Chapter 3 The Integral Applied Calculus 228
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Section 8: Differential Equations
A differential equation is an equation involving the derivative of a function. They allow us to
express with a simple equation the relationship between a quantity and it's rate of change.
Example 1
A bank pays 2% interest on its certificate of deposit accounts, but charges a $20 annual fee.
Write an equation for the rate of change of the balance,
)(tB
.
If the balance
)(tB
has units of dollars, then
)(tB
has units of dollars per year. When we think
of what is changing the balance of the account, there are two factors:
1) The interest, which increases the balance, and
2) The fee, which decreases the balance.
Considering the interest, we know each year the balance will increase by 2%, but 2% of what?
Each year that will change, since we earn interest on whatever the current balance is. We can
represent the amount of increase as 2% of the balance:
)(02.0 tB
dollars/year.
The fee already has the units of dollars/year. Since everything now has the same units, we can
put the two together, and create the equation:
20)(02.0)(
tBtB
The result is an example of a differential equation. Notice this particular equation involves both
the derivative and the original function, and so we can't simplify find
)(tB
using basic
integration.
Algebraic equations contain constants and variables, and the solutions
of an algebraic equation are typically numbers. For example, x = 3
and x = –2 are solutions of the algebraic equation x
2
= x + 6.
Differential equations contain derivatives or differentials of functions.
Solutions of differential equations are functions. The differential
equation y' = 3x
2
has infinitely many solutions, and two of those
solutions are the functions y = x
3
+ 2 and y = x
3
– 4.
You have already solved lots of differential equations: every time you
found an antiderivative of a function f(x), you solved the differential
equation y' = f(x) to get a solution y. The differential equation
y' = f(x) , however, is just the beginning. Other applications generate
different differential equations, like in the bank balance example
above.
Chapter 3 The Integral Applied Calculus 229
Checking Solutions of Differential Equations
Whether a differential equation is easy or difficult to solve, it is important to be able to check
that a possible solution really satisfies the differential equation.
A possible solution of an algebraic equation can be checked by putting the solution into the
equation to see if the result is true: x = 3 is a solution of 5x + 1 = 16 since 5(3) + 1 = 16 is
true. Similarly, a solution of a differential equation can be checked by substituting the function
and the appropriate derivatives into the equation to see if the result is true: y = x
2
is a solution
of xy' = 2y since y' = 2x and x(2x) = 2(x
2
) is true.
Example 2
Check (a) that y = x
2
+ 5 is a solution of y'' + y = x
2
+ 7 and
(b) that y = x + 5/x is a solution of y' +
y
x
= 2.
(a) y = x
2
+ 5 so y' = 2x and y'' = 2. Substituting these functions for y and y'' into the
differential equation y'' + y = x
2
+ 7, we have y'' + y = (2) + (x
2
+ 5) = x
2
+ 7, so y = x
2
+ 5 is
a solution of the differential equation.
(b) y = x + 5/x so y' = 1 – 5/x
2
. Substituting these functions for y and y ' in the
differential equation y' +
y
x
= 2 , we have
y' +
y
x
= (1 – 5/x
2
) +
1
x
(x + 5/x) = 1 – 5/x
2
+ 1 + 5/x
2
= 2, the result we wanted to verify.
Separable Differential Equations
A differential equation is called separable if the variables can be separated algebraically so that
all the x's and dx are one one side of the equation, and all the y's and dy are on the other side of
the equation. In other words, so the equation has the form
dyygdxxf )()(
.
Once separated, separable differential equations can be solved by integrating both sides of the
equation.
Example 3
Find the solution of
y
x
y
2
16
Rewriting y' is a helpful first step.
y
x
dx
dy
2
16
Chapter 3 The Integral Applied Calculus 230
Now we can multiply both sides by dx and by 2y to separate the variables.
Integrating each side,
dxxydy 162
2
2
1
2
3 CxxCy
Notice that we can combine the two constants to create a new, consolidated constant C, so we
usually only bother to put a constant on the right side.
Cxxy
22
3
As expected, there is a whole family of solutions to this differential equation.
An initial value problem is a differential equation that provides additional information about the
initial, or starting, value of the function. This allows us to then solve for the constant and find a
single solution.
Example 4
Find the solution of
y
x
y
2
16
, which satisfies
3)2( y
In the previous example we found the general solution,
Cxxy
22
3
.
Substituting in the initial condition, y = 3 when x = 2,
C 2)2(33
22
, so
C 2129
, giving
5C
.
The solution is
53
22
xxy
. Sometimes it is desirable to solve for y.
53
2
xxy
, but since the initial condition had a positive y value, we isolate the solution
53
2
xxy
Example 5
A bank pays 2% interest on its certificate of deposit accounts, but charges a $20 annual fee. If
you initially invest $3,000, how much will you have after 10 years?
You may recognize this as the example from the beginning of the section, for which we set up
the equation
20)(02.0)(
tBtB
, or more simply,
2002.0 B
dt
dB
We can separate this equation by multiply by dt and dividing by the entire expression on the
right.
Chapter 3 The Integral Applied Calculus 231
dt
B
dB
2002.0
Integrating the left side of this equation requires substitution. Let
2002.0 Bu
, so
dBu 02.0
. Making the substitution,
2002.0ln
02.0
1
ln
02.0
11
02.0
1
02.0
102.0/
2002.0
Bu
u
du
uu
du
B
dB
Integrating on the right side of the differential equation is comparably easier:
Ctdt
Together, this gives