Ordinary Differential Equations:
Worked Examples with Solutions
Edray Herber Goins
Talitha Michal Washington
July 31, 2016
2
Contents
I
First Order Differential Equations
1 What is a Differential Equation?
1.1 Limits and Continuity . . . . . .
1.2 Differential Calculus . . . . . . .
1.3 Integral Calculus . . . . . . . . .
1.4 Slope Fields . . . . . . . . . . . .
1.5 Classical Mechanics . . . . . . . .
5
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
7
. 7
. 9
. 11
. 14
. 15
2 Separable Equations and Population Dynamics
2.1 Separable Equations . . . . . . . . . . . . . . . .
2.2 Autonomous Equations . . . . . . . . . . . . . .
2.3 Population Growth Models . . . . . . . . . . . .
2.4 Newton’s Law of Cooling . . . . . . . . . . . . .
2.5 Radioactive Decay . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
23
23
26
29
37
39
3 Exact Equations and Integrating Factors
3.1 Exact Equations . . . . . . . . . . . . . .
3.2 Equidimensional Equations . . . . . . . .
3.3 Bernoulli Equations . . . . . . . . . . . .
3.4 Integrating Factors . . . . . . . . . . . . .
3.5 Linear Equations . . . . . . . . . . . . . .
3.6 Loan Calculations . . . . . . . . . . . . .
3.7 Diffusion Problems . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
41
41
43
45
46
47
48
50
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
Part I
First Order Differential Equations
5
Chapter 1
What is a Differential Equation?
1.1
Limits and Continuity
Problem 1. Let d ≥ 1 be a rational number, and define the sequence of rational numbers xn by
d−1 1
for n = 1, 2, . . . . Assuming that this sequence converges, show
x1 = 1 and xn+1 = 1 +
4 xn
the following:
d−1
for n = 1, 2, . . .
4
√
1+ d
(b.) lim xn =
n→∞
2
Solution. First note that x1 ≥ 1 so all terms in the sequence xn − 1 ≥ 0. This gives
1
d−1
d−1 1
d−1
1≥
= 1+
−1≥ 1+
=⇒
− 1 = xn+1 − 1.
xn
4
4
4 xn
(a.) 0 ≤ xn − 1 ≤
As for the value of such a limit, we have the identity
1
d−1
;
lim xn+1 = 1 +
n→∞
4
lim xn
n→∞
so that if we denote L = limn→∞ xn we have
L=1+
d−1 1
4 L
=⇒
L2 − L −
d−1
= 0.
4
√
1± d
By the quadratic formula, we have two possibilities:
. Since d ≥ 1, we cannot have the −
2
√
1+ d
sign or else L would be negative. Hence L =
.
2
π
radians, and define the five points A = (0, 0),
2
B = (1, 0), C = (cos θ, sin θ), D = (cos θ, 0), E = (cos2 θ, cos θ sin θ). Note that B and C lie on a
circle of radius 1, while D and E lie on a circle of radius cos θ.
Problem 2. Let θ be an angle between 0 and
7
Let a(θ) be the region bounded by the line AB, the line AC, and the arc BC. Let b(θ) be the
subregion of a(θ) which is the triangle ADC. Let c(θ) be the subregion of b(θ) which is bounded
by the line AD, the line AE, and the arc DE. Show the following:
(a.)
area of c(θ)
= cos2 θ
area of a(θ)
(b.)
area of b(θ)
cos θ sin θ
=
area of a(θ)
θ
(c.) cos θ <
sin θ
1
<
θ
cos θ
Solution. The area of the sector s of a circle of radius r and an angle θ is s =
θ
area of a(θ) = ,
2
area of b(θ) =
cos θ sin θ
,
2
area of c(θ) =
1 2
θ r , so we have
2
θ cos2 θ
.
2
Since c(θ) is a subregion of b(θ) is a subregion of a(θ), we have
θ cos2 θ
cos θ sin θ
θ
<
<
2
2
2
=⇒
cos θ <
sin θ
1
<
.
θ
cos θ
Problem 3. Using the previous exercise, show that we have the following limits.
(a.) lim cos x = 1
x→0
(b.) lim
sin x
=1
x
(c.) lim
1 − cos x
1
=
x2
2
x→0
x→0
Solution. First, note that cos x is a continuous function so limx→0 cos x = cos 0 = 1. Now using the
previous exercise (with x in place of θ) we have
sin x
1
≤ lim
=1
x→0 x
x→0 cos x
1 = lim cos x ≤ lim
x→0
sin x
= 1.
x→0 x
=⇒
lim
Finally, to work out the third limit, we note that
1 − cos x
1 − cos2 x
1
=
=
2
2
x
x (1 + cos x)
1 + cos x
sin x
x
2
;
so that we have the limit
1 − cos x
1
lim
= lim
·
x→0
x→0 1 + cos x
x2
8
sin x
lim
x→0 x
2
=
1 2 1
·1 = .
2
2
Problem 4. Let F (x) be a function satisfying the recursive formula F (x + 1) = x F (x) where F (1)
is nonzero. Show that F (x) has a singularity at the nonpositive integers 0, −1, −2, . . . .
Solution. Let k be a nonnegative integer; then −k is a nonpositive integer. Using the recursive
formula,
F x − (k − 1)
F (x)
F (x + 1)
F (x − k) =
=
=
(x − k)
(x − 1) · · · (x − k)
x (x − 1) · · · (x − k)
so that we have the limit
(−1)k
1
F (x + 1)
=
F (1) · lim
x→0 x
x→0 x(x − 1) · · · (x − k)
k!
lim F (x) = lim F (x − k) = lim
x→−k
x→0
which does not converge.
Problem 5. Let n be a positive real number. Show that
(a.) lim xn log x = 0
x→0
(b.) lim x−n log x = 0
x→∞
Solution. We show the first limit. Substitute x = e−t/n so that as t = −n log x. Hence, as x → 0,
the variable t → ∞. This gives
lim xn log x = −
x→0
1
· lim t e−t = 0
n t→∞
The value is zero because the exponential function f (t) = e−t dies faster than the linear function
g(t) = t grows. As for the second limit, the substitution x = 1/t means t → 0 as x → ∞ so that
lim x−n log x = − lim tn log t = 0.
x→∞
1.2
t→0
Differential Calculus
Problem 6. Let y = f (x) be a function which is smooth i.e. a function where derivatives of all
dn y
order exist. Given a nonegative integer n, define f (n) = n . Show that for smooth functions f (x)
dx
and g(x),
(f · g)(0) =
f (0) g (0)
(1)
(f · g)
=
f (1) g (0) + f (0) g (1)
(2)
(2)
(0)
(f · g)
= f g + 2 f (1) g (1) + f (0) g (2)
In general, we can write
(f · g)(n) =
n
X
Cn−k,k f (n−k) g (k) .
k=0
What are the coefficients Cn−k,k ?
9
Solution. This exercise shows Leibnitz’s Rule: the coefficients
n
n!
Cn−k,k =
=
(n − k)! k!
k
are the Binomial Coefficients. One verifies the formulas above by repeated use of the product
rule.
Problem 7. Define the function y = cos (log x). Find positive constants a, b, and c such that
a x2 y 00 + b x y 0 + c y = 0.
1
Solution. The derivative of y = cos log x is y 0 = − sin log x· , so that x y 0 = − sin log x. The second
x
derivative is
1 1
1
00
y = − cos log x ·
− sin log x · − 2
=⇒
x2 y 00 = sin log x − cos log x
x x
x
and so x2 y 00 = −x y 0 − y. Hence, we may choose a = b = c = 1.
Problem 8. For real numbers x1 , x2 , . . . , xn and positive integers C1 , C2 , . . . , Cn ; define the polynomial
g(x) = (x − x1 )C1 (x − x2 )C2 · · · (x − xn )Cn .
Show the identity
C1
g 0 (x)
C2
Cn
=
+
+ ··· +
.
g(x)
x − x1 x − x2
x − xn
Solution. The logarithm of g(x) is the sum
log g(x) = C1 log(x − x1 ) + C2 log(x − x2 ) + · · · + Cn log(x − xn )
which in turn has the derivative
g 0 (x)
d
1
1
1
=
log g(x) = C1
+ C2
+ · · · + Cn
.
g(x)
dx
x − x1
x − x2
x − xn
Problem 9. Denote the curve
C = (x, y) ∈ R2 y 3 = x5 − 1 .
Without solving for y, show that show that
(a.) (1, 0) ∈ C has a vertical tangent.
(b.) (0, −1) ∈ C has a horizontal tangent.
10
(c.) The tangent line at every other point on C has positive slope.
Solution. The curve C is defined by y 3 − x5 + 1 = 0, so we can find the derivative via implicit
differentiation:
dy
dy
5 x4
3 y2
− 5 x4 = 0
=⇒
=
.
dx
dx
3 y2
When x = 1 and y = 0, the slope dy/dx = ∞ so the curve has a vertical tangent. In fact, when
y = 0, then x5 = 1 and the only real number where this is true is x = 1. That is, (1, 0) is the only
point on C with a vertical tangent. When x = 0 and y = −1, the slope dy/dx = 0 so the curve
has a horizontal tangent. In fact, when x = 0, then y 3 = −1 and the only real number where this
is true is y = −1. That is, (0, −1) is the only point on C with a horizontal tangent. Otherwise
both x4 and y 2 are nonzero and hence must be positive, so that dy/dx is positive as well. Hence
the slope is always positive in these cases.
Problem 10. Let y = y(x) be any function which satisfies the implicit relation x = tan y.
(a.) Show that y = y(x) must be an increasing function without relative extrema.
(b.) It is true that y(0) = 0 and y 0 (0) = 1?
Solution. First we make an observation:
1 + x2 = 1 + tan2 y =
1
cos2 y + sin2 y
=
= sec2 y.
2
cos y
cos2 y
Now we implicitly differentiate the expression tan y − x = 0:
sec2 y
dy
−1=0
dx
=⇒
dy
1
= cos2 y =
.
dx
1 + x2
Hence dy/dx 6= 0 so that it has no relative extrema. In fact, dy/dx > 0 so that it is always
increasing. Note that y 0 (0) = 1/(1 + 02 ) = 1 must indeed be true. It need not be true that
y(0) = 0, but it must be true that y(0) satisfy tan y(0) = 0. This occurs only when y(0) = n π
where n is an integer.
1.3
Integral Calculus
Problem 11. Compute the value of the following integrals:
Z 1
dx
√
(a.)
1 − x2
−1
Z 2
dx
√
(b.)
2 x − x2
0
Z 5
dx
p
(c.)
(x − 3)(5 − x)
3
11
Solution. If x = a + b sin θ then dx = b cos θ dθ. This yields
Z 1
Z π/2
Z π/2
dx
cos θ dθ
√
p
=
=
dθ = π.
1 − x2
−1
−π/2
−π/2
1 − sin2 θ
With the substitution x = 1 + sin θ, as x ranges from 0 to 2, the function sin θ ranges from -1 to
+1, so the variable θ ranges from −π/2 to π/2. Through the identity 2 x − x2 = 1 − (x2 − 2 x + 1) =
1 − (x − 1)2 = 1 − sin2 θ = cos2 θ we may evaluate the integral as
Z 2
Z π/2
Z π/2
dx
cos θ dθ
√
√
=
=
dθ = π.
2 x − x2
cos2 θ
0
−π/2
−π/2
Again, with the substitution x = 4 + sin θ, as x ranges from 3 to 5, the function sin θ ranges from
-1 to +1, so the variable θ ranges from −π/2 to π/2. Through the identity (x − 3)(5 − x) =
(1 + sin θ)(1 − sin θ) = 1 − sin2 θ = cos2 θ we may evaluate the integral as
Z 5
Z π/2
dx
p
dθ = π.
=
(x − 3)(5 − x)
3
−π/2
Problem 12. Given that a x2 +b x+c is a quadratic polynomial with leading term a < 0, discriminant
b2 − 4 a c > 0, and roots α < β, show that
Z β
dx
π
√
= 1/2 .
2
|a|
ax + bx + c
α
Solution. We’ll make the substitution
√
b2 − 4 a c
b
+
sin θ
x=−
2a
2 |a|
√
=⇒
dx =
b2 − 4 a c
cos θ dθ.
2 |a|
First, using the quadratic formula to find the roots of a x2 + b x + c, we define
√
√
b
b
b2 − 4 a c
b2 − 4 a c
−
,
β=−
+
α=−
2a
2 |a|
2a
2 |a|
so that as x ranges from α to β, the function sin θ ranges from -1 to +1, and hence the variable θ
π
π
ranges from − to + . We also have the identity (remember that a < 0):
2
2
"
2 #1/2
2
1/2
2 − 4ac
p
√
√
b
b − 4 a c b2 − 4 a c
b
2
2
= −a
a x + b x + c = −a
− x+
−
sin θ
4 a2
2a
4 a2
4 a2
2
1/2
√
b − 4ac
dx
2
= −a
cos
θ
= |a|1/2
4 a2
dθ
so that the integral is
Z
β
α
dx
1
√
= 1/2
2
|a|
ax + bx + c
12
Z
π/2
dθ =
−π/2
π
.
|a|1/2
Problem 13. Consider the identity
1 1 1 1
1
1− + − + −
+ ··· =
3 5 7 9 11
1
Z
dx
.
+1
x2
0
Compute the value of this sum/integral.
Solution. We’ll substitute x = tan θ. As x ranges from 0 to 1, the variable θ ranges from 0 to π/4.
Recall that the derivative of tan θ is sec2 θ; hence dx = sec2 θ dθ. Also, we have the identity
sin2 θ
sin2 θ + cos2 θ
1
+
1
=
=
= sec2 θ
2
2
cos θ
cos θ
cos2 θ
x2 + 1 = tan2 θ + 1 =
