085I0 . Assuming exponential decay, I (d ) = I0e kd , we have the equation . I (25) = I0e25k = 085I0 from which we can find k= To find d, we use the equation I0e-0.0065d = from which we determine the depth to be d= ! 17. Tripling Time Here k = ln2 . We can find the tripling time by solving for t in the equation 10 y0e (ln 2) 10 t = 3 y0 , giving 1 ln(300,000) 1940 feet. 0.0065 1 I0 , 300,000 ln 0.85 -0.0065 . 25
F ln 2 I t = ln 3 or H 10 K
t= 10 ln 3 1585 hours. . ln 2
! 18.
Extrapolating the Past If P0 is the initial number of bacteria present (in millions), then we are given P0e6k = 5 and 8 P0e9 k = 8 . Dividing one equation by the other we obtain e3k = , from which we find 5 k= ln 8 5 3 .
SECTION 2.3
Growth and Decay Phenomena
103
Plugging this value into the first equation gives P0e2 lna8 5f = 5 , in which we can solve for P0 = 5e-2 lna8 5f 195 million bacteria. . ! 19. Unrestricted Yeast Growth From Problem 2, we are given k= ln 2 = ln 2 1
with the initial population of P0 = 5 million. The population at time t will be 5e(ln 2)t million, so at t = 4 hours, the population will be 5e ! 20. Unrestricted Bacterial Growth From Problem 2, we are given k = ln2 so the population equation is 12 P(t ) = P0et (ln 2) 12 . In order to have five times the starting value, we require P0et (ln 2) 12 = 5P0 , from which we can find t = 12 ! 21. Growth of Tuberculosis Bacterial We are given the initial number of cells present is P0 = 100 , and that P(1) = 150 (1.5 times 3 larger), then 100 e k (1) = 150 , which yields k = ln . Therefore, the population P(t ) at any time t 2 is P(t ) = 100et lna3 2f 100e0.405t cells. ! 22. Cat and Mouse Problem (a) For the first 10 years, the mouse population simply had exponential growth M (t ) = M0ekt . Because the mouse population doubled to 50,000 in 10 years, the initial population must ln2 . For the first 10 years, the mouse population (in have been 25,000, hence k = 10 thousands) was M (t ) = 25et ln(2) 10 . ln 5 27.9 hours. ln 2
4 ln 2
= 5 16 = 80 million.
104
CHAPTER 2
Linearity and Nonlinearity
Over the next 10 years, the differential equation was
dM = kM - 6 , where M (0) = 50 ; t dt
now measures the number of years after the arrival of the cats. Solving this differential equation yields M (t ) = ce kt + 6 . k 6 . The number of mice (in k
Using the initial condition M (0) = 50 , we find c = 50 - thousands) t years after the arrival of the cats is M (t ) = 50 - where the constant k is given by k =
F H
6 kt 6 e + k k
I K
ln 2 0.069 . Plugging this in we get 10
M (t ) = -37e0.069t + 87 . (b) (c) M (10) = 87 - 37e0.069(10) 132 thousand mice. . Starting with M0 = 50 thousand mice a 10% harvest would give M (t ) = M0ekt where we know M (1) = 0.9 M0 . Hence, we have M0e k (1) = 0.9 M0 , which gives k = ln(0.9) -0105. . Thus the mouse population (in thousands) would decline according to
. . M (t ) = M0e-0105t = 50e-0105t
where t is the number of years after the arrival of the cats. After 10 years, the mouse population would have fallen to M (10) = 50e-0.105(10) 17.5 thousand mice. ! 23. Banker's View of e The amount of money in a bank account that collects compound interest with continuous compounding is given by A(t ) = A0ert 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(10) = $1 e0.10(10) $2.72 .
SECTION 2.3
Growth and Decay Phenomena
105
! 24.
Rule of 70 The doubling time is given in Problem 2 by td = ln 2 0.70 70 = r r 100r
where 100r is an annual interest rate (expressed as a percentage). The rule of 70 makes sense. ! 25. Power of Continuous Compounding The future value of the account will be A(t ) = A0ert , If A0 = $0.50 , r = 0.06 and t = 160 , then the value of the account after 160 years will be A(160) = 05e(0.06)(160) $7382.39 . . ! 26. Credit Card Debt 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 ) = A0ert = $5000e0.1995t . After t = 4 years, she owes
. A(4) = $5000e01995(4) $11,10547 . .
Hence, she pays $11,10547 - $5000 = $6,10574 interest for borrowing this money. . . ! 27. Compound Interest Thwarts Hollywood Stunt The growth rate is A(t ) = A0ert . In this case A0 = 3, r = 0.08 , and t = 320 . Thus, the total bottles of whiskey will be A(320) = 3e(0.08)(320) 393,600,000,000 . That's 393.6 billion bottles of whiskey! ! 28. It Ain't Like It Use to Be The growth rate is A(t ) = A0ert , where A0 = 1, t = 50 , and A(50) = 18 (using thousands of dolln 18 0.0578 , or 5.78%. lars). Hence, we have 18 = e50r , from which we can find r = 50
106
CHAPTER 2
Linearity and Nonlinearity
! 29.
How to Become a Millionaire (a) From Equation (11) we see that the equation for the amount of money is A(t ) = A0ert + d rt e -1 . r
b
g
In this case, A0 = 0 , r = 0.08 , and d = $1000. The solution becomes A(t ) = (b) A(40) = $1,000,000 = deposit d = $3399.55. (c) 2500 40r e - 1 . To solve this equation for t we require a computer. r Using Maple, we find the interest rates r = 0.090374 ( 9.04% ) . You can confirm this A(40) = $1,000,000 = result using direct substitution. ! 30. Living Off Your Money A(t ) = A0ert - d rt e - 1 . Setting A(t ) = 0 and solving for t gives r d 1 t = ln r d - rA0 1000 0.08t e -1 . 0.08
b
g
d e0.08(40) - 1 . Solving for d, we get that the surviving annual 0.08
b
g
b
g
b
g
FG H
IJ K
Notice that when d = rA0 this equation is undefined, as we have division by 0; if d < rA0 , this equation is undefined because we have a negative logarithm. For the physical translation of these facts, you must return to the equation for A(t ) . If d = rA0 , you are only withdrawing the interest, and the amount of money in the bank remains constant. If d < rA0 , then you aren't even withdrawing the interest, and the amount in the bank increases and A(t ) never equals zero. ! 31. How Sweet It Is From the equation of Problem 30, we have A(t ) = $1,000,000e0.08t - Setting A(t ) = 0 , and solving for t, we have t= the time that the money will last. ln 5 201 years, . 0.08 100,000 0.08t e -1 . 0.08
b
g
SECTION 2.3
Growth and Decay Phenomena
107
! 32.
The Real Value of the Lottery Following the hint, we let A = 010 A - 50,000 . . Solving this equation with initial condition A(0) = A0 yields A(t ) = A0 - 500,000 e0.10t + 500,000 . Setting A(20) = 0 and solving for A0 we get A0 =
a
f
500,000 e2 - 1 $432,332 . e2
b
g
! 33.
Continuous Compounding (a) After one year compounded continuously the value of the account will be S (1) = S0er . With r = 0.08 (8%) interest rate, we have the value S (1) = S0e0.08 $1.0833S0 . This is equivalent to an annual compound of 8.33%. (b) If we set the annual yield from a single compounding with interest reff , S0 1 + reff equal to the annual yield from continuous compounding with interest r, S0e , we have S0 1 + reff = S0er . Solving for reff yields reff = er - 1. (c) rdaily = 1 +
r
a
f
a
f
FG H
0.008 365
IJ K
365
- 1 = 0.083277539 (i.e., 8.328%) effective annual interest rate
! 34. ! 35.
Good Test Equation for Computer or Calculator Student Project. Your Financial Future We can write annuity equation as
A' = 0.08 * ( A + 5000) , A(0) = 0 .
The exact solution is A = 5000(e0.08t - 1)
108
CHAPTER 2
Linearity and Nonlinearity
and this formula would represent the value of the annuity for the first 20 years (we plot the solution for t = 0 , 1, 2, ... 20). We have also rounded the solution to thousands. Annuity Problem t 1 2 3 4 5 6 7 8 9 10 A(t) $416,435 $867,554 $1356,246 $1885,639 $2459,124 $3080,372 $3753,362 $4482,404 $5272,1661 $6127,705 t 11 12 13 14 15 16 17 18 19 20 A(t) $7054,498 $8058,482 $9146,085 $10324,271 $11600,585 $12983,199 $14480,966 $16103,479 $17861,126 $19765,162
After 20 years at 8% the account has grown to $19765,162. Experiment will show that the interest rate is more important than the annual deposit. ! 36. Suggested Journal Entry Student Project
SECTION 2.4
Linear Models: Mixing and Cooling
109
2.4
! 1.