so that the integral becomes
Z
1
0
π/4
Z
dx
=
2
x +1
0
Problem 14. Show that when a > b > 0,
Z ∞
−∞
(1 +
sec2 θ dθ
=
sec2 θ
dx
a2 x2 ) (1
+
b2 x2 )
Z
π/4
dθ =
0
=
π
.
4
π
.
a+b
Solution. First consider the partial fraction expansion
1
(1 +
a2 x2 ) (1
+
b2 x2 )
=
A
B
+
2
2
1+a x
1 + b2 x2
where A and B satisfy the equations b2 A + a2 B = 0 and A + B = 1. That is, A = a2 /(a2 − b2 )
and B = b2 /(b2 − a2 ). The integral of interest can now be expressed as the sum of two integrals:
Z ∞
Z ∞
Z ∞
dx
dx
dx
=A
+B
.
2
2
2
2
2
2
2 2
−∞ (1 + a x ) (1 + b x )
−∞ 1 + a x
−∞ 1 + b x
Now let’s focus on the integral
R∞
−∞ dx/(1
+ a2 x2 ). We’ll make the substitution x =
ranges from −∞ to ∞, the variable θ ranges from −π/2 to π/2. This gives
Z
∞
−∞
dx
=
1 + a2 x2
Z
π/2
−π/2
1
sec2 θ
dθ =
a 1 + tan2 θ
Z
π/2
−π/2
1
tan θ. As x
a
1
π
dθ = .
a
a
Finally, we can evaluate the original integral. It is
Z ∞
dx
Aπ Bπ
aπ
bπ
(a − b) π
π
=
+
= 2
+ 2
= 2
=
.
2
2
2
2
2
2
2
a
b
a −b
b −a
a −b
a+b
−∞ (1 + a x ) (1 + b x )
13
Problem 15. Given that a x2 + b x + c is a positive definite quadratic polynomial (that is, a > 0
and b2 − 4 a c < 0), show that
Z
∞
−∞
dx
2π
.
=√
a x2 + b x + c
4 a c − b2
Solution. We’ll make the substitution
√
b
4 a c − b2
x=−
+
tan θ
2a
2a
√
=⇒
dx =
4 a c − b2
sec2 θ dθ.
2a
As x ranges from −∞ to ∞, the variable θ ranges from −π/2 to +π/2. Using the identity tan2 θ+1 =
sec2 θ, we also have the identity
"
#
b 2 4 a c − b2
2
ax + bx + c = a x +
+
2a
4 a2
√
4 a c − b2
4 a c − b2
4 a c − b2 dx
4 a c − b2
2
2
tan
θ
+
=
sec
θ
=
=a
4 a2
4 a2
4a
2
dθ
so that the integral is
Z
∞
−∞
1.4
dx
=
2
ax + bx + c
Z
π/2
√
−π/2
2
2π
dθ = √
.
4 a c − b2
4 a c − b2
Slope Fields
For the following differential equations, draw a slope field for the given differential equation and
state whether you think that the solutions are converging or diverging.
dy
Problem 16.
= −x y + 0.1 y 3 .
dx
Solution. Figure 1.1 contains a plot of the slope field. Convergence of the solution depends on the
initial condition. The graph shows that if |y(0)| ≥ 2.4 then y → ±∞, yet if |y(0)| ≤ 2.2 then y → 0.
Figure 1.2 contains a closer analysis of the direction field. Using this, we see that the solutions
diverge y → ±∞ if |y(0)| > 2.37, yet the solutions converge y → 0 if |y(0)| < 2.37.
Problem 17.
dy
= x2 + y 2 .
dx
Solution. Figure 1.3 contains a plot of the slope field. It is clear that the solutions are diverging y → ∞
regardless of the initial condition.
14
dy
= −x y + 0.1 y 3
dx
Figure 1.1: Direction Field for
4.8
4
3.2
2.4
1.6
0.8
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
-0.8
1.5
Classical Mechanics
Problem 18. At t = 0, a ball is propelled downward from an initial height of 1000 m with an initial
speed of 25 m/s. Calculate the time t that the ball hits the ground.
Solution. The height of the ball at time t is given by the formula
x(t) = x0 + v0 t −
1 2
gt
2
where x0 = 1000 m is the initial height, v0 = −25 m/s is the initial speed (note negative because
the ball is propelled downward!), and g = 9.81 m/s2 is the acceleration due to gravity. We want to
find the time t when x(t) = 0 i.e.
0 = 1000 − 25 t −
1
9.8 t2
2
=⇒
t = 11.9556
or
− 17.0525.
Time can only be positive, so we find t = 12 s .
Problem 19. A balloon is ascending at a rate of 15 m/s at a height of 100 m about the ground, when
a package is dropped from the gondola. How long will it take the package to reach the ground?
Ignore air resistance.
15
Figure 1.2: Solutions of
dy
= −x y + 0.1 y 3 , with y(0) = 2.37
dx
3.5
3.25
3
2.75
2.5
2.25
0
0.08
0.16
0.24
0.32
0.4
0.48
0.56
0.64
0.72
0.8
0.88
0.96
1.04
Solution. Denote the height of the package at time t by
x(t) = x0 + v0 t −
1 2
gt
2
where x0 = 100 m is the initial height, v0 = 15 m/s is the initial velocity (note positive because the
balloon is ascending!) and g = 9.81 m/s2 is the acceleration due to gravity. We want to find the
time t when x(t) = 0 i.e.
0 = 100 + 15 t −
1
9.81 t2
2
=⇒
t = 6.29616
or
− 3.23806.
Time can only be positive, so we find t = 6.3 s .
Problem 20. The velocity v(t) of a falling object satisfies the initial value problem
where v(0) = 0.
dv
v
= 9.8 −
dt
5
(a.) Find the time that must elapse for the object to reach 98% of its terminal velocity.
(b.) How far does the object fall in the time found in part (a)?
Solution. The velocity of the falling object at t seconds is v(t) = 49 1 − e−t/5 . The limiting
16
Figure 1.3: Direction Field for
dy
= x2 + y 2
dx
0.75
0.5
0.25
-1.25
-1
-0.75
-0.5
-0.25
0
0.25
0.5
0.75
1
1.25
-0.25
-0.5
-0.75
velocity is vL = 49 m/sec, so the velocity reaches 98% of this value when
49 1 − e−t/5 = 0.98 · 49
1 − e−t/5 = 0.98
e−t/5 = 0.02
et/5 = 50
=⇒
t = 5 ln 50 = 19.56 sec.
The distance the object has fallen after t seconds is x(t) = 49 t+245 e−t/5 −245. Hence the distance
after t = 19.56 seconds is x = 49 · 19.56 + 245 · 0.02 − 245 = 718.346 m.
Problem 21. A skydiver of mass 60 kg free-falls from an airplane at an altitude of 5000 meters. He
is subjected to an air resistance force that is proportional to his speed. Assume that the constant
of proportionality is 10 (kg/sec). Find and solve the differential equation governing the altitude of
the skydiver at time t seconds after the start of his free-fall. Assuming that he does not employ his
parachute, find his limiting velocity and how much time will elapse before he hits the ground.
dx
the velocity of the
Solution. Let x(t) denote the altitude of the skydiver at time t, and v(t) =
dt
skydiver at time t. Then when the skydiver initially jumps from the plane i.e. at t0 = 0 we have
x(t0 ) = 5000 m and v(t0 ) = 0 m/s. The force due to gravity is −m g = −588.6 kg m/s2 , and the
d2 x
dx
force due to air resistance is −10 v. Hence the equation of motion is m 2 = −m g − 10 .
dt
dt
17
First we solve for the velocity. We have the initial value problem
60
dv
= −588.6 − 10 v
dt
where
v(0) = 0.
dv
dt
t
= − , so that ln |v + 58.86| = − + C1 . The constant is
v + 58.86
6
6
t
C1 = ln 58.86, so we have ln |v + 58.86| = − + ln 58.86, which gives v(t) = 58.86 e−t/6 − 58.86. As
6
t → ∞, the limiting velocity is v(t) → −58.86 m/s .
Second we solve for the altitude. We have the initial value problem
This equation is separable:
dx
= 58.86 e−t/6 − 58.86
dt
where
x(0) = 5000.
We integrate term by term to find x(t) = −58.86 · 6 e−t/6 − 58.86 t + C2 . Using the initial conditions,
we find that the constant is C2 = 5353.16, so we have x(t) = 5353.16 − 58.86 t + 353.16 e−t/6 . To
find out how much time will elapse before the skydiver hits the ground, we want to find the time t
such that t = 0, i.e., 0 = 5353.16 − 58.86 t + 353.16 e−t/6 so that t = 90.9473 s .
Problem 22. A sky diver weighing 180 lb (including equipment) falls vertically downward from an
altitude of 5,000 ft and opens the parachute after 10 sec of free fall. Assume that the force of air
resistance is 0.75 |v| when the parachute is closed and 12 |v| when the parachute is open, where
the velocity v is measured in ft/sec. Note: The force due to gravity is m g = 180 lb where the
gravitational acceleration is g = 32 ft/sec2 . The mass of the sky diver is m = 180/g = 5.63 lb.
(a.) Find the speed of the sky diver when the parachute opens.
(b.) Find the distance fallen before the parachute opens.
(c.) What is the limiting velocity vL after the parachute opens?
(d.) Determine how long the sky diver is in the air after the parachute opens.
(e.) Plot the graph of velocity versus time from the beginning of the fall until the skydiver reaches
the ground.
Solution. Let v = v(t) denote the speed of the sky diver in the downward direction. The force due
to air resistance is −γ v, where
(
0.75 lb/sec
γ=
12 lb/sec
when the parachute is closed,
when the parachute is open.
Now we find an expression for the speed of the sky diver. Newton’s Second Law of Motion states
dv
F = m a, which is equivalent to the differential equation m
= m g − γ v. We solve this equation
dt
18
using integrating factors. After dividing through by m, we multiply by µ(t) = eγt/m :
dv
+ γ v = mg
dt
dv
γ
+ v=g
dt
m
dv
γ γt/m
+ e
v = g eγt/m
dt
m
d h γt/m i
e
v = g eγt/m
dt
m
eγt/m
=⇒
eγt/m v(t) =
m g γt/m
e
+C
γ
for some constant C. Say that we have the initial condition v(t0 ) = v0 . If we set t = t0 , we have
mg
mg
v(t0 ) =
eγt0 /m .
+ C e−γt0 /m
=⇒
C = v(t0 ) −
γ
γ
Hence the speed of the sky diver is
v(t) =
mg
mg
mg
+ C e−γt/m =
+ v0 −
e−γ(t−t0 )/m .
γ
γ
γ
Initially the sky diver is in free fall, so v0 = 0 ft/sec when t0 = 0 sec. The speed of the sky diver
when the parachute opens is
180
180
v(10) =
+ 0−
e−0.75(10−0)/5.63 = 176.74 ft/sec.
0.75
0.75
Let x = x(t) denote the distance the sky diver has fallen by time t. Since we have chosen
dx
v = v(t) as the velocity in the downward direction, we have the initial value problem
= v where
dt
x(t0 ) = x0 . We have the differential equation
dx
mg
mg
mg
m g m −γ(t−t0 )/m
−γ(t−t0 )/m
=
+ v0 −
e
=⇒ x(t) =
t − v0 −
e
+C
dt
γ
γ
γ
γ
γ
for some constant C. Upon setting t = t0 we have
mg
mg m
x0 =
t0 − v0 −
+C
=⇒
γ
γ
γ
C = x0 −
mg
mg m
t0 + v0 −
.
γ
γ
γ
Hence the distance the sky diver has fallen is
mg
mg m x(t) = x0 +
(t − t0 ) + v0 −
1 − e−γ(t−t0 )/m .
γ
γ
γ
Initially we have v0 = 0 ft/sec and x0 = 0 ft when t0 = 0 sec, so the distance the sky diver has fallen
when the parachute opens is
180
180 5.63 (10 − 0) + 0 −
1 − e−0.75(10−0)/5.63 = 1074.47 ft.
x(10) = 0 +
0.75
0.75 0.75
19
After the parachute opens, the differential equation which governs the motion of the sky diver
remains the same – except that γ = 12 lb/sec. Hence the speed of the sky diver is
mg
mg
v(t) =
e−γ(t−t0 )/m
+ v0 −
γ
γ
where now v0 = 176.74 ft/sec when t0 = 10 sec. Hence the limiting velocity after the parachute
opens is
mg
180
vL = lim v(t) =
=
= 15 ft/sec.
t→∞
γ
12
After the parachute opens, the distance the sky diver has fallen is
mg
mg m
−γ(t−t0 )/m
1−e
x(t) = x0 +
(t − t0 ) + v0 −
γ
γ
γ
where now v0 = 176.74 ft/sec and x0 = 1074.47 ft when t0 = 10 sec. Say that T = t − t0 is the time
that the sky diver is in the air after the parachute opens i.e., x(t) = 5000 ft. If T is large enough,
then e−γT /m ≈ 0, so we can ignore it. Hence we have the equation
mg
mg m
x(T + t0 ) ≈ x0 +
T + v0 −
γ
γ
γ
180 5.63
180
T + 176.74 −
5000 ≈ 1074.47 +
12
12
12
3849.71 ≈ 15 T
so that T = 256.65 sec.
Figure 1.4: Plot of v(t) vs. t
175
150
125
100
75
50
25
-50
-25
0
25
50
75
100
-25
20
125
150
175
200
225
250
Figure 1.4 has a plot of the function:

180
180


e−0.75t/5.63
+ 0−


0.75
 0.75
v(t) =



180
180 −12(t−10)/5.63


e
+ 176.74 −
12
12
if t ≤ 10 sec,
if t ≥ 10 sec.
Problem 23. For small, slowly falling objects, the assumption that the drag force is proportional to
the velocity is a good one. For larger, more rapidly falling objects, it is more accurate to assume
that the drag force is proportional to the square of the velocity.
(a.) Write a differential equation for the velocity of a falling object of mass m if the drag force is
proportional to the square of the velocity.
(b.) Determine the limiting velocity after a long time.
(c.) If m = 10 kg, find the drag coefficient so that the limiting velocity is 49 m/sec.
(d.) Using the data in part (c), draw a slope field.
Solution. Let v = v(t) denote the velocity of the falling object in the downward direction at time t.
The force due to gravity on this mass is m g. The force due to air resistance is proportional to v 2 , so
this force is −k v 2 for some positive constant k. Newton’s Second Law of Motion states that F = m a
dv
is the sum of the forces on this object, so we have the differential equation m
= m g − k v 2 . Let
dt
vL = lim v(t) denote the limiting velocity after a long time. It satisfies the equation m g−k vL 2 = 0,
t→∞
r
r
mg
mg
so that vL =
. Hence v →
. Gravitational acceleration is g = 9.8 m/sec2 . If the
k
k
limiting velocity is to be vL = 49 m/sec for a mass m = 10 kg, then we have m g − k vL 2 = 0, so
that
mg
10 · 9.8 kg m/sec2
2
k= 2 =
=
kg/m.
2
2
2
vL
49
m /sec
49
dv
Hence k = 2/49 = 0.041 kg/m. We consider the differential equation m
= m g − k v 2 , which we
dt
dv
can express as
= 9.8 − 0.0041 v 2 . A plot of the direction field can be found in Figure 1.5, along
dt
with the equilibrium solution v = 49 m/sec.
21
Figure 1.5: Slope Field for
dv
= 9.8 − 0.0041 v 2
dt
50
25
0
25
50
75
-25
22
100
125
Chapter 2
Separable Equations and Population
Dynamics
2.1
Separable Equations
Problem 1. Find the general solution of the following differential equation. If possible, find an
explicit solution:
dy
= 1 + y 2 ex .
dx
Solution. This is a separable equation:
dy
= 1 + y 2 ex
dx
1
dy = ex dx
1 + y2
Z
Z
1
dy
=
ex dx
1 + y2
=⇒
arctan y = ex + C.
Hence the solution to the differential equation is y(x) = tan (ex + C) .
Problem 2. Find the exact solution of the following initial value problem. Indicate the interval of
existence.
dy
= 1 + y2
where
y(0) = 1.
dx
Solution. This equation is separable:
dy
= 1 + y2
dx
dy
= dx
1 + y2
arctan y = x + C
=⇒
23
y(x) = tan (x + C) .
π
When x = 0 we want y = 1, so we choose C such that tan C = 1. This happens when C =
π π 4
π
(modulo multiples of π) so y(x) = tan x +
. Now tan C is defined on the interval − ,
4
2 2
π
π
π
π
– notice that C =
is in this interval – so we must have − < x +
<
which implies
4
4
2
2
3π
3π π
π
−
< x < . Hence the interval of existence is − ,
.
4
4
4 4
dy
Problem 3. Solve the initial value problem
= 2 (1+x) (1+y 2 ) subject to y(0) = 0 and determine
dx
where the solution attains its minimum value.
Solution. This differential equation is a separable equation:
dy
= 2 (1 + x) 1 + y 2
dx
1 dy
= 2 + 2x
1 + y 2 dx
d
[arctan y] = 2 + 2 x
dx
arctan y = x2 + 2 x + C
=⇒
for some constant C. Hence y(x) = tan x2 + 2 x + C is the general solution. When x = 0, we
find that 0 = y(0) = tan C, so that C must be an integer multiple of π. Using the Angle Addition
Formula for tangent:
tan θ + tan C
tan (θ + C) =
= tan θ
1 − tan θ · tan C
we have the explicit solution y(x) = tan x2 + 2 x . This solution attains its minimum value when
its derivative vanishes. But y 0 = 2 (1 + x) (1 + y 2 ), and 2 (1 + y 2 ) > 0, so this happens when
1 + x = 0. Hence the solution attains its minimum value when x = −1.
Problem 4. Consider the differential equation
dy
1 − 2x
=
.
dx
y
(a.) Find the solution y = y(x) corresponding to y(1) = −2.
(b.) Plot the graph of the solution.
(c.) Determine the interval in which the solution is defined.
Solution. This differential equation is a separable equation:
dy
1 − 2x
=
dx
y
y dy = 1 − 2 x dx
Z
Z
y dy =
1 − 2 x dx
=⇒
24
y2
= x − x2 + C
2
for some constant C. Upon evaluating at the point (x0 , y0 ) = (1, −2) we find that C = 2. Hence
p
the solution to the initial value problem is y(x) = − 2 x − 2 x2 + 4. Figure 2.1 has a plot of the
p
solution. The solution y = − 2 (2 − x) (1 + x) is defined when the factors 2 − x and 1 + x have
the same sign, so the solution is defined when −1 < x < 2.
√
Figure 2.1: Plot of y(x) = − 2 x − 2 x2 + 4
0.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-0.5
-1
-1.5
-2
-2.5
dy
x
= −4 where y(0) = y0 . Determine how the
dx
y
interval in which the solution exists depends on the initial value y0 .
Problem 5. Solve the given initial value problem
Solution. This differential equation is a separable equation:
4x
dy
=−
dx
y
Z
y dy = −4 x dx
Z
y dy = − 4 x dx
=⇒
y2
= −2 x2 + C
2
y0 2
for some constant C. Upon evaluating at the point (t0 , y0 ) = (0, y0 ), we find that C =
. Hence
2
we find the expression
y2
y0 2
= −2 x2 +
2
2
=⇒
25
p
y(x) = ± y0 2 − 4 x2 .
This solution exists when |x| <
1
|y0 |. In particular, no solution exists if y0 = 0, so we must also
2
have y0 6= 0.
2.2
Autonomous Equations
dy
In the three exercises below, an autonomous differential equation is given in the form
= f (y).
dx
Perform each of the following tasks without the aid of technology.
(a.) Sketch a graph of f (y).
(b.) Use the graph of f to develop a phase line for the autonomous equation. Classify each
equilibrium point as either unstable or asymptotically stable.
(c.) Sketch the equilibrium solutions in the xy plane. These equilibrium solutions divide the xy
plane into regions. Sketch at least one trajectory in each of these regions.
Problem 6.
dy
= (y + 1) (y − 4).
dx
Solution. Figure 2.2 has a graph of f (y) = (y + 1) (y − 4). The equilibrium solutions are y = −1
and y = 4. Figure 2.3 has a graph of the phase line. We use the First Derivative Test to check for
df
the stability: since f (y) = (y + 1) (y − 4) = y 2 − 3 y − 4 we have
= 2 y − 3. When y = −1 we
dy
df
df
have
= −5 is negative, so that y = −1 is stable . When y = 4 we have
= 5 is positive, so
dy
dy
that y = 4 is unstable . Figure 2.4 has a graph of the trajectories in the various regions.
Problem 7.
dy
= 6 + y − y2.
dx
Solution. We have the factorization f (y) = 6 + y − y 2 = − (y + 2) (y − 3). Figure 2.5 has a graph
of f (y). The equilibrium solutions are y = −2 and y = 3. Figure 2.6 has a graph of the phase
line. We use the First Derivative Test to check for the stability: since f (y) = 6 + y − y 2 we have
df
df
= 1 − 2 y. When y = −2 we have
= 5 is positive, so that y = −2 is unstable . When y = 3
dy
dy
df
= −5 is negative so that y = 3 is stable . Figure 2.7 has a graph of the trajectories in
we have
dy
the various regions.
Problem 8.
dy
= y (y − 1) (y − 2).
dx
Solution. Figure 2.8 has a graph of f (y) = y (y − 1) (y − 2). The equilibrium solutions are y = 0,
y = 1, and y = 2. Figure 2.9 has a graph of the phase line. We use the First Derivative Test to check
for the stability: since f (y) = y 3 −3 y 2 +2 y we have f 0 (y) = 3 y 2 −6 y +2. The values of interest are
26
Figure 2.2: Plot of f (y) = (y + 1) (y − 4)
f 0 (0) = 2 > 0, f 0 (1) = −1 < 0 and f 0 (2) = 2 > 0. Hence y = 0 and y = 2 are unstable equilibria ,
whereas y = 1 is a stable equilibrium. Figure 2.10 has a graph of the trajectories in the various
regions.
dy
Problem 9. Draw a slope field for the differential equation
= 3 − 2 y. Use this to determine the
dx
behavior of y as x → ∞. Does this behavior depends on the initial condition of y at x = 0?
Solution. Figure 2.11 contains a plot of the slope field for the differential equation. It is clear that
3
y → as x → ∞, regardless of the initial condition of y.
2
Problem 10. Sometimes a constant equilibrium solution has the property that solutions lying on
27
Figure 2.3: Phase Line for
dy
= (y + 1) (y − 4)
dx
one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side
depart from it. In this case the equilibrium solution is said to be semistable.
dy
(a.) Consider the equation
= k (1 − y)2 , where k is a positive constant. Show that yL = 1 is
dx
the only critical point.
(b.) Sketch f (y) versus y. Show that y is increasing as a function of x for y < 1 and also for
y > 1. Note: The phase line has upward-pointing arrows both below and above y = 1. Thus
solutions below the equilibrium solution approach it, and those above it grow farther away.
Therefore yL = 1 is semistable.
dy
= k (1 − y)2 subject to the initial condition y(0) = y0 and confirm the conclusions
dx
reached in part (b).
(c.) Solve
Solution. Denote the function f (y) = k (1 − y)2 . The critical points yL correspond to f (yL ) = 0,
so yL = 1 is the only critical point. Figure 2.12 contains a plot of f (y) = k (1 − y)2 versus y. Say
that y = y(x) is a solution to the differential equation y 0 = f (y). When y 6= 1 we have f (y) > 0, so
the derivative y 0 (x) is positive. Hence y is an increasing function of x whenever y 6= 1.
To solve the differential equation, divide both sides by (1 − y)2 :
dy
= k (1 − y)2
dx
(y − 1)−2 dy = k dt
Z
Z
−2
(y − 1) dy = k dt
=⇒
− (y − 1)−1 = k t + C
for some constant C. When t = 0, we have C = − (y0 − 1)−1 . We may use this to explicitly solve
for y as a function of time:
−
1
1
= kt−
y−1
y0 − 1
1
(y0 − 1) k t − 1
=−
y−1
y0 − 1
y−1=
y0 − 1
1 + (1 − y0 ) k t
=⇒
y(t) = 1 +
28
y0 − 1
y0 + (1 − y0 ) k t
=
.
1 + (1 − y0 ) k t
1 + (1 − y0 ) k t
Figure 2.4: Plot of solutions to
dy
= (y + 1) (y − 4)
dx
If y0 < 1 then (1 − y0 ) > 0 so that the denominator for y = y(t) does not vanish for t ≥ 0. As t
increases without bound, we see that y → 1. On the other hand, if y0 > 1 then (1 − y0 ) < 0 so that
the denominator for y = y(t) vanishes when t = 1/ ((y0 − 1) k) > 0. Hence there is a finite value of
t for which y → ∞. We conclude that solutions y < 1 approach the limiting solution yL = 1, yet
solutions y > 1 grow farther away.
2.3
Population Growth Models
Problem 11. The size P (t) at t months of a certain population satisfies the differential equation
dP
= 0.5 p − 450.
dt
(a.) Find the time at which the population becomes extinct if P (0) = 850.
(b.) Find the time of extinction if P (0) = P0 , where 0 < P0 < 900.
29
Figure 2.5: Plot of f (y) = 6 + y − y 2
(c.) Find the initial population P0 if the population is to become extinct in 1 year.
Solution. The solution so the differential equation is P (t) = 900 + C et/2 for some constant C. If
P (0) = 850, then C = −50. The population becomes extinct when 0 = P (t) = 900 − 50 et/2 . This
900
= 18, so that the population becomes extinct when t = 2 ln 18 = 5.78074 months.
gives et/2 =
50
More generally, if P (0) = P0 , then C = P0 −900. The population becomes extinct when 0 = P (t) =
900
, so that the population becomes extinct when
900 + (P0 − 900) et/2 . This gives et/2 =
900 − P0
900
t = 2 ln
months. If the population is to become extinct in 1 year, then P (12) = 0. This
900 − P0
gives 900 + (P0 − 900) e12/2 = 0, so that P0 = 900 (e6 − 1)/e6 = 897.769.
Problem 12. Suppose that you have a closed system containing 1000 individuals. A flu epidemic
starts. Let N (t) represent the number of infected individuals in the closed system at time t. Assume
30
Figure 2.6: Phase Line for
dy
= 6 + y − y2
dx
that the rate at which the number of infected individuals is changing jointly proportional to the
number of infected individuals and the number of noninfected individuals. Furthermore, suppose
that when 100 individuals are infected, the rate at which individuals are becoming infected is 90
individuals per day. If 20 individuals are infected at time t = 0, when will 90% of the population
be infected?
Solution. Assuming that there are only healthy individuals and sick individuals, if N (t) is the
number of sick, then 1000 − N (t) is the number of healthy. Then there is a constant of propordN
tionality k such that
= k N (t) (1000 − N (t)). When N (t) = 100, we want N 0 (t) = 90. Hence,
dt
k = 90/ (100 · 900) = 1/1000 day−1 . The differential equation is
dN
1
=
N (1000 − N )
dt
1000
=⇒
N (t) =
1
1/1000 + C e−t
where the time t is measured in days. If N (0) = 20, then we have C = 49/1000. Hence N (t) =
1000
.
1 + 49 e−t
We want to find the number of days t such that N (t) = 90% · 1000 = 900 individuals:
1000
= 900
1 + 49 e−t
=⇒
et = 441
=⇒
t = 6.08904
Hence 90% of the population will be infected after approximately 6 days .
Problem 13. A population of bacteria, growing according to the Malthusian model, doubles itself
in 10 days. If there are 1,000 bacteria present initially, how long will it take the population to reach
10,000?
Solution. Let P (t) denote the number of bacteria present at time t days. The Malthusian model
states that there are constant P0 and r such that P (t) = P0 ert . Initially, there are 1000 bacteria
present, so P0 = 1000. The population doubles itself in 10 days, so when t = 10 we have P (t) =
ln 2
= 0.0693147. The time t it will take to reach
2000. Hence 2000 = 1000 e10r so that r =
10
ln 10
a population of 10,000 satisfies 10000 = 1000 ert , so t =
= 33.2193. Hence it will take
r
33.2 days .
31
Figure 2.7: Plot of solutions to
dy
= 6 + y − y2
dx
Problem 14. Suppose a population is growing according to the logistic equation
dP
P
= rP 1−
.
dt
K
Prove that the rate at which the population is increasing is at its greatest when the population is
at one-half of its carrying capacity.
Solution. The rate at which the population P is increasing is
P
r 2
f (P ) = r P 1 −
= rP −
P .
K
K
The rate at which the rate is increasing is
h
df
df dP
r ih
r 2i
=
= r−2 P rP −
P = r2 P
dt
dP dt
K
K
32
2P
P
1−
1−
.
K
K
Figure 2.8: Plot of f (y) = y (y − 1) (y − 2)
0.75
0.5
f'(y)<0
f'(y)>0
f'(y)>0
0.25
-0.25
0
0.25
0.5
0.75
y=0
1
1.25
1.5
1.75
2
y=1
y=2
-0.25
-0.5
0.75
-0.75
0.5
f'(y)>0
0.25
-0.25
0
f'(y)<0
dy
Figure 2.9: Phase Line for
= y (y − 1) (y − 2)
dx
0.25
0.5
y=0
0.75
1
y=1
1.25
1.5
1.75
f'(y)>0
2
y=2
-0.25
K
rK
, and K; which gives the values f (P ) = 0,
, and
2
4
0, respectively. When P = 0, K the rate f (P ) = 0; but when P = K/2 the rate f (P ) > 0 is
K
increasing. Hence the rate at which the populaiton is increasing at the greatest is when P = ,
2
i.e., when the population is one-half its carrying capacity.
We find relative extrema when P = 0,
-0.5
-0.75
Problem 15. A population is observed to obey the logistic equation with the eventual population
20,000. The initial population is 1,000, and 8 hours later, the observed population is 1,200. Find
the reproductive rate and the time required for the population to reach 75% of its carrying capacity.
Solution. Let P (t) denote the size of the population at time t hours. This function satisfies the
differential equation
dP
P
K P0
20000
= rP 1−
=⇒
P (t) =
=
,
dt
K
P0 + (K − P0 ) e−rt
1 + 19 e−rt
33
Figure 2.10: Plot of solutions to
dy
= y (y − 1) (y − 2)
dx
3
2.5
2
1.5
1
0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
where r is the reproductive rate, P0 = 1, 000 is the initial population size, and K = 20, 000 is the
eventual population size. When t = 8 we know that P (t) = 1200. We find that
1200 =
20000
1 + 19 e−8r
e8r =
=⇒
57
47
=⇒
r = 0.024113 .
The time it will take to reach 75% of its carrying capacity is
75% · 20000 =
20000
1 + 19 e−rt
ert = 57
=⇒
=⇒
t = 167.671
Hence the time is 167.7 hours .
Problem 16. At a given level of effort, it is reasonable to assume that the rate at which fish are
caught depends on the population P : The more fish there are, the easier it is to catch them. Thus
we assume that the rate at which fish are caught is given by E P , where E is a positive constant,
with units of 1/time, that measures the total effort made to harvest the given species of fish. To
include this effect, the logistic equation is replaced by
dP
P
=r 1−
P − E P.
dt
K
(a.) Show
thatif E < r, then there are two equilibrium points, namely P1 = 0 and P2 =
E
K 1−
> 0.
r
34
Figure 2.11: Slope Field for
dy
= 3 − 2y
dx
2
1.5
1
0.5
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
-0.5
-1
(b.) Show that P = P1 is an unstable equilibrium and P = P2 is a stable equilibrium.
(c.) A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the
product of the effort E and the asymptotically stable population P2 . Find Y as a function of