Linear Models: Mixing and Cooling
Mixing Details Separating variables, we find dx 2 = dt x t - 100 from which we get
ln x = 2 ln t - 100 + c .
We can solve for x(t ) using properties of the logarithm, getting
x = ece2 ln t -100 = Celn(t -100) = C(t - 100)2
2
where C = ec > 0 is an arbitrary positive constant. Hence, the final solution is
x(t ) = C(t - 100)2 = c1(t - 100)2
where c1 is an arbitrary constant. ! 2. English Brine (a) Salt inflow is
a2 lbs galfa3 gal minf = 6 lbs min .
Salt outflow is
F Q lbs galI a3 gal minf = Q lbs min . H 300 K 100
The differential equation for Q(t ) , the amount of salt in the tank, is dQ = 6 - 0.01Q . dt Solving this equation with initial condition Q(0) = 50 yields Q(t ) = 600 - 550e-0.01t . (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 ) = (c) Q(t ) 11 = 2 - e-0.01t . 300 6
As t , e-0.01t 0 . Hence Q(t ) 600 lbs of salt in the tank.
110
CHAPTER 2
Linearity and Nonlinearity
(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. 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.
Q(t ) 600 500 400 300 200 100 0 100 200 300 400 t 0 100 200 300 400 t 1 conc (t ) 2
(e)
Number of lbs of salt in the tank ! 3. Metric Brine (a) The salt inflow is
Concentration of salt in the tank
. a01 kg literfa4 liters minf = 0.4 kg min . The outflow is 4 Q kg min . Thus, the differential equation for the amount of salt is 100 dQ = 0.4 - 0.04Q . dt Solving this equation with the given initial condition Q(0) = 50 gives Q(t ) = 10 + 40e-0.04t . (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 t = (c) (d) . b g Qbt g = 01 - 0.4e 100
-0.04 t
.
As t , e-0.04t 0 . Hence Q(t ) 10 kg of salt in the tank. 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. .
SECTION 2.4
Linear Models: Mixing and Cooling
111
! 4.
Salty Goal The salt inflow is given by
a2 lb galfa3 gal minf = 6 lbs min .
The outflow is 3 Q . Thus, 20 dQ 3 = 6- Q . dt 20 Solving this equation with the given initial condition Q(0) = 5 yields the amount Q(t ) = 40 - 35e-3t 20 . To determine how long this process should continue in order to raise the amount of salt in the tank to 25 lbs, we set Q(t ) = 25 and solve for t to get t= ! 5. Mysterious Brine Input in lbs min is 2x (where x is the unknown concentration of the brine). Output is 2 Q lbs min . 100 The differential equation is given by dQ = 2 x - 0.01Q , dt which has the general solution Q(t ) = 200 x + ce-0.01t . Because the tank had no salt initially, Q(0) = 0 , which yields c = -200 x . Hence, the amount of salt in the tank at time t is
Q(t ) = 200 x 1 - e-0.01t .
20 7 ln 56 minutes. . 3 3
b
g
We are given that Q(120) = (14)(200) = 280 , . from which we solve for x, to get x 2.0 lb gal.
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! 6.
Correcting a Goof Input in lbs min is 0 (she's not adding any salt). Output is 0.03Q lbs min . The differential equation is dQ = -0.03Q , dt which has the general solution Q(t ) = ce-0.03t . Using the initial condition Q(0) = 20, we get the particular solution Q(t ) = 20e-0.03t . Because she wants to reduce the amount of salt in the tank to 10 lbs, we set Q(t ) = 10 = 20e-0.03t . Solving for t, we get t= 100 ln 2 23 minutes. 3
! 7.
Cleaning Up Lake Erie (a) The inflow of pollutant is
b40 mi yrg(0.01%) = 0.004 mi
3
3
yr ,
and the outflow is mi b40 mi yrg V (t ) mi 100
3 3 3
= 0.4V (t ) mi3 yr .
Thus, the DE is dV = 0.004 - 0.4V dt with the initial condition
V (0) = (0.05% ) 100 mi3 = 0.05 mi3 .
b
g
(b)
Solving the IVP in part (a) we get the expression V (t ) = 0.01 + 0.04e-0.4t where V is the volume of the pollutant in cubic miles.
SECTION 2.4
Linear Models: Mixing and Cooling
113
(c)
A pollutant concentration of 0.02% corresponds to
0.02% 100 mi3 = 0.02 mi3
b
g
of pollutant. Finally, setting V (t ) = 0.02 gives the equation 0.02 = 0.01 + 0.04e -0.4 t , which yields
t = 2.5 ln(4) 35 years. .
! 8.
Cascading Tanks (a) The inflow of salt into tank A is zero because fresh water is added. The outflow of salt is
F Q lbs galI a2 gal minf = 1 Q H 100 K 50
A
A
lb min .
Tank A initially has a total of 05(100) = 50 pounds of salt, so the initial-value problem is . dQA Q = - A , QA (0) = 50 lbs. 50 dt (b) Solving for QA gives QA (t ) = ce-t 50 and with the initial condition QA (0) = 50 gives QA (t ) = 50e-t 50 . (c) The input to the second tank is
A
F Q lb galI a2 gal minf = 1 Q H 100 K 50
B
A
lb min = e-t 50 lb min .
The output from tank B is
F Q lb galI a2 gal minf = 1 Q H 100 K 50
Thus the differential equation for tank B is dQB 1 = e- t 50 - QB dt 50 with initial condition QB (0) = 0 . (d) Solving the initial-value problem in (c), we get
B
lbs min .
QB (t ) = te- t 50 pounds.
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! 9.
More Cascading Tanks (a) Input to the first tank is
a0 gal alch galfa1 gal minf = 0 gal alch min.
Output is 1 x0 gal alch min . 2 The tank initially contains 1 gallon of alcohol, or x0(0) = 1 . Thus, the differential equation is given by dx0 1 = - x0 . dt 2 Solving, we get x0(t ) = ce- t 2 . Plugging in x0(0) , we get c = 1 , so the first tank's alcohol content is x0(t ) = e- t 2 . (b) The first step of a proof by induction is to check the initial case. In our case we check n = 0 . For n = 0 , t 0 = 1, 0! = 1 , 20 = 1, and hence the given equation yields x0(t ) = e- t 2 . This result was found in part (a). The second part of an induction proof is to show that the statement holds for case n, and then prove the statement holds for case n + 1. Hence, we assume xn (t ) = t ne - t 2 , n! 2n
which means the concentration flowing into the next tank will be
xn (because the 2 1 1 volume is 2 gallons). The input of the next tank is x n and the output xn+1(t ) . The 2 2 differential equation for the (n + 1) tank will be dx n +1 1 t n e -t 2 + x n +1 = , xn+1(0) = 0. dt 2 n !2 n +1
Solving this IVP, we find xn +1(t ) = t n +1e- t 2 (n + 1)!2n +1
which is what we needed to verify. The induction step is complete.
SECTION 2.4
Linear Models: Mixing and Cooling
115
(c)
To find the maximum of xn (t ) , we take its derivative, getting xn =
b g
t n -1e - t 2 t ne-t 2 - . n n - 1 !2 n !2 n +1
Setting this value to zero, the equation reduces to 2nt n -1 - t n = 0 , and thus has roots
t = 0 , 2n. When t = 0 the function is a minimum, but when t = 2n , we apply the first
derivative test and conclude it is a local maximum point. Plugging this into xn (t ) yields the maximum value xn (2n) Mn = (2n)n e-n nne- n . = n !2 n n!
We can also see that xn (t ) approaches 0 as t and so we can be sure this point is a global maximum of xn (t ) . (d) Direct substitution. Stirling's approximation for n! into the formula for M n in part (c) gives M n 2 n ! 10. Drug Metabolism The drug concentration C(t ) satisfies dC = a - bC dt where a and b are constants, with C(0) = 0 . Solving this IVP gives C(t ) = a 1 - e-bt . b
b g
-1/ 2
.
b
g
As t , we have e-bt 0 (as long as b is positive), so the limiting concentration of C(t ) is a . Notice that b must be positive or for large t we would have C(t ) < 0, which makes no sense, b a because C(t ) is the amount in the body. To reach one-half of the limiting amount of we set b a a = 1 - e-bt 2b b and solve for t, getting t = ! 11. ln 2 . b
b
g
Another Solution Method Separating variables, we get dT = - kdt . T-M
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Solving this equation yields ln T - M = - kt + c or T - M = ece- kt . Eliminating the absolute value, we can write T - M = ece- kt = Ce- kt where C is an arbitrary constant. Hence, we have T (t ) = M + Ce- kt . Finally, using the condition T (0) = T0 gives T (t ) = M + T0 - M e- kt . ! 12. Still Another Approach If y(t ) = T (t ) - M , then dy dT = , and dt dt T (t ) = y(t ) + M . Hence the equation becomes dy = -k( y + M - M ) dt or dy = - ky , dt a decay equation. ! 13. Using the Time Constant (a)
T (t ) = T0e- kt + M 1 - e- kt , from Equation (8). In this case, M = 95 , T0 = 75, and k =
a
f
b
g
1 , 4
yielding the expression
T (t ) = 75e-t 4 + 95 1 - e- t 4
b
g
where t is time measured in hours. Plugging in t = 2 in this case (2 hours after noon), yields T (2) 82.9 F.
SECTION 2.4
Linear Models: Mixing and Cooling
117
(b)
Setting T (t ) = 80 and simplifying for T (t ) yields t = -4 ln which translates to 1:09 P.M. 3 115 hours, . 4
! 14.
A Chilling Thought (a)
T (t ) = T0e- kt + M 1 - e- kt , from Equation (8). In this problem, T0 = 75, M = 10 , and 1 T = 50 (taking time to be in hours). Thus, we have the equation 50 = 10 + 60e- k 2 , 2
FI HK
b
g
from which we can find the rate k constant = -2 ln 2 081 . . 3
After one hour, the temperature will have fallen to 4 2 T (1) 10 + 60e2 lna2 3f = 10 + 60 = 36 36.7 F. 9 3 (b) Setting T (t ) = 15 gives the equation 15 = 10 + 60e2t lna2 3f . Solving for t gives t=- ! 15. Temperature in the Fridge
T (t ) = T0e- kt + M 1 - e- kt . Taking t in hours, we measure T
ln 12 306 hours (3 hrs, 3.6 min). . 2 ln 2 3
bg
b
g
F 1I H 2K
and T (1). We then subtract M
from each value and divide, getting the equation T1 -M = e- k 2 T 1 -M 2 from which we can find k = -2 ln
bg bg
T1 -M . T 1 -M 2
bg bg
Once we have k, we solve for T0 in the equation for T (1) or T T0 = (T (1) - M )e k + M .
F 1I , getting the value H 2K
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Finally setting M = 70 , T freezing!). ! 16. Warm or Cold Beer? Again, we use
F 1I = 45 , and T(1) = 55, we get k = 1022 . Hence T 28.3 F (below . H 2K
0
T (t ) = M + T0 - M e- kt . In this case, M = 70 , T0 = 35. If we measure t in minutes, we have T (10) = 40 , giving 40 = 70 - 35e-10k . Solving for the decay constant k, we find k=- ln 10
a
f
b g 0.0154 .
6 7
Thus, the equation for the temperature after t minutes is T (t ) 70 - 35e-0.0154t . Substituting t = 20 gives T (20) 44.3 F. ! 17. Case of the Cooling Corpse
T (t ) = T0e- kt + M 1 - e- kt . We know that M = 50 and T0 = 98.6 F. The first measurement takes
b
g
place at unknown time t1 so T t1 = 70 = 50 + 48.6e- kt1 or 48.6e- kt1 = 20 . The second measurement is taken two hours later at t1 + 2 , yielding 60 = 50 + 48.6e- k at1 +2f or 48.6e- k at1 +2f = 10 . Dividing the second equation by the first equation gives the relationship 1 ln2 e-2 k = from which k = . Using this value for k the equation for T t1 gives 2 2
af
af
70 = 50 + 48.6e- t ln 2 2 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. ! 18. Professor Farlow's Coffee
T (t ) = T0e- kt + M 1 - e- kt . For this problem, M = 70 and T0 = 200 F. The equation for the
b
g
coffee temperature is T (t ) = 70 + 130e- kt .
SECTION 2.4
Linear Models: Mixing and Cooling
119
Measuring t in hours, we are given T so the rate constant is k = -4 ln Hence T (t ) = 70 + 130e-3.8t . Finally, setting T (t ) = 90 yields
. 90 = 70 + 130e-38t ,
F 1I = 120 = 70 + 130e H 4K
5 38 . . 13
-k 4
;
from which we find t 0.49 hours, or 29 minutes and 24 seconds. ! 19. The Coffee and Cream Problem The basic law of heat transfer states that if two substances at different temperatures are mixed together, then the heat (calories) lost by the hotter substance is equal to the heat gained by the cooler substance. The equation expressing this law is M1S1 t1 = M2 S2 t2 where M1 and M2 are the masses of the substances; S1 and S2 are the specific heats, and t1 and 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) is the same as the specific heat of cream. Defining C(0) = initial temperature of the coffee R = room temperature (temp of the cream) T = temperature of the coffee after the cream is added we have M1(C(0) - T ) = M2 (T - R). If we assume the mass of the cream is effect the answer), we have 10(C(0) - T ) = T - R . 1 the mass Mg of the coffee (the exact fraction does not 10
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The temperature of the coffee after John initially adds the cream is T= 10C(0) + R . 11
After that John and Maria's coffee cools according to the basic law of cooling, or John:
F 10C(0) + R I e H 11 K
kt ,
Maria: C(0)e kt
where we measure t in minutes. At time t = 10 the two coffees will have temperature John:
F 10C(0) + R I e H 11 K
10k
, Maria: C(0)e10k .
Maria then adds the same amount of cream to her coffee, which means John and Maria's coffees now have temperature John:
F 10C(0) + R I e H 11 K
10k
, Maria:
FG 10C(0)e H 11
10k
+R
IJ . K
Multiplying each of these temperatures by 11, subtracting 10C(0)e10k and using the fact that Re10k > R , we conclude that John drinks the hotter coffee. ! 20. A Real Mystery
T (t ) = T0e- kt + M 1 - e- kt
b
g
T (t ) = 40 + 30e- kt .
While the can is in the refrigerator T0 = 70 and M = 40 , yielding the equation
Measuring time in minutes, we have T (15) = 40 + 30e-15k = 60 , which gives k = - ln
2 3
15
0.027. Letting t1 denote the time the can was removed from the
refrigerator, we know that the temperature at that time is given by T t1 = 40 + 30e- kt1 , which would be the W0 for the warming equation W (t ) , the temperature after the can is removed from the refrigerator
W t = 70 + W0 - 70 e - kt
af
bg
b
g
(the k of the can doesn't change). Plugging in W0 where t = t1 and simplifying, we have
W (t ) = 70 + 30 e- kt1 - 1 e- kt .
b
g
SECTION 2.4
Linear Models: Mixing and Cooling
121
The initial time for this equation is t1 (the time the can was taken out of the refrigerator), so the time at 2 P.M. will be 60 - t1 minutes yielding the equation in t1 : W 60 - t1 = 60 = 70 + 30 e- kt1 - 1 e- k a60-t1 f . This simplifies to 1 = e-60k - e- k a60- t1 f , 3 which is relatively easy to solve for t1 (knowing that k 0.027 ). The solution is t1 37 ; hence the can was removed from the refrigerator at 1:37 P.M. ! 21. Computer Mixing y + 1 y = 2, y(0) = 0 1+ t 22. y + 1 y = 2, y(0) = 0 1- t
a
f
b
g
When the inflow is greater than the outflow, the amount of dissolved substance keeps growing without end.
y 4
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 is also when the tank is emptied.
1 y
0
0
3
t
0 0 1
t
! 23.
Suggested Journal Entry Student Project
122
CHAPTER 2
Linearity and Nonlinearity
2.5
!