the effort E; the graph of this function is known as the yield-effort curve.
(d.) Determine E so as to maximize Y and thereby find the maximum sustainable yield Ym .
Solution. Consider the function
P
r 2
r
K (E − r)
f (P ) = r 1 −
y − E P = − P + (r − E) P = − P P +
.
K
K
K
r
K (E − r)
E
= K 1−
are equilibrium points. (If E < r then
Hence P = 0 and P = −
r
r
E
2r
1−
> 0.) We have the derivative f 0 (P ) = −
P + (r − E), which shows f 0 (P1 ) = r − E > 0
r0
K
yet f (P2 ) = E − r < 0. Hence P = P1 is an unstable equilibrium whereas P = P2 is a stable
E
K
equilibrium. The sustainable yield is Y (E) = E K 1 −
= − E 2 + K E. This function has
r
r
2K
r
a maximum when Y 0 (E) = −
E + K = 0, so that E = . Hence the maximum sustainable
r
2
Kr
1
Kr
yield is Ym =
.
1−
=
2
2
4
35
Figure 2.12: Plot of f (y) = k (1 − y)2
1
0.75
0.5
0.25
0
0.25
0.5
0.75
1
y=1
1.25
1.5
1.75
2
2.25
-0.25
-0.5
Problem 17. Some diseases (such as typhoid fever) are spread largely by “carriers”, individuals who
can transmit the disease but who exhibit no overt symptoms. Let S and I, respectively, denote
the proportion of susceptibles and carriers (i.e., infected) in the population, respectively. Suppose
dI
= −β I. Suppose
that carriers are identified and removed from the population at a rate β, so
dt
dS
also that the disease spreads at a rate proportional to the product of S and I; thus
= −α S I.
dt
dI
(a.) Determine I(t) at any time t by solving
= −β I subject to the initial condition I(0) = I0 .
dt
dS
= −α S I subject to the
(b.) Use the result of part (a) to find S(t) at any time t by solving
dt
initial condition S(0) = S0 .
(c.) Find the proportion of the population that escapes the epidemic by finding the limiting value
of S(t) as t → ∞.
Solution. To determine I(t), we solve a separable equation:
dI
= −β I
dt
1
dI = −β dt
I
Z
Z
1
dI = − β dt
I
ln |I| = −β t + C1
=⇒
36
I(t) = C e−βt
where C = ±eC1 is a constant. When t = 0 we have C = I0 , so that I(t) = I0 e−βt . To determine
S(t), we solve another separable equation:
dS
= −α S · I0 e−βt
dt
1
dS = −α I0 e−βt dt
S
Z
Z
1
dS = − α I0 e−βt dt
S
α I0 −βt
e
+ C1
ln |S| =
β
=⇒
α I0 −βt
S(t) = C exp
e
β
where C = ±eC1 is a constant. When t = 0, we have
α I0
C = S0 exp −
β
1 − e−βt
S(t) = S0 exp −α I0
.
β
=⇒
α I0
.
As time increases without bound, we have lim S(t) = S0 exp −
t→∞
β
2.4
Newton’s Law of Cooling
Problem 18. Newton’s law of cooling asserts that
the rate at which an object cools is proportional to the difference between the object’s
temperature (T ) and the temperature of the surrounding medium (T∞ ).
Let’s suppose that the ambient temperature is T∞ = 70◦ F and that the rate constant is 0.05
(min)−1 . Write a differential equation for the temperature of the object at any time.
Solution. Let T (t) denote the temperature of the object in ◦ F at any time t in minutes. Newton’s
dT
dT
law of cooling states that
is proportional to T − 70◦ F, so say that
= r (T − 70◦ F) for some
dt
dt
constant r. We know that |r| = 0.05 (min)−1 , so we determine the sign of r. If T > 70◦ F then the
dT
object will cool i.e.,
< 0. Hence r must be a negative constant. The differential equation must
dt
dT
= −0.05 (T − 70).
be
dt
Problem 19. Assuming Newton’s Law of Cooling, show that T (t) = T∞ + (T0 − T∞ ) ert , where T0
is the temperature of the body at time t = 0 and r is the proportionality constant.
Solution. Newton’s law of cooling is equivalent to the differential equation T 0 ∝ (T − T∞ ), i.e.,
T 0 = r (T − T∞ ). Note that if T > T∞ then the object will cool, i.e., T 0 < 0, whereas if T < T∞
37
then the object will warm up, i.e., T 0 > 0. Hence r is a negative constant of proportionality. This
equation is separable:
dT
= r (T − T∞ )
dt
dT
= −k dt
T − T∞
ln |T − T∞ | − ln |T0 − T∞ | = r dt
T − T∞ = rt
ln T0 − T∞ T − T∞ rt
T0 − T∞ = e
T − T∞
= ±ert
T0 − T∞
=⇒
When t = 0 we see that the left-hand side is 1, so we must choose the positive sign on the right-hand
T (t) − T∞
side. Hence
= ert , so that T (t) = T∞ + (T0 − T∞ ) ert .
T0 − T∞
Problem 20. A murder victim is discovered at midnight and the temperature of the body is recorded
at 31◦ C. One hour later, the temperature of the body is 29◦ C. Assume that the surrounding air
temperature remains constant at 21◦ C. Calculate the victim’s time of death. Note: The “normal”
temperature of a living human being is approximately 37◦ C.
Solution. Assume that the temperature of the body at time t is
T (t) = T∞ + (T0 − T∞ ) e−kt ,
where T∞ = 21◦ C and T0 = 37◦ C. We denote t0 = 0 the time when the victim died, t1 the time
in minutes that has elapsed since the body was recorded to be T (t1 ) = 31◦ C, and t2 = 60 + t1 to
be the time when the body was discovered to be T (t2 ) = 29◦ C. We want to solve for t1 . First we
solve for k. Consider the two equations
T (t1 ) − T∞ = (T0 − T∞ ) e−k t1
=⇒
−k t2
T (t2 ) − T∞ = (T0 − T∞ ) e
T (t1 ) − T∞
= ek(t2 −t1 ) .
T (t2 ) − T∞
Upon taking the logarithms of both sides, we can solve for the proportionality constant:
k=
1
T (t1 ) − T∞
1
31 − 21
ln
=
ln
= 0.00371906 minutes−1 .
t2 − t1
T (t2 ) − T∞
60
29 − 21
Now we solve for t1 . We have
T (t1 ) − T∞ = (T0 − T∞ ) e−k t1
=⇒
ek t1 =
Hence
T0 − T∞
T (t1 ) − T∞
=⇒
t1 =
1
T0 − T∞
ln
.
k
T (t1 ) − T∞
37 − 21
31 − 21
ln
= 126.377 minutes.
31 − 21
29 − 21
In other words, at midnight the victim had been dead for 126 minutes, so the victim died at
9:54 PM .
t1 = 60 · ln
38
2.5
Radioactive Decay
Problem 21. The half-life of a radioactive material is the time required for an amount of this
material to decay to one-half its original value. Show that for any radioactive material that decays
dQ
according to the equation
= −r Q, the half-life τ and the decay rate r satisfy the equation
dt
r τ = ln 2.
Solution. First, we solve the differential equation:
dQ
= −r Q
dt
1
dQ = −r dt
Q
Z
Z
1
dQ = − r dt
Q
ln |Q| = −r t + C1
=⇒
Q(t) = Q0 e−rt
for some constant Q0 = ±eC1 . The half-life τ is that time such that Q(τ ) = 0.5 Q0 . We have the
equation Q0 e−rτ = 0.5 Q0 so that erτ = 2. Upon taking logarithms, we find that r τ = ln 2.
Problem 22. Radium-266 has a half-life of 1620 years. Using the previous problem, find the time
period during which a given amount of this material is reduced by one-quarter.
Solution. Let Q(t) denote the amount of some given initial amount Q0 of Radium-266 after t
years. We know that Q(t) = Q0 e−rt for some decay rate r. The half-life is τ = 1620 yr, so
ln 2
we have r τ = ln 2, so that r =
= 0.000428 yr−1 . We wish to find the time t such that
τ
1
4
Q(t) = Q0 − Q0 = 0.75 Q0 . We have the equation Q0 e−rt = 0.75 Q0 , so that ert = . Hence
4
3
ln(4/3)
t=
= 672.361 yr.
0.000428
Problem 23. The isotope Iodine 131 is used to destroy tissue in an overactive thyroid gland. It has
a half-life of 8.04 days. If a hospital receives a shipment of 500 mg of 131 I, how much of the isotope
will be left after 20 days?
Solution. Let N (t) denote the amount of 131 I; then N (t0 ) = 500 mg when t0 = 0 days. This amount
dN
satisfies the differential equation
= −r N for some decay rate r which we will find later. This
dt
equation is separable, so we find that N (t) = 500 e−rt . Now we solve for the decay rate r. From the
1
definition of half-life, when t1 = 8.04 days we have N (t1 ) = N (0), so 250 = 500 e−8.04 r . Taking
2
39
ln 2
logarithms gives ln 250 = ln 500 − 8.04 r, so that r =
= 0.0862123 days−1 . Hence the amount
8.04
left after t2 = 20 days is
N (t2 ) = 500 e−r 20 = 500 e−0.0862123·20 = 5.60829 mg .
Problem 24. Suppose that the ratio of 14 C to carbon in the charcoal on a cave wall is 0.617 times a
similar ratio in living wood in the area. Use the Libby half-life to estimate the age of the charcoal.
Solution. We use the formula found in the previous problem, where R(t) = R0 e−rt . We want
ln 0.617
to solve for t when R(t) = 0.617 · R0 . We have 0.617 · R0 = R0 e−r t , so that t = −
=
r
3878.99 years. Hence the age of the charcoal is approximately 3879 years .
40
Chapter 3
Exact Equations and Integrating
Factors
3.1
Exact Equations
Verify that the following equations are exact,
and then find the solution.
2
2
Problem 1. 2 x y + 2 y dx + 2 x y + 2 x dy = 0.
Solution. Denote the functions