Nonlinear Models: Logistic Equation
Equilibrium Points
Note; Problems 16 are all autonomous equations, so lines of constant slope (isoclines) are horizontal lines. 1. y = ay + by 2 , (a > 0, b > 0) We find the equilibrium points by solving y = ay + by 2 = 0, a getting y = 0 , - . By inspecting b y = y(a + by) , we see that solutions have positive slope ( y > 0) when y > 0 or y < - ( y < 0) for - a and negative slope b
a < y < 0 . Hence, the equilibrium solution y(t ) 0 is unstable, and the equilibrium b a solution y(t ) - is stable. b
y
y= 0 y = a/b
unstable equilibrium t
stable equilibrium
2.
y = ay - by 2 , (a > 0, b > 0) We find the equilibrium points by solving y = ay - by 2 = 0, getting y = 0 , a . By inspecting b 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 > ( y > 0) for 0 < y < solution y(t )
a and positive slope b
a . Hence, the equilibrium solution y(t ) 0 is unstable, and the equilibrium b
a is stable. b
y stable equilibrium t unstable equilibrium
y = a/b y= 0
3.
y = -ay + by 2 , (a > 0, b > 0) We find the equilibrium points by solving y = -ay + by 2 = 0, getting y = 0 , a . By inspecting b y = y(-a + by) , we see that solutions have positive slope when y < 0 or y > a a and negative slope for 0 < y < . b b a Hence, the equilibrium solution y(t ) 0 is stable, and the equilibrium solution y(t ) is b
y unstable equilibrium t stable equilibrium
unstable.
y = a/b y= 0
124
CHAPTER 2
Linearity and Nonlinearity y
4.
y = -ay - by 2 , (a > 0, b > 0) We find the equilibrium points by solving y = -ay - by 2 = 0, a getting y = 0 , - . By inspecting b y = - y(a + by) , we see that solutions have negative slope when y > 0 or y < - -
y= 0 y = a/b stable equilibrium
t unstable equilibrium
a and positive slope for b
a < y < 0 . Hence, the equilibrium solution y(t ) 0 is stable, and the equilibrium solution b a y(t ) - is unstable. b
y
5.
y = e y - 1 Solving for y in the equation y = e y - 1 = 0, we get y = 0 , hence we have one equilibrium (constant) solution y(t ) 0. Also y > 0 for y positive, and y < 0 for y negative. This says that y(t ) 0 is an unstable equilibrium point.
y= 0
t unstable equilibrium
6.
y = y - y
2 y= 1 y= 0 2 stable equilibrium
y unstable equilibrium t
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 defined when y 0 .
2
DE not defined 2
SECTION 2.5
Nonlinear Models: Logistic Equation
125
!
Nonautonomous Sketching
For nonautonomous equations, the lines of constant slope are not horizontal lines as they were in the autonomous equations in Problems 16. 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 analysis of the sign y . As can be seen from this figure, the equilibrium y 0 is stable at t > 0 and unstable at t < 0 .
4 y Isocline of slope 0. y=t
Isocline. of slope 0. y=0 4
4
t
4
126
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Linearity and Nonlinearity
8.
y = ( y - t )2 In this equation we observe that y = 0 when y = t . We can draw isoclines y - t = c and elements with 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 y y = t +1
2 unstable solution stable solution 2
2
t
9.
y = sin( yt )
5
y
Isoclines of horizontal slopes (dashed) are hyperbolas yt = n for n = 0, 1, 2, ... . On the computer drawn graph you can sketch the hyperbolas for isoclines and verify the alternating occurrence of positive and negative slopes between them as specified by the DE. Only y = 0 is an equilibrium (unstable for t < 0 , stable for t > 0 ). ! 10. Inflection Points y = r 1 -
5 y=0 stable solution t
FG H
y y L dy from the DE. dt
IJ K
We differentiate with respect to t (using the chain rule), and then substitute for This gives
d 2 y d dy 1 dy y dy 2ry y = = ry - + r 1- = - + r ry 1 - . 2 dt dt L dt L dt L L dt Setting d2y L . Values y = 0 and y = L are equilibrium = 0 and solving for y yields y = 0 , L, 2 dt 2 L points; y = is an inflection point. See text Figure 2.5.8. 2
FG IJ H K
FG IJ H K
FG H
IJ K
FG H
IJ FG K H
IJ K
SECTION 2.5
Nonlinear Models: Logistic Equation
127
11.
y = -r 1 -
FG H
y y T dy from the DE. dt
IJ K
We differentiate with respect to t (using the chain rule), and then substitute for This gives
d 2 y d dy 1 dy y dy 2ry y y. = = - ry - - r 1- =- - r r 1- T dt T dt T T dt 2 dt dt Setting d2y T . Values y = 0 and y = T are equilibrium = 0 and solving for y yields y = 0 , T, 2 dt 2 T points; y = is an inflection point. See text Figure 2.5.9. 2 y = cos( y - t ) We differentiate y with respect to t (using the chain rule), and then substitute for DE. This gives d 2 y d dy dy = = - sin y - t = - sin y - t cos y - t . 2 dt dt dt dt dy from the dt
FG IJ H K
FG IJ H K
FG H
IJ K
FG H
IJ FG KH
IJ K
12.
FG IJ H K
b g
b g b g
Setting
d2y n = 0 and solving for y yields y - t = n , y - t = + n for n = 0, 1, 2, ... . 2 2 dt
Note the inflection points change with t in this nonautonomous case. See text Figure 3.5.3, graph for (2) this section to see that the inflection points occur only when y = -1 , so they lie along the lines y = t + m where m is an odd integer. ! 13. Logistic Equation y = r 1 - (a)
F H
y y L
I K
We rewrite the logistic DE by separation of variables and partial fractions to obtain
F 1 + I dy = rdt . GH y 1 - JK
1 L y L
Integrating gives ln y - ln 1 - y = rt + c . L
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CHAPTER 2
Linearity and Nonlinearity
If y0 > L , we know by qualitative analysis (see text Figure 2.5.8) that y > L for all y y future time. Thus ln 1 - = ln - 1 in this case, and the implicit solution (8) becomes L L
F H
I K
y
y L
-1
= Cert , with C =
y0 L
y0 -1
.
Substitution of this new value of C and solving for y gives y0 > L gives y(t ) = 1+
e
L
L y0
- 1 e-rt
j
which turns out (surprise!) to match (10) for y0 < L . You must show the algebraic steps to confirm this fact. (b) The derivation of formula (10) required by 1 - y , which is undefined if y = L . Thus, L 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 y = r 1 - that case to (c)
F H
y y is fine if y L and reduces in L
I K
dy = 0 , so y = a constant (which must be L if y(0) = L ). dt
The solution formula (10) states y(t ) = 1+
e
L
L y0
- 1 e-rt
j
.
If 0 < y0 < L , the denominator is greater than 1 and as t increases, y(t ) approaches L from below. If y0 > L , the denominator is less than 1 and as t increases, y(t ) approaches L from above. If y0 = L , 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 ) = 1+
e
L
L y0
- 1 e-rt
j
(10)
SECTION 2.5
Nonlinear Models: Logistic Equation
129
gives (with the help of Maple) y = Setting y = 0 , we get 2 or L
e
L y0
- 1 r 2e-rt 2
j
oe
L y0
- 1 e-rt - 1 +
L y0
j
e
1+
e
- 1 e-rt
j
3
L y0
- 1 e-rt
j t.
FG L - 1IJ e Hy K
0
- rt
- 1+
LM FG L - 1IJ e OP = 0 N Hy K Q
- rt 0 - rt
Solving for t, we get t *
FG L - 1IJ e = 1 . Hy K 1 F L I = lnG - 1J . Plugging this value into the analytical solution K r Hy
0 0
for the logistic equation, we get y(t *) = At t * the rate y is
L . 2
r 1-
! 14. Fitting the Logistic Law The logistic equation is
FG H
IJ F L I = r F L I = rL . LKH 2 K 2 H 2 K 4
2
L
y = ry 1 -
F H
y . L
I K
If initially the population doubles every hour, we have td = which gives the growth rate r = ln2 =1 r
1 14 . We are also given L = 5 109 . The logistic curve after . ln 2 5 109 5 109 = 4.9 109 . 9 1 + 510 - 1 e-1.4(4) 1 + 4e-5.6 109
4 hrs is calculated from the analytic solution formula, y(t ) = 1+
e
L
L y0
- 1 e-rt
j
=
d
i
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CHAPTER 2
Linearity and Nonlinearity
! 15.