P (x, y) = 2 x y 2 + 2 y 

Q(x, y) = 2 x2 y + 2 x
=⇒

∂P


 ∂y = 4 x y + 2


 ∂Q = 4 x y + 2
∂x


Hence the equation is indeed exact . We find a solution in the form F (x, y) = C for the differential
equation. Define the functions
Z
G(x, y) = P (x, y) dx = x2 y 2 + 2 x y,
Z Z
∂G
H(y) =
Q(x, y) −
dy =
2 x2 y + 2 x − 2 x2 y + 2 y dy = 0;
∂y
and set F (x, y) = G(x, y) + H(y). Hence the solution is x2 y 2 + 2 x y = C.
Problem 2.
x
(x2
+
y 2 )3/2
dx +
y
(x2
+ y 2 )3/2
dy = 0.
Solution. Denote the functions
P (x, y) =
x
(x2 +
y 2 )3/2
and
41
Q(x, y) =
y
(x2 + y 2 )3/2
.
We compute the partial derivatives
−3/2 i
−5/2
∂P
∂ h
3
3xy
;
=
x x2 + y 2
= − · x x2 + y 2
· 2y = −
2
∂y
∂y
2
(x + y 2 )5/2
−3/2 i
−5/2
∂Q
∂ h
3
3xy
.
=
y x2 + y 2
= − · y x2 + y 2
· 2x = −
2
∂x
∂x
2
(x + y 2 )5/2
Hence the equation is indeed exact. We find a solution in the form F (x, y) = C for the differential
equation. Define the functions
Z
Z
1
x
dx = − p
G(x, y) = P (x, y) dx =
,
3/2
2
2
2
x + y2
(x + y )
#
Z "
Z y
∂G
y
Q(x, y) −
dy =
−
dy = 0;
H(y) =
∂y
(x2 + y 2 )3/2 (x2 + y 2 )3/2
and set F (x, y) = G(x, y) + H(y). Hence the solution is F (x, y) = C1 in terms of some constant
C1 , which we can write as x2 + y 2 = C for some constant C = C1 −2 .
Problem 3. p
x
y
dx + p
dy = 0.
x2 + y 2
x2 + y 2
x
y
Solution. Denote P (x, y) = p
and Q(x, y) = p
. We have the partial derivative
x2 + y 2
x2 + y 2
−1/2
−3/2
∂P
∂
x
xy
=
x x2 + y 2
= − x2 + y 2
· 2 y = −p
3
∂y
∂y
2
x2 + y 2
∂P
∂Q
xy
=
= −p
. Hence the equation indeed is exact. It is easy
3
∂y
∂x
x2 + y 2
p
∂F
∂F
= P and
= Q. Hence the solution to the
to verify that for F (x, y) = x2 + y 2 we have
∂x
∂y
differential equation is x2 + y 2 = C for some constant C.
and so by symmetry
Problem 4. Find the function y = y(x) satisfying the differential equation (2 x − y) dx+(2 y − x) dy =
0 as well as the initial condition y(1) = 3. For what values x is this function valid?
Solution. First we verify that this differential equation is exact. Denote the functions

P (x, y) = 2 x − y 

Q(x, y) = 2 y − x
=⇒


42

∂P


 ∂y = −1


 ∂Q = −1
∂x
Now we compute F (x, y) = G(x, y) + H(y) in terms of the functions
Z
G(x, y) =
P (x, y) dx = x2 − x y,
Z
Z
Z ∂G
dy = [(2 y − x) − (−x)] dy = 2 y dy = y 2 .
H(y) =
Q(x, y) −
∂y
Hence F (x, y) = G(x, y) + H(y) = x2 − x y + y 2 , so that the desired solution y = y(x) satisfies the
implicit relation x2 −x y +y 2 = C for some constant C. Since y(1) = 3, we see that C = F (1, 3) = 7.
p
1
Hence y(x) =
x + 28 − 3 x2 . This solution is valid when 28 − 3 x2 ≥ 0, which is the same
2
r
28
.
as saying |x| ≤
3
Problem 5. Find the value of b for which the given equation is exact, and then solve it using that
value of b.
x y 2 + b x2 y dx + (x + y) x2 dy = 0.
Solution. Denote the functions
P (x, y) = x y 2 + b x2 y



2
3
2
Q(x, y) = (x + y) x = x + x y
=⇒



∂P
2


 ∂y = 2 x y + b x

 ∂Q

= 3 x2 + 2 x y
∂x
The differential equation is exact when these derivatives are equal. They are equal when b = 3. We
solve the differential by constructing a function F (x, y) = G(x, y) + H(y). We define the functions
Z
1 2 2
x y + x3 y,
2
Z Z
3
∂G
H(y) =
Q(x, y) −
dy =
x + x2 y − x2 y + x3 dy = 0.
∂y
G(x, y) =
P (x, y) dx =
The solution of the original differential equation is F (x, y) = C, which we can write in the form
x2 y 2 + 2 x3 y = C .
3.2
Equidimensional Equations
Problem 6. Find the general solution of the equidimensional equation below.
x2 + y 2 dx − 2 x y dy = 0.
43
Solution. We make the substitution y = t x to find a separable differential equation:
x2 + y 2 dx − 2 x y dy = 0
dy
x2 + y 2
=
dx
2xy
dt
1 + t2
x+t=
dx
2t
=⇒
dx
2tx
=
.
dt
1 − t2
We find the general solution to this equation:
dx
2tx
=
dt
1 − t2
2t
1
dx =
dt
x
1 − t2
ln |x| = − ln1 − t2 + C1
for some constant C = ±eC1 . Substituting t =
=⇒
x=C
1
1 − t2
y
we find the equation x2 − y 2 = C x .
x
Problem 7. Find the general solution to the homogeneous equation below.
(x + y) dx + (y − x) dy = 0.
Solution. We make the substitution y = t x to find a separable differential equation:
(x + y) dx + (y − x) dy = 0
x+y
dy
=
dx
x−y
dt
1+t
x+t=
dx
1−t
=⇒
dx
1−t
=
x.
dt
1 + t2
We find the general solution to this equation:
1−t
dx
=
x
dt
1 + t2
1
1−t
dx =
dt
x
1 + t2
1 ln 1 + t2 + C1
2
This can be simplified a bit. Recall the formula
ln |x| = arctan t −
tan (A − B) =
tan A − tan B
1 + tan A tan B
=⇒
=⇒
where C = tan C1 is a constant.
44
p
t = tan ln |x| 1 + t2 − C1 .
p
x tan ln x2 + y 2 − y
p
=C
y tan ln x2 + y 2 + x
3.3
Bernoulli Equations
Problem 8. Show that the change of variable, z = y 1−n , will transform the Bernoulli equation
dy
dz
= P (x) y + Q(x) y n into the linear equation
= (1 − n) P (x) z + (1 − n) Q(x). Here, we
dx
dx
assume that n 6= 1.
dz
dy
dy
1
dz
= (1 − n) y −n
, and so
=
yn
.
dx
dx
dx
1−n
dx
1 − n n dz
dy
= P (x) y + Q(x) y n transforms to
y
=
Note that y = z y n . Hence the equation
dx
dx
dz
P (x) z y n + Q(x) y n , so that
= (1 − n) P (x) z + (1 − n) Q(x) as claimed.
dx
Solution. If we substitute z = y 1−n , then we have
Use the technique of the previous exercise to transform the Bernoulli equations into a linear
equation. Find the general solution of the resulting linear equation.
Problem 9.
dP
= a P − b P 2 , assuming that a and b are constants.
dx
1
dP
1 dz
1
, P 2 = 2 , and
=− 2
. Hence
z
z
dx
z dx
dP
a
b
dz
1 dz
the equation
= a P − b P 2 transforms to − 2
= − 2 , which simplifies to
+ az = b .
dx
z dx
z z
dx
In order to solve this equation, we multiply by the integrating factor µ(x) = eax :
d dz
dµ
ax dz
µ(x) z = µ(x)
+
z=e
+ a z = b eax .
dx
dx dx
dx
Solution. We set n = 2 and z = P −1 . Then we have P =
Upon integrating we find µ(x) z =
b ax
b
e + C1 , and soz(x) = + C1 e−at for some constant C1 .
a
a
Hence P (x) =
1
a
=
for some constant C = C1 /a.
z(x)
b + C e−ax
Problem 10. x2
dy
+ 2 x y − y 3 = 0.
dx
dy
1
dz
Solution. We sat n = 3 and z = y −2 . Then we have y = z −1/2 and
= − z −3/2
. We
dx
2
dx
substitute these values into the differential equation:
dy
+ 2 x y = y3
dx
h
i h
i3
dz
+ 2 x z −1/2 = z −1/2
dx
x2
x
2
1
− z −3/2
2
−
x2 dz
+ 2xz = 1
2 dx
=⇒
45
dz
4
2
− z = − 2.
dx x
x
This differential equation is linear, so we may solve it using integrating factors. We compute
Z
4
1
µ(x) = exp −
dx = exp (−4 ln |x|) = 4 .
x
x
Multiply the differential equation above by this function:
dz
4
2
− z=− 2
dx x
x
1 dz
4
2
− 5z=− 6
4
x dx x
x
2
d 1
·z =− 6
4
dx x
x
=⇒
1
2
·z =
+C
4
x
5 x5
for some constant C. Hence, the solution to the differential equation is
r
2 + 5 C x5
1
5x
z(x) =
=⇒
y(x) = p
=±
.
5x
2
+
5 C x5
z(x)
3.4
Integrating Factors
Problem 11. The differential equation 3 (y + 1) dx − 2 x dy = 0 is not exact. Verify that µ(x, y) =
y+1
integrating factor, then solve the resulting exact equation.
x4
Solution. The resulting differential equation is
3 (y + 1)2
2 (y + 1)
dx −
dy = 0.
x4
x3
First we verify that this equation is indeed an exact equation. Denote


∂P 6 (y + 1)
3 (y + 1)2 



P (x, y) =


∂y
x4
x4
=⇒



2 (y + 1) 