Logistic Model with Harvesting y = ry 1 - (a)
FG H
y -h t L
IJ b g K
y' 0.2 r =1 L =1
Graphs of y versus y for different values of harvesting h are shown. Feasible harvests are those values of h that keep the slope y positive for some 0< y < L. Because the curve y versus y is always a maximum at y= L , 2
0.2
0.4
0.6
0.8
h=0 1 y h = 0.1
0.2
y = y(1 - y ) - h
h = 0.2 h = 0.25
we find the value of h that gives y
hmax harvesting. By setting y
F L I = 0 , we find h = rL . H 2K 4 F H
y y-h L
F L I = 0 ; this will be the maximum sustainable value H 2K
(b)
As a preliminary to graphing, we find the equilibrium solutions under harvesting by setting y = 0 in the equation y = r 1 - getting r 1- Solving for y, we get y1 , y2 = L L2 - 2
I K
F H
y y - h = 0 or y 2 - rLy + hL = 0 . L
I K
b g
2 hL 2 r
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 equilibrium solution. As we say in part (a), harvesting (fish/day or some similar unit) must satisfy rL h< . 2
SECTION 2.5 y 2 y
Nonlinear Models: Logistic Equation
131
stable equilibrium
y= 1 semistable equilibrium y = 0.5 t
unstable equilibrium 0 2
y= 0 t
0
0
2
Straight logistic
y = y(1 - y)
Logistic with maximum 1 sustainable harvesting
y = y(1 - y) - 0.25
Note that the equilibrium value with harvesting h = 0.25 is lower than the equilibrium value without harvesting. Note further that maximum harvesting has changed the phase line and the direction of solutions below equilibrium. The harvesting graph implies that fishing is fine when the population is above equilibrium, but wipes out the population when it is below equilibrium. ! 16. Semistable Equilibrium
y = (1 - y)2
2 y
We draw upward arrows on the y-axis for y 1 to indicate the solution is increasing. When y =1
1
semistable equilibrium
y= 1
6
t
we have a constant solution. Because the slope lines have positive slope both above and below the constant solution y(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 value below 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 customarily noted with half-filled circles.
132
CHAPTER 2
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! 17.
Gompertz Equation (a)
dy = y(1 - b ln y) dt
Letting z = ln y and using the chain rule we get dz dz = dt dy Hence, the Gompertz equation becomes dz = a - bz . dt
FG H
IJ F dy I = FG 1 IJ F dy I . K H dt K H y K H dt K
(b)
Solving this DE for z we find z(t ) = Ce-bt + Substituting back z = ln y gives y(t ) = ea beCe
- bt
a . b
. a . b
Using the initial condition y(0) = y0 , we finally get C = ln y0 - (c) ! 18.
t
From the solution in part (b), lim y(t ) = ea b when b > 0 , y(t ) when b < 0 .
Fitting the Gompertz Law (a) From Problem 17, y(t ) = ea bece where c = ln y0 -
- bt
a . In this case y(0) = 1 , y(2) = 2 . We note y(24) y(28) 10 means b
the limiting value ea b has been reached. Thus ea b = 10 , so a = ln 10 2.3. b The constant c = ln 1 - a = 0 - 2.3 = -2.3 . Hence, b y(t ) = 10e-2.3e and y(2) = 10e-2.3e
-2 b - bt
=2.
SECTION 2.5
Nonlinear Models: Logistic Equation
133
Solving for b: -2.3e-2b = ln 2 -1609 . 10 1609 . 0.6998 e-2b - 2.3 -2b = ln(0.6998) -0.357 . b 01785
and a = 2.3b gives a 0.4105 . (b) The logistic equation y = ry 1 - has solution y(t ) =
y
F H
y L
I K
.
- rt
10
logistic Gompertz
y = 10
FL I 1 + G - 1J e Hy K
0
L
0
y(2) = 2 y(0) = 1 12
24 t
We have L = 10 and y0 = 1 , so y(t ) = and y(2) = Solving for r 9e -2 r = 10 -1 = 4 2 4 -2r = ln -0.8109 9 -0.8109 r= 0.405. -2 10 =2. 1 + 9e-2r 10 1 + 9e-rt
134
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Linearity and Nonlinearity
! 19.
Autonomous Analysis (a) (b) y = y2 One semistable equilibrium at y(t ) 0 is stable from below, unstable from above.
4 semistable equilibrium t 4 4 y
4
20.
(a) (b)
y = - y(1 - y)
The equilibrium solutions are y(t ) 0, y(t ) 1 . The solution y(t ) 0 is stable. The solution y(t ) 1 is unstable.
4 y
y=1 y=0
unstable equilibrium 4 stable equilibrium t 4
4
21.
(a) (b)
y = - y 1 -
F H
y y 1- , y = - y(1 - y)(1 - 0.5 y) L M
IF KH
I K
The equilibrium points are y = 0 , L, M. y = 0 is stable. y = M is stable if M > L and unstable if M < L . y = L is stable if M < L and unstable if M > L .
y
stable equilibrium unstable equilibrium stable equilibrium
y=M y=L y=0 t
SECTION 2.5
Nonlinear Models: Logistic Equation 4 y
135
22.
(a)
y = y - y Note that the DE is only defined for y 0.
y=1 y=0
unstable equilibrium 2 stable equilibrium 2 t
(b)
The constant solution y(t ) 0 is stable, the solution y(t ) 1 is unstable.
4
23.
(a) (b)
y = k (1 - y )2 , k > 0
The constant solution y(t ) 1 is semi-stable (unstable above, stable below).
4 y
y=1 4
semistable equilibrium t 4
4
24.
(a) (b)
y = y 2 4 - y2
b
g
y stable equilibrium semistable equilibrium t 2 unstable equilibrium 4
The equilibrium solution y(t ) 2 is stable, the solution y(t ) -2 is unstable and the solution y(t ) 0 is semistable.
4 y=2 y=0 2 y = 2
136
CHAPTER 2
Linearity and Nonlinearity
! 25.
Hubbert Peak (a) From even a hand-sketched logistic curve you can graph its slope y and find a roughly bell-shaped curve for y (t ) . Depending on the scales used, it may be steeper or flatter than the bell curve shown in Fig. 1.3.5.
1 inflection point y' y, y' 2 y
0
t
(b)
For a pure logistic curve, the inflection point always occurs when y =
L . However, if 2
we consider models different from the logistic model that still show similar solutions between 0 and the nonzero equilibrium, it is possible for the inflection point to be closer to 0. When this happens oil recovery reaches the maximum production rate much earlier. Of course the logistic model is a crude model of oil production. For example it doesn't take into consideration the fact that when oil prices are high, many oil wells are placed into production. If the inflection point is lower than halfway on an approximately logistic curve, the peak on the y curve occurs sooner and lower creating an asymmetric curve for y . 0 Lower inflection point inflection point y' 1 y y = 0.5 t 2 y, y'
SECTION 2.5
Nonlinear Models: Logistic Equation
137
(c)
These differences may or may not be significant to people studying oil production; it depends on what they are looking for. The long-term behavior, however, is the same; the peak just occurs sooner. After the peak occurs, if the model holds, it is downhill insofar as oil production is concerned. Typical skew of peak position is presented on these figures:
10 y 4 y
4
t 10
4
t 10
Typical curves with different inflection points ! 26. Stefan's Law Again T = k M 4 - T4
y curves for corresponding ones
b
g
0<T < M ,
y
The equation tells us that when
stable equilibrium y=3
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 of Stefan's equation for M = 3, k = 0.05 .
t
0
2
dT = 0.05 34 - T 4 dt
b
g
To > M gives solutions falling to M. To < M gives solutions rising to M. The one actions match intuition and experiment.
138
CHAPTER 2
Linearity and Nonlinearity
! 27.
Chemical Reactions
x = k (100 - x)(50 - x)
200 y
The solutions for the given initial conditions are shown on the graph. Note that all behaviors are at equilibrium or flown off scale before t = 0.3 ! Conclusions: Any causes x(t ) to increase without bound. On the other hand, for any x(0) (0,100)
0 0 unstable equilibrium
stable equilibrium t 0.5
the solution will approach an equilibrium value of 50, which implies the tiniest amount is sufficient to start the reaction. If you are looking for a different scenario, you might consider some other modeling options that appear in Problem 28.
200 y 200 y
0
0
t 0.5
0
0
t 0.5
Phase portrait The curve for x(0) = 150 is almost vertical. (a) (b) (c) ! 28.
Products amount
A solution starting at x(0) = 0 increases and approaches 50. A solution starting at x(0) = 75 decreases and approaches 50. A solution starting at x(0) = 150 increases without bound.
General Chemical Reaction of Two Substances (a), (b) We consider the four cases when the exponent is an even positive integer and an odd positive integer. In each case, we analyze the sign of the derivative for different values of x. For convenience we pick a = 1, b = 2 .
SECTION 2.5
Nonlinear Models: Logistic Equation
139
dx = k (1 - x )even (2 - x)even . dt We have drawn a graph of dx versus x. dt
dx dt
By drawing arrows to the right when dx is positive dt and arrows to the left when dx is negative, dt
1 semistable
2 semistable
3 x
Both even exponents
we have a horizontal phase line for x(t ) . We also see that both equilibrium solutions
x(t ) 1 , x(t ) 2 are unstable; although both are semistable; stable from below and
unstable from above. dx = k (1 - x)even (2 - x)odd . dt Here x(t ) 1 is unstable although it is stable from below. The solution x(t ) 2 is stable.
1 semistable stable 2 3 x 1 dx dt
1
Even and odd exponents dx = k (1 - x)odd (2 - x)even . dt Here x(t ) 2 is semistable, stable from above and unstable from below. The solution x(t ) 1 is stable.
semistable 1 stable 1 2 3 x 1 dx dt
Odd and even exponents
140
CHAPTER 2
Linearity and Nonlinearity
dx = k (1 - x)odd (2 - x)odd . dt Here the smaller of the two solutions, x(t ) 1 , is stable; the larger solution,
x(t ) 2 , is unstable.
1
dx dt
1 stable
2 unstable
3 x
1
Both odd exponents ! 29. Useful Transformation
y = ky(1 - y)
Letting z= yields y 1- y
dz dz = dt dy Substituting for
FG IJ F dy I = 1 F dy I . H K H dt K (1 - y) H dt K
2
dy from the original DE yields a new equation dt (1 - y)2 dz = ky(1 - y) , dt
which gives the result dz ky = = kz . dt 1 - y Solving this first-order equation for z = z(t ), yields z(t ) = cekt and plugging this in the transfory mation z = , we get 1- y y = cekt . 1- y Finally solving this for y gives y(t ) = where 1 1 = - 1. c y0 1 1+
1 e - kt c
,
SECTION 2.5
Nonlinear Models: Logistic Equation
141
! 30.
Solving the Threshold Equation y = -ry 1 -
F H
y T
I K
dy dy d dy = =- . dt d dt d
Introducing backwards time = -t , yields
Hence, if we run the threshold equation dy y = -ry 1 - dt T backwards, we get - dy y = - ry 1 - . d T
F H
I K
FG H
IJ K
Equivalently it also yields the first-order equation dy y , = ry 1 - d T which is the logistic equation with L = T and t = . We know the solution of this logistic equation to be y( ) = 1+ T
T y0
F H
I K
e
- 1 e-r
j
.
We can now find the solution of the threshold equation by replacing by -t , yielding y(t ) = ! 31. 1+
e
T
T y0
- 1 ert
j
.
Limiting Values for the Threshold Equation y = -ry 1 - (a)
F H
y T
I K
y(t ) = 1+
For 0 < y0 < T as t the denominator of
e
T
T y0
- 1 ert
j
goes to plus infinity and so y(t ) goes to zero.
142
CHAPTER 2
Linearity and Nonlinearity
(b)
For y0 > T the denominator of y(t ) = will reach zero when 1+ 1+
e
T
T y0
- 1 ert
j
FG T - 1IJ e Hy K
0
rt
= 0.
Solving for t gives the location of a vertical asymptote on the ty graph y0 1 t * = ln . r y0 - T ! 32. Pitchfork Bifurcation y = y - y 3 = y - y 2 (a)
FG H
IJ K
d
i b g b g
y = y - y2 > 0
For 0 the only real root of y - y 2 = 0 is y = 0 . Because y = y - y 2 < 0 for y > 0 and
b
g
for y > 0 , the equilibrium solution y = 0 is stable. (b) When > 0 the equation y = y - y2 = 0 has roots y = 0, . The points y = are stable, but y = 0 is unstable as illustrated by the graph of y = - y 1 -
stable equilibria 2 1 1 unstable equilibria 1 2
(c)
y 1
b
g
b
y2
g.
Pitchfork bifurcation at (0, 0) ! 33. Another Bifurcation y = y 2 + by + 1 (a) We find the equilibrium points of the equation by setting y = 0 and solving for y. Doing this we get y= -b b2 - 4 . 2
SECTION 2.5
Nonlinear Models: Logistic Equation
143
We see that for -2 < b < 2 there are no (real) solutions, and thus no equilibrium solutions. For b = -2 we have the equilibrium solution +1, and for b = +2 we have equilibrium 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.
3 y
b = 3
unstable equilibrium
1
stable equilibrium t
3 3 y
b = 2
semistable equilibrium 1 t
3 3 y
b = 1
3
y
b=0
1 no equilibrium points 3
t
1 no equilibrium points 3
t
144
CHAPTER 2
Linearity and Nonlinearity 3 y
b =1
1 no equilibrium points 3 3 y
t
b=2
1
t semistable equilibrium
3 3 y
b=3
1
t unstable equilibrium
3
stable equilibrium
(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 unstable and the smaller one is stable. For b < 2 there are no equilibria.
(e)
The bifurcation diagram shows the location of equilibrium points for y versus the parameter value b. Solid circles represent stable equilibria; open circles represent unstable equilibria.
y 4 semistable unstable stable 4 2 2 semistable stable 4 2 unstable b 2 4 t
Equilibria of y = y 2 + by + 1 versus b
SECTION 2.5
Nonlinear Models: Logistic Equation
145
! 34.
Computer Lab: Bifurcation y = ky 2 + y + 1 (a) Set y = 0 to find the equilibria use an open-ended graphic DE solver shown in sample graphs, as in Problems 33 and 35. Setting y = 0 yields two equilibria,
ye = -1 1 - 4 k 2k
1 1 1 for k < ; none for k > ; one for k = . The following phase-plane graphs illustrate the 4 4 4 bifurcation. (b) Sample figures illustrates solution behavior.
2 y 2 y
2
2
t
2
2
t
2
2
k = 05 No equilibrium point .
k = -05 . Two semistable equilibrium points
5
y
5
5
t
5
k = 0.25 One semistable equilibrium point
146
CHAPTER 2
Linearity and Nonlinearity
35.
y = y 2 + y + k (a) Setting y = 0 yields two equilibria, ye = -1 1 - 4k , 2
1 1 1 for k < ; none for k > ; one for k = . The following phase-plane graphs illustrate the 4 4 4 bifurcation. (b)
4 y
k=1
t 4 4 no equilibrium points 4 4 y
k = 1/4
4
t 4 semistable equilibrium
4 4 y
k = 1
unstable equilibrium t 4 4 stable equilibrium 4
SECTION 2.5
Nonlinear Models: Logistic Equation
147
! 36.
Computer Lab: Growth Equation y = ry 1 -
F H
y L
I K
We graph the direction field of this equation for L = 1 , r = 05 , 1, 2, and 5. We keep L fixed . because 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.
2 y
r = 0.5
stable equilibrium
unstable equilibrium 1 2 y 3
t
r=1
stable equilibrium
unstable equilibrium 1 2 y 3
t
r=2
stable equilibrium
unstable equilibrium 1 3
t
148
CHAPTER 2 2
Linearity and Nonlinearity y
r=3
stable equilibrium
unstable equilibrium unstable equilibrium 1 3
t
Solutions of y = ry(1 - y) for different values of r 37. y = -r 1 -
F H
y y T
I K
The parameter r governs the steepness of the solution curves; the higher r the more steeply y leaves the threshold level T. 38. y = r 1 -
F H
ln y y L
I K
Equilibrium at y = e L ; higher r values give steeper slopes. 39. y = re- t y For larger or for larger r, solution curves fall more steeply. Unstable equilibrium r = 1 , = 1
2 2 t 100 y
100
! 40.
Suggested Journal Entry Student Project
SECTION 2.6
Systems of DEs: A First Look
149
2.6
! 1.
Systems of DEs: A First Look
Predicting System Behavior (a) x = y y = x - 3y This (linear) system has one equilibrium point at the origin, x, y = 0, 0 , as do all linear systems. The - and h-nullclines are respectively, as shown in part (b). (b)
2 y
a f a f
h -nullcline
-nullcline
2 2
x
2
(c)
A few solutions along with the vertical and horizontal nullclines are drawn.
2
y
x 2 2
2
(d)
The equilibrium point (0, 0) is unstable. All solutions tend quickly to y = x then move gradually towards + or - 3
asymptotically along that line. Whether the motion is left or right depends on the initial conditions. 2. (a) x = 1 - x - y y = x - y2 Setting x = 0 and y = 0 gives
- nullcline 1 - x - y = 0 h - nullcline x - y2 = 0.
150
CHAPTER 2
Linearity and Nonlinearity
From the intersection of the two nullclines we find two equilibrium points shown in the following figures. We can locate them graphically far more easily than algebraically! (b)
-nullcline
2 y y
(c)
h-nullcline
2
2
2
x
2
2
x
2
2
(d)
The lower equilibrium point at
LM 1 (1 + N4
is unstable and the upper equilibrium at
5)2 ,
1 ( -1 - 5) 2
OP Q
LM 1 (1 - N4
is stable.