 ∂Q = 6 (y + 1)
Q(x, y) = −
3
x
∂x
x4
We seek a function F (x, y) = G(x, y) + H(y) such that
∂F
= P (x, y)
∂x
and
∂F
= Q(x, y).
∂y
Define the functions
(y + 1)2
,
x3
Z Z ∂G
2 (y + 1) 2 (y + 1)
H(y) =
Q(x, y) −
dy =
−
+
dy = 0.
∂y
x3
x3
Z
G(x, y) =
Hence F (x, y) = −
constant C.
P (x, y) dx = −
(y + 1)2
= C, so the corresponding solution is (y + 1)2 + C x3 = 0 for some
x3
46
3.5
Linear Equations
Problem 12. For the following differential equation, find an integrating factor and solve the given
equation.
dy
= e2x + y − 1.
dx
Solution. This is a linear equation because it can be expressed in the form y 0 − y = e2x − 1, so
µ(x) = e−x is an integrating factor. We see that the solution is y = C ex + 1 + e2x where C is a
constant.
Problem 13. Find the function y(x) satisfying
y(0) = 2.
dy
− 2 y = e2x subject to the initial condition
dx
Solution. This is a linear differential equation, so an integrating factor is µ(x) = e−2x . We multiply
both sides of the original differential equation by this function:
dy
− 2 y = e2x
dx
e−2x
dy
− 2 e−2x y = 1
dt
d −2x e
y =1
dx
e−2x y(x) = x + C
=⇒
for some constant C. This constant is uniquely determined by the initial condition: C = y(0) = 2
so that y(t) = (x + 2) e2t .
dy
Problem 14. Show that all solutions of 2
+ x y = 2 approach a limit as x → ∞, and find the
dx
limiting value. Hint: Find the general solution then use l’Hôpital’s rule.
Solution.
2This
is a linear differential equation, and an integrating factor is the function µ(x) =
1
x
exp
. We multiply both sides of the original differential equation by this function:
2
4
dy
+ xy = 2
dx
dy x x2 /4
2
+ e
y = ex /4
dx 2
d h x2 /4 i
2
e
y = ex /4
dx
2
ex
2 /4
=⇒
x2 /4
e
Z
y(x) =
x
2 /4
et
dt + C
0
for some constant C. Hence the general solution to the differential equation is y(x) = e
C e−x
2 /4
−x2 /4
Z
x
es
2 /4
0
. In order to compute the limiting value as x increases without bound, we consider the
47
ds+
limit
Z
f (x)
lim y(x) = lim
x→∞
x→∞ g(x)
in terms of
f (x) =
x
2 /4
et
dt
and g(x) = ex
2 /4
,
0
which is independent of C. Since lim f (x) = lim g(x) = ∞, we may compute the limit of the
x→∞
x→∞
ratio using l’Hôpital’s Rule. We compute the derivatives
f 0 (x) = ex
2 /4
and g 0 (x) =
x x2 /4
e
2
f (x)
f 0 (x)
2
= lim 0
= lim
= 0.
x→∞ g(x)
x→∞ g (x)
x→∞ x
=⇒
lim
Hence every solution y(x) → 0 as x → ∞.
3.6
Loan Calculations
Problem 15. Clarissa wants to buy a new car. Her loan officer tells her that her annual rate is
8%, compounded continuously, over a four-year term. Clarissa informs her loan officer that she can
make equal monthly payments of $225. How much can Clarissa afford to borrow?
Solution. Let S(t) denote the amount of the loan at time t years. Clarissa makes yearly payments
dS
of $225 · 12 = $2700 to pay off the loan, so this function satisfies the differential equation
=
dt
0.08 S − 2700. Note that S(0) is the amount that Clarissa borrows. First we solve this differential
equation by multiplying by the integrating factor µ(t) = e−0.08 t :
d µ(t) S = µ(t)
dt
dS
− 0.08 S
dt
= −2700 µ(t)
S(t) = 33750 + C e0.08 t .
=⇒
Now at the end of four years, this loan amount should be zero. Hence, we want S(4) = 0, so that
we have the equation 33750 + C e0.08·4 = 0, implying that C = −24507.53. Finally, the amount
that Clarissa can borrow is S(0) = 33750 + C = $9242.47 .
Problem 16. Jamal wishes to invest an unknown sum in an account where interest is compounded
continuously. Assuming that Jamal makes no additional deposits or withdrawals, what annual
interest rate will allow his initial investment to double in exactly five years?
Solution. Let S(t) denote the sum of money in the account at time t years. Then if r is the annual
dS
interest rate, this function satisfies the differential equation
= r S, so that S(t) = S0 ert . We
dt
want S(5) = 2 S0 , so we find
2 S0 = S0 e5r
=⇒
2 = e5r
=⇒
5 r = ln 2
Hence the annual interest rate should be 13.9% .
48
=⇒
r=
ln 2
= 0.13862
5
Problem 17. Don and Heidi would like to purchase a new home. They’ve examined their budget
and determined that they can afford monthly payments of $1000. If the annual interest is 7.25%,
and the term of the loan is 30 years, what amount can they afford to borrow?
Solution. Let S(t) denote the amount of the loan at time t years. Don and Heidi make annual
payments of $1, 000 · 12 = $12, 000, so that the function S(t) satisfies the differential equation
dS
= 0.0725 S − 12000. Note that P (0) is the amount that Don and Heidi borrow. First we solve
dt
this differential equation by multiply by the integrating factor µ(t) = e−0.0725 t :
dS
d µ(t) S = µ(t)
− 0.0725 S = −12000 µ(t)
dt
dt
so that S(t) = 165517.24 + C e0.0725 t . Now at the end of 30 years this loan amount should be zero.
Hence we want 0 = S(30) = 165517.24 + C e0.0725·30 , so that C = −18804.11. Finally, the amount
that Don and Heidi can borrow is S(0) = 165517.24 + C = $146,713.13 .
Problem 18. A certain college graduate borrows $8000 to buy a car. The lender charges interest at
an annual rate of 10%. Assuming that interest is compounded continuously and that the borrower
makes payments continuously at a constant annual rate D, determine the payment rate D that is
required to pay off the loan in 3 years. Also determine how must interest is paid during the 3-year
period.
Solution. Let S(t) denote the amount of the loan at t years. Initially S(0) = $8000, and we wish
to have S(3) = $0. If r = 10% is the annual percentage rate of the loan, the rate of change of this
loan is given by the initial value problem
dS
− 0.10 S = −D
dt
where
S(0) = $8000.
We solve this problem using integrating factors. Multiplying this equation by µ(t) = e−0.10t we
find that
dS
− 0.10 S = −D
dt
e−0.10t
dS
− 0.10 e−0.10t S = −D e−0.10t
dt
d −0.10t e
S = −D e−0.10t
dt
=⇒
e−0.10t S(t) = 10 D e−0.10t + C
for some constant C. Upon substituting t = 0, it is easy to see that C = $8000 − 10 D, so the
solution to the initial value problem is the function
S(t) = 10 k + C e0.10t = $8000 e0.10t − 10 D e0.10t − 1 .
Upon substituting t = 3, we wish to have S(3) = $0, so we find that
8000 e0.3 − 10 D e0.3 − 1 = 0
D=
=⇒
49
8000 e0.3
= $3086.64/year.
10 (e0.3 − 1)
The total amount borrowed over the 3-year period is S(0) − S(3) = $8000. The total amount paid
for the loan over this period is 3 D = $9259.92. Hence the total amount paid towards interest
during the 3-year period is 3 D − $8000 = $1259.92.
3.7
Diffusion Problems
Problem 19. A tank contains 100 gal of pure water. A salt solution with concentration 3 lb/gal
enters the tank at a rate of 2 gal/min. Solution drains from the tank at a rate of 2 gal/min. Use
qualitative analysis to find the eventual concentration of the salt solution in the tank.
Solution. Let Q(t) denote the quantity of salt in the tank at time t minutes. We list the relevant
information in the following table.
In
Out
Flow Rate
2 gal/min
2 gal/min
Concentration
3 lb/gal
Q(t)/100 lb/gal
Rate
6 lb/min
Q(t)/50 lb/min
The amount of salt satisfies the initial value problem
dQ
Q
=6−
dt
50
where
Q(0) = 0.
This is an autonomous equation, so over time we find the equilibrium point QL = lim Q(t) that
t→∞
QL
satisfies 6 −
= 0, so that QL = 300 lb. Hence the eventual concentration is c∞ = QL /100 =
50
3 lb/gal .
Problem 20. A 50 gal tank initially contains 20 gal of pure water. Salt-water solution containing 0.5
lb of salt for each gallon of water begins entering the tank at a rate of 4 gal/min. Simultaneously,
a drain is opened at the bottom of the tank, allowing the salt-water solution to leave the tank at
rate of 2 gal/min. What is the salt content (lb) in the tank at the precise moment that the tank is
full of salt-water solution?
Solution. Let Q(t) denote the quantity of salt in pounds in the tank at time t minutes, and V (t) =
20+2 t as the volume of water in the tank at time t. We make a chart with the relevant information.
In
Out
Flow Rate
4 gal/min
2 gal/min
Concentration
0.5 lb/gal
Q(t) lb/ V (t) gal
50
Rate
2 lb/min
Q(t)/(10 + t) lb/min
We have the initial value problem
Q
dQ
=2−
dt
10 + t
where
Q(0) = 0.
We multiply by µ(t) = 10 + t to find the equation
d dQ dµ
µ(t) Q = µ
+
Q = (10 + t)
dt
dt
dt
dQ
Q
+
dt
10 + t
= 20 + 2 t.
Upon integrating we find µ(t) Q(t) = 20 t + t2 + C for some constant C. When t = 0 we want
20 t + t2
Q(t) = 0, so C = 0. This gives Q(t) =
. The tank is full of salt-water solution when
10 + t
V (t) = 50, so 50 = 20 + 2 t yields t = 15 min. Hence we find that the amount of salt is Q(15) =
20 · 15 + 152
= 21 lb .
10 + 15
Problem 21. Suppose that a solution containing a drug enters a bodily organ at the rate a cm3 /s,
with drug concentration κ g/cm3 . Solution leaves the organ at a slower rate of b cm3 /s. Further, the
faster rate of infusion causes the organ’s volume to increase with time according to V (t) = V0 + r t,
with V0 its initial volume. If there is no initial quantity of the drug in the organ, show that the
concentration of the drug in the organ is given by
"
(b+r)/r #
aκ
V0
c(t) =
1−
.
b+r
V0 + r t
Solution. Let Q(t) denote the quantity of the drug (g) in the organ at t seconds. Note that
c(t) = Q(t)/V (t) is the concentration of the drug in the organ – assuming that we have instantaneous
mixing. We make a chart with the relevant information.
In
Out
Flow Rate
a cm3 /s
b cm3 /s
Concentration
κ g/cm3
c(t) g/cm3
Rate
a κ g/s
b Q(t)/V (t) g/s
Hence the differential equation is
dQ
bQ
= aκ −
dt
V0 + r t
where
Q(0) = 0.
We choose the integrating factor
Z
b
b
V0
µ(t) = exp
dt = exp
ln
+ t + C1 = (V0 + r t)b/r .
V0 + r t
r
r
b