5)2 ,
1 ( 5 - 1) 2
OP Q
Most trajectories spiral counterclockwise toward the first quadrant equilibrium point. However, if the initial condition is somewhat left or below the 4th quadrant equilibrium, they shoot down towards - . We suspect a dividing line between these behaviors, and we will find it in Chapter 6. 3. (a) x = 1 - x - y y = 1 - x2 - y2 Setting x = 0 and y = 0 gives
- nullcline x + y =1 2 + y 2 = 1. h - nullcline x
From the intersection of the two nullclines we find two equilibrium points (0, 1), (1, 0).
SECTION 2.6
Systems of DEs: A First Look 2 y
151
(b)
2
y
(c)
2
2
x
2
2
x
h-nullcline
2
-nullcline
2
(d)
The equilibrium at (1, 0) is unstable; the equilibrium at (0, 1) is stable. Most trajectories seem to be attracted to the stable equilibrium, but those that approach the lower unstable equilibrium from below or form the right will turn down toward the lower right.
4.
x = x + y y = 2 x + 2 y (a) This (singular) system has an entire line of equilibrium points on the line x + y = 0. (b) The direction field and the line of unstable equilibrium points are shown at the right. (c) (d) We superimpose on the direction field a few solutions. 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 straight lines (of slope 2), towards + if the IC is above the line x + y = 0 and toward - if the IC is below x + y = 0 . x = 4 - x - y y = 3 - x2 - y2 Setting x = 0 and y = 0 gives the intersection of the nullclines:
2 2 2 x 2 y
5.
(a)
- nullcline y=4-x 2 + y 2 = 3. h - nullcline x
We find no equilibria because the nullclines do not intersect.
152
CHAPTER 2
Linearity and Nonlinearity 4 y
(b)
h-nullcline
4
-nullcline
x 4
4
(c)
4
y
x 4 4
4
(d) 6. (a)
There are no equilibria--all solutions head down to lower right. x = y y = 5x + 3 y This (linear) system has one equilibrium point at x, y = 0, 0 as do all linear systems. The 64-dollar question is: Is it stable? The - and h-nullclines: y = 0 , 5x + 3 y = 0 , are shown following and indicate that the origin (0, 0) is unstable. Hence, points starting near the origin will leave the origin. We will see later other ways for showing that (0, 0) is unstable.
y
a f a f
(b)
The direction field and a few solutions are drawn. Note how the solutions cross the vertical and horizontal nullclines.
2
2
h-nullcline
-nullcline
2
x
2
SECTION 2.6
Systems of DEs: A First Look
153
(c)
2
y
2
2
x
2
(d)
We see from the preceding figure that solutions come from infinity along a line, and then if they are not exactly on the line head off either upwards and to the left or downwards and go to the left on another line. Whether they go up or down depends on whether they initially lie above or below the line. It appears that points that start exactly on the line will go to (0, 0). We will see later in Chapter 6 when we study linear systems using eigenvalues and eigenvectors that the solutions come from infinity on one eigenvector and go to infinity on another eigenvector. x = 1 - x - y y = x - y Setting x = y = 0 and finding the intersection of the nullclines:
7.
(a)
- nullcline y = 1 - x h - nullcline y =x
we find one equilibrium point rium. (b)
2 y
F 1 , 1I . The arrows indicate that it is a stable equilibH 2 2K
(c)
2 y
2
2
x
2
2
x
2
2
(d)
The equilibrium is stable; all other solutions spiral into it.
154
CHAPTER 2
Linearity and Nonlinearity
8.
(a)
x = x + 2 y y = x This (linear) system has one equilibrium point at the origin (0, 0), as do all linear systems. The - and h-nullclines: x + 2 y = 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 system is unstable.
y y
(b)
2
(c)
h-nullcline
2
2
2
x
2
2
x
-nullcline
2 2
(d)
The equilibrium point (0, 0) is unstable. Other solutions come from upper left, lower right, heading toward origin but veers off towards in upper right or lower left.
! 9.
Sharks and Sardines with Fishing (a) With fishing the equilibrium point of the system x = x(a - by - f ) y = y(-c + dx - f ) is ~ = c+ f = c + f xe d d d a- f a f ~ = ye = - . b b b With fishing we increase the equilibrium of the prey xe by rium of the predator ye by f . b f and decrease the equilibd
SECTION 2.6
Systems of DEs: A First Look
155
Using the parameters from Example 3 we set a = 2 , b = 1 , c = 3 , d = 05 ; . the new equilibrium point of the fished model is ~ c+ f = c + f =6+2f xe = d d d a- f a f ~ = ye = - =2- f . b b b
4
y
0
0
x 15
Shark (y) and sardine (x) trajectories
The trajectories are closed curves representing periodic motion of both sharks and sardines. The trajectories look like the trajectories of the unfished case in Example 3 except the equilibrium point has moved to the right (more prey) and down (fewer predators). (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 increases from 0 to 2, the equilibrium point moves along the line from the unfished equilibrium at (6, 2) to (10, 0). Hence, the fishing of each population at the same rate benefits the sardines (x) and drives the sharks (y) to extinction. This is illustrated in the figure. (c)
2 4 6 8 1 (sardines) (10, 0) 10 12 x 2 (6, 2) (7, 1.5) y (sharks)
You should fish for sardines when the sardine population is increasing and sharks when the shark population is increasing. In both cases, more fishing tends to move the populations closer to equilibrium while maintaining higher populations in the low parts of the cycle.
(d)
If we look at the insecticide model and assume both the good guys (predators) and bad guys (prey) are harvested at the same rate, the good guys will also be diminished and the bad 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 not allow either population to go below zero, as the x- and y-axes are solutions and the solutions move along the axes, thus it is impossible for other solutions to cross either of
156
CHAPTER 2
Linearity and Nonlinearity
these axes. You might continue this exploration with the IDE tool, Lotka-Volterra with Harvest, as in Problem 24. ! 10. Finding the Model x = ax - bx 2 - dxy - fx y = -cy + dxy x = ax - bx 2 - cxy - dxz y = ey - fy 2 + gxy z = -hz + kxz ! 13. Host-Parasite Models (a) A suggested model is H = aH - c H 1+ P P = -bP + dHP 11. x = ax + bxy y = cy - dxy + eyz z = fz - gx 2 - hyz 12.
where a, b, c, and d are positive parameters. Here a species of beetle (parasite) depends on a certain species of tree (host) for survival. Note that if the beetle were so effective as to wipe out the entire population of trees, then it would die out itself, which is reflected in our model (note the differential equation in P). On the other hand, in the absence of the beetle, the host tree may or may not die out depending on the size of the parameters a and c. We would probably pick a > c , so the host population would increase in the absence of the parasite. Note too that model says that when the parasite (P) population H gets large, it will not destroy the host entirely, as becomes small. The modeler 1+ P might want to estimate the values of parameters a, b, c, d so the solution fits observed data. The modeler would also like to know the qualitative behavior of P, H in the PH
a
f
plane. (b) Many bacteria are parasitic on external and internal body surfaces; some invading inner tissue causing diseases such as typhoid fever, tuberculosis, and pneumonia. It is important to construct models of the dynamics of these complex organisms.
SECTION 2.6
Systems of DEs: A First Look
157
! 14.
Competition (a) x = x (4 - 2 x - y ) y = y (4 - x - 2 y ) Setting x = 0 and y = 0 we find
y
- nullclines 2 x + y = 4, x = 0 h - nullclines x + 2 y = 4, y = 0 .
Equilibrium points: (0, 0), (0, 2), (2, 0), 4 4 . The directions of the solution , 3 3
F H
I K
0
5
x
curves are shown in the figure. (b) It can be seen from the figure, that the equilibrium points (0, 0), (0, 2) and (2, 0) are 4 4 is stable because all solution curves nearby point unstable. Only the point , 3 3
F H
I K
toward it. (c) (d) Some solution curves are shown in the figure. Because all the solution curves eventually reach the stable equilibrium at two species described by this model can coexist. 15. (a) x = x(1 - x - y) y = y (2 - x - y ) Setting x = 0 and y = 0 we find
y
F 4 , 4 I , the H 3 3K
- nullclines x + y = 1, x = 0 h - nullclines x + y = 2, y = 0 .