ln r.) Note that µ(t) = V (t)b/r . We have the product
r
d dQ dµ
dQ
bQ
µ(t) Q = µ
+
Q = µ(t)
+
= a κ (V0 + r t)b/r
dt
dt
dt
dt
V0 + r t
(We chose the constant C1 =
51
so upon integrating we find
Z
µ(t) Q(t) =
a κ (V0 + r t)b/r dt =
aκ
(V0 + r t)(b+r)/r + C2 .
b+r
We want Q(0) = 0, so the constant
C2 = −
aκ
(V0 )(b+r)/r
b+r
=⇒
µ(t) Q(t) =
i
aκ h
(V0 + r t)(b+r)/r − (V0 )(b+r)/r .
b+r
Hence the concentration is
"
(b+r)/r #
V0
µ(t) Q(t)
aκ
Q(t)
1−
.
=
=
c(t) =
V (t)
b+r
V0 + r t
(V0 + r t)(b+r)/r
Problem 22. Consider two tanks, labeled tank A and tank B for reference. Tank A contains 100
gal of solution in which is dissolved 20 lb of salt. Tank B contains 200 gal of solution in which is
dissolved 40 lb of salt. Pure water flows into tank A at a rate of 5 gal/s. There is a drain at the
bottom of tank A. Solution leaves tank A via this drain at a rate of 5 gal/s, and flows immediately
into tank B at the same rate. A drain at the bottom of tank B allows the solution to leave tank
B at a rate of 2.5 gal/s. What is the salt content in tank B at the precise moment that tank B
contains 250 gal of solution?
Solution. Let QA (t) denote the quantity of salt in tank A at time t seconds, and QB (t) as the
quantity of salt in tank B at time t seconds. Then tank A contains a constant volume of 100 gal
of solution, whereas tank B has a volume of V (t) = 200 + 2.5 t. We make a chart with the relevant
information.
Tank A
In
Out
Tank B
In
Out
Flow Rate
5 gal/s
5 gal/s
Flow Rate
5 gal/s
2.5 gal/s
Concentration
0 lb/gal
QA (t) lb/ 100 gal
Concentration
QA (t) lb/ 100 gal
QB (t) lb/ V (t) gal
Rate
0 lb/s
QA (t)/20 lb/s
Rate
QA (t)/20 lb/s
QB (t)/(80 + t) lb/s
The differential equations are
dQA
QA
=−
,
dt
20
dQB
QA
QB
=
−
,
dt
20
80 + t
52
QA (0) = 20,
QB (0) = 40.
It is clear that QA (t) = 20 e−t/20 , so that QB (t) satisfies the differential equation
dQB
QB
+
=
dt
80 + t
e−t/20 . Upon multiplying by the integrating factor µ(t) = 80 + t we find that
dQB
dµ
dQB
QB
d = (80 + t) e−t/20 .
µ(t) QB = µ
+
QB = (80 + t)
+
dt
dt
dt
dt
80 + t
We integrate to find
Z
µ(t) QB (t) =
(80 + t) e−t/20 dt = −20 (100 + t) e−t/20 + C.
When t = 0 we have µ(t) = 80 and we want QB (t) = 40, so that C = 5200. This implies QB (t) =
5200
100 + t −t/20
− 20
e
. Tank B contains 250 gal of solution when V (t) = 250 = 200 + 2.5 t, so
80 + t
80 + t
t = 20 s. The amount of salt in tank B at this time is
QB (20) =
100 + 20 −20/20
5200
− 20
e
= 52 − 24 e−1 = 43.1709 lb .
80 + 20
80 + 20
Problem 23. Consider a lake of constant volume V containing at time t an amount Q(t) of pollutant,
evenly distributed throughout the lake with a concentration c(t), where c(t) = Q(t)/V . Assume
that water containing a concentration k of pollutant enters the lake at a rate r, and that water
leaves the lake at the same rate. Suppose that pollutants are also added directly to the lake at
a constant rate P . Note that the given assumptions neglect a number of factors that may, in
some cases, be important — for example, the water added or lost by precipitation, absorption,
and evaporation; the stratifying effect of temperature differences in a deep lake; the tendency
of irregularities in the coastline to produce sheltered bays; and the fact that pollutants are not
deposited evenly throughout the lake but (usually) as isolated points around its periphery. The
results below must be interpreted in the light of the neglect of such factors as these.
(a.) If at time t = 0 the concentration of pollutant is c0 , find an expression for the concentration
c(t) at any time. What is the limiting concentration as t → ∞?
(b.) If the addition of pollutants to the lake is terminated (k = 0 and P = 0 for t > 0), determine
the time interval T that must elapse before the concentration of pollutants is reduced to 50%
of its original value; to 10% of its original value.
Solution. Say that the volume V of the lake is measured in petalitres (i.e., litres×1015 = km3 ×103 ),
the time t measured in years, and the quantity Q(t) of pollutant is measured in metric tons (i.e.,
kg × 103 ). Then the concentration c(t) of pollutant in the lake is Q(t)/V metric tons per petalitre.
First we find an initial value problem for c(t). There is a water solution containing a concentration of k metric tons per petalitre entering the lake at a rate of r petalitres per year; and pollutants
are also being added at a constant rate of P metric tons per year. Hence pollutants are being added
to the lake at (r k + P ) metric tons per year. There is a water solution containing a concentration
of c(t) metric tons per petalitre leaving the lake at a rate of r petalitres per year. Hence pollutants
are leaving the lake at r · c(t) metric tons per year. The total rate of change for the quantity Q(t)
53
dQ
of pollutant is
= (r k + P ) − r · c(t). Upon substituting Q(t) = V · c(t) in terms of the constant
dt
V , we find the initial value problem
dc
rk +P
r
=
− c
dt
V
V
where
c(0) = c0 .
Next we find the solution to this initial value problem. Multiply by µ(t) = ert/V :
ert/V
r
rk +P
dc
+ c=
dt V
V
r rt/V
r k + P rt/V
dc
+ e
c=
e
dt V
V
i rk +P
d h rt/V
e
c(t) =
ert/V
dt
V
=⇒
ert/V c(t) =
r k + P V rt/V
· e
+C
V
r
for some constant C. This constant can be determined by substituting t = 0 into this solution:
c0 =
rk +P
+C
r
=⇒
C = c0 − k −
P
.
r
In particular, the solution to the initial value problem is the function
c(t) =
P
k+
r
P
+ c0 − k −
r
e−rt/V .
As time increases without bound, the exponential becomes small. Hence the limiting concentration
P
is lim c(t) = k + .
t→∞
r
If k = 0 and P = 0, then the concentration of pollutant at time t is c(t) = c0 e−rt/V . The time
interval T that must elapse before this concentration is reduced to 0.50 c0 satisfies the equation
1
V
c0 = c0 e−rT /V so that T = ln 2 · . The time interval T that must elapse before this concen2
r
V
1
c0 = c0 e−rT /V so that T = ln 10 · .
tration is reduced to 0.10 c0 satisfies the equation
10
r
Problem 24. A pond initially contains 1,000,000 gal of water and an unknown quantity of an
undesirable chemical. Water containing 0.01 gram of this chemical per gallon flows into the pond
at a rate of 300 gal/hr. The mixture flows out at the same rate, so the amount of water in the
pond remains constant. Assume that the chemical is uniformly distributed throughout the pond.
(a.) Let Q(t) be the amount of the chemical in the pond at time t. Write down an initial value
problem for Q(t).
(b.) Solve the problem in part (a) for Q(t). How much chemical is in the pond after 1 year?
(c.) At the end of 1 year the source of the chemical in the pond is removed; thereafter pure water
flows into the pond, and the mixture flows out at the same rate as before. Write down the
initial value problem that describes this new situation.
54
(d.) Solve the initial value problem in part (c). How much chemical remains in the pond after 1
additional year (2 years from the beginning of the problem)?
(e.) How long does it take for Q(t) to be reduced to 10 g?
(f.) Plot Q(t) versus t for 3 years.
Solution. Let Q(t) denote the quantity of the chemical in grams at any time t in hours. The
concentration of chemical flowing in to the pond is 1/102 g/gal, whereas the concentration of
chemical flowing out of the pond is Q(t)/106 g/gal. These concentrations are flowing at a rate of
300 gal/hr. We summarize this in the following table:
Flowing In
Flowing Out
Concentration of Chemical
10−2 g/gal
Q(t) · 10−6 g/gal
Rate of Flow
300 gal/hr
300 gal/hr
Since the change of concentration is equal to the rate in minus the rate out, we have the differential
dQ
equation
= 300 10−2 − Q 10−6 . To solve the differential equation, multiply both sides by
dt
4
−10 :
dQ
= 3 1 − 10−4 Q
dt
dQ
−104
= 3 Q − 104
dt
−
104 dQ
=3
Q − 104 dt
1
3
dQ = − 4 dt
Q − 104
10
Z
Z
3
1
dQ = −
dt
4
Q − 10
104
3t
lnQ − 104 = − 4 + C1
10
=⇒
4
Q(t) = 104 + C e−3t/10
for some constant C = ±eC1 . When t = 0 we have Q(0) = 0, so C = −104 . Hence we find that the
4
quantity of chemical at any time t in hours is Q(t) = 104 1 − e−3t/10 . In one year, there are
t = 1 year ×
365 days 24 hours
×
= 8760 hours,
1 year
1 day
4
so that we have Q = 104 1 − e−3·8760/10 = 9277.77 grams.
Now let Q(t) denote the amount in grams of the chemical in the pond in t hours after the source
of the chemical is removed. We now have the following table:
55
Flowing In
Flowing Out
Concentration of Chemical
0 g/gal
Q(t) · 10−6 g/gal
Rate of Flow
300 gal/hr
300 gal/hr
At the time the chemical is removed, we have 9277.77 g of chemical in the pond. This gives the
initial value problem
dQ
3
=− 4Q
where
Q(0) = 9277.77.
dt
10
This is a separable equation:
dQ
3
=− 4Q
dt
10
1
3
dQ = − 4 dt
Q
10
Z
Z
3
1
dQ = −
dt
Q
104
ln |Q| = −
3t
+ C1
104
=⇒
4
Q(t) = C e−3t/10
for some constant C = ±eC1 . When t = 0 we have Q(0) = 9277.77, so C = 9277.77. Hence we find
4
that the quantity of chemical at any time t in hours is Q(t) = 9277.77 e−3t/10 . In one year, there
4
are t = 8760 hours, so that we have Q = 9277.77 e−3·8760/10 = 670.066 grams. The time it takes
104
4
ln 927.777. This works out to
to reduce to 10 g satisfies 10 = 9277.77 e−3t/10 , so that t =
3
t=
1 day
1 year
104
ln 927.777 hours ×
×
= 2.599 years.
3
24 hours 365 days
Figure 3.1 has a plot of the function:

4
−3t/104

1
−
e
10


Q(t) =


9277.77 e−3(t−8760)/104
if t ≤ 8760 hours,
if t ≥ 8760 hours.
The vertical lines denote 1 year, 2 years, and 3 years, respectively.
56
Figure 3.1: Plot of Q(t) vs. t
1.5⋅104
1⋅104
5000
-5000
0
5000
1⋅104
1.5⋅104
-5000
-1⋅104
57
2⋅104
2.5⋅104
3⋅104