Equilibrium points: (0, 0), (0, 2), (1, 0). The directions of the solution curves are shown in the figure. (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. (c) (d) Some solution curves are shown in the figure. Because all the solution curves eventually reach the stable equilibrium at (0, 2), the x species always die out and the two species described by this model cannot coexist.
0 3 x
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CHAPTER 2
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16.
(a)
x = x (4 - x - 2 y ) y = y(1 - 2 x - y) Setting x = 0 and y = 0 we find
y
- nullclines x + 2 y = 4, x = 0 h - nullclines 2 x + y = 1, y = 0 .
Equilibrium points: (0, 0), (0, 1), (4, 0). The directions of the solution curves are shown in the figure. (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. (c) (d) Some solution curves are shown in the figure. Because all the solution curves eventually reach the stable equilibrium at (4, 0), the y species always die out and the two species described by this model cannot coexist. 17. (a) x = x (2 - x - 2 y ) y = y (2 - 2 x - y ) Setting x = 0 and y = 0 we find
y 0 5 x
- nullclines x + 2 y = 2, x = 0 h - nullclines 2 x + y = 2, y = 0.
Equilibrium points: (0, 0), (0, 2), (2, 0), 2 2 . The directions of the solution , 3 3
F H
I K
0
3
x
curves are shown in the figure. (b) It can be seen from the figure, that the equilibrium points (0, 0) and
F 2 , 2I H 3 3K
are
unstable; the points (0, 2) and (2, 0) are stable because all nearby arrows point toward them. (c) (d) Some solution curves are shown in the figure. 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 2 2 the unstable equilibrium point . Which species dies out is determined by the , 3 3
F H
I K
initial conditions.
SECTION 2.6
Systems of DEs: A First Look
159
! 18.
Simpler Competition Setting x = 0 , we find the -nullclines are the vertical line x = 0 and the horizontal line y = Setting y = 0 , we find the h-nullclines are the horizontal y = 0 and vertical line x = equilibrium points are (0, 0) and a . b
F c , a I . By observing the signs of x , y we find H d bK
x > 0, y > 0 when x < x < 0, y < 0 when x > x < 0, y > 0 when x > x > 0, y < 0 when x < c a , y< d b c a , y> d b c a , y< d b c a , y> d b
c . The d
Hence, both equilibrium points are unstable. We can see from the following direction field (with a = b = c = d = 1) that one of two species, depending on the initial conditions, goes to infinity and the other toward extinction. Equation used in the drawing on the right are: x' = x(1 - y) y' = y(1 - x).
y 4
2
2
4
x
One can get the initial values for these curves directly from the graph. ! 19. Nullcline Patterns (ae) When the -nullcline lies above the h-nullcline, there are three equilibrium points in the d d a and are unstable and first quadrant: (0, 0), 0, , 0 . The points (0, 0), 0, f f b a , 0 is stable. Hence, only population x survives. b
F I H K
FG H
IJ K
F I H K
FG H
IJ K
160
CHAPTER 2
Linearity and Nonlinearity
(a)(b)
y a/c
(c)(d)
y a/c
-nullcline
d/f h-nullcline
-nullcline
d/f
h-nullcline d/e a/b x d/e a/b x
Nullclines and equilibria
Sample trajectories when the
-nullcline is above the h-nullcline
(e) 20. (ae) Population x survives. When the h-nullcline lies above the -nullcline, there are three equilibrium points in the d a a . The points (0, 0), first quadrant: (0, 0), , 0 and 0, , 0 are unstable and f b b d 0, stable. Hence, only population y survives. f
FG H
IJ K
F I H K
FG H
IJ K
F I H K
(a)(b)
d/f
y
(c)(d)
d/f
y
a/c
-nullcline
h-nullcline
a/c
-nullcline
h-nullcline
a/b
d/e
x
a/b
d/e
x
Nullclines
Sample trajectories when the h-nullcline is above the -nullcline
(e) 21. (ae)
Population y survives. When the two nullclines intersect as they do in the figure, then there are four equilibrium d a points in the first quadrant: (0, 0), , and xe , ye , where xe , ye is the , 0 , 0, f b
F I FG H K H
IJ K
a
f
a
f
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 that xe , ye unstable. Hence, the two populations can coexist.
a
f
is stable and the others
SECTION 2.6
Systems of DEs: A First Look
161
(a)(b)
y a/c
(c)(d)
-nullcline
y a/c
-nullcline
d/f h-nullcline h-nullcline x x a/b d/e
d/f
a/b
d/e
Nullclines and equilibria
Typical trajectories when the nullclines intersect and the slope of the vertical nullcline is more negative.
(e) 22. (ae)
The two populations coexist When the two nullclines intersect as they do in the figure, then there are four equilibrium d a points in the first quadrant: (0, 0), , and xe , ye , where xe , ye is the , 0 , 0, f b
F I FG H K H
IJ K
a
f
a
f
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
F a , 0I Hb K
y d/f
and 0,
FG H
d f
IJ K
are stable and the
other two unstable. Hence, only one of the two populations survives, and which survives depends on the initial conditions. (a)(b)
y d/f h-nullcline a/c a/c
(c)(d)
h-nullcline
-nullcline
x d/e a/b d/e
-nullcline
a/b x
Nullclines and equilibria
Typical trajectories when the nullclines intersect and the slope of the h-nullcline is more negative.
(e)
Only one of the two populations survives, and which survives depends on the initial conditions, as indicated by shading. For initial conditions in the upper region y survives; for initial conditions in the lower region, x survives.
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Linearity and Nonlinearity
! 23.
Unfair Competition x = ax(1 - bx) - cxy y = dy - exy Setting x = y = 0 , we find three equilibrium points: 1 a0, 0f , FH b , 0IK , and FH d , a e - bd IK . e ce The point (0, 0) corresponds to both populations becoming extinct, the point to the second population becoming extinct, and the point
F d , a e - bd I corresponds to either a H e ce K
1 d > , e.g., b e
F 1 , 0I corresponds Hb K
stable coexistent point or an unstable point. If we take the special case where x = x(1 - x) - xy y = 05y - xy .
where a = b = c = e = 1 , d = 05 , we have the equilibrium points (0, 0), (1, 0), and (0.5, 0.5). If . we draw two nullclines; v-nullcline: y = 1 - x , 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.
1.5 y
h-nullcline
1.5
y
-nullcline
0 x 1.5 0 x 1.5
0
d/e
1/b
0
d/e
1/b
Nullclines and equilibria for
1 d > b e
Sample trajectories for 1 d 1 d = or < . b e b e
1 d > b e
The reader must check separately the cases where ! 24. Computer Lab: Parameter Investigation
Hold three of the parameters constant and observe how the fourth parameter affects the behavior of the two species. See if the behavior makes sense in your understanding of the model. Keep in mind 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 the prey is absent (large aF if the given prey is the only source of food for the predator), cR is a
SECTION 2.6
Systems of DEs: A First Look
163
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. Even if you are not a biology major, you may still ponder the relative sizes of the four parameters in the two predatorprey systems: foxes and rabbits, and ladybugs and aphids. You can use these explanations to reach the same conclusions as in Problem 9. ! 25. Computer Lab: Competition Outcomes (a) Using the IDE software, hold all parameters fixed except one and observe how the last parameters affects the solution. See if the behavior of the two species makes sense in your understanding of the model. Play a mind game and predict if there will be coexistence between the species, whether one becomes extinct, and so on, before you make 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 the way the parameters in the IDE software affect the two isoclines, called the N1 and N 2 isoclines in the IDE software. By changing the parameters K1 and K2 in the IDE software, you simply move the respective isoclines in a parallel direction. The parameters B1 , B2 change the slopes of the nullclines. And finally, the parameters r1, r2 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 like the following four pictures, with the basins of attraction colored for each equilibrium.
3 y 3 y
0
0
5
x
0
0
5
x
Case 1: Population x dies out x = x(1 - x - y ) y = y(2 - x - y)
Case 2: Population y dies out x = x( 4 - x - 2 y ) y = y(1 - 2 x - y )
164
CHAPTER 2
Linearity and Nonlinearity 2 y 2 y
0
0
2
x
0
0
2
x
Case 3: Populations coexist x = x(2 - 2 x - y) y = y(2 - x - 2 y )
Case 4: One of the populations dies out x = x(2 - x - 2 y) y = y(2 - 2 x - y )
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 the species that dies out depends on the initial conditions, and in Case 1 and 2 one species will die out regardless of the initial condition. Note too that in all four cases if one population 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
y y
Case 2
x
x
Case 3 ! 26. Suggested Journal Entry Student Project
Case 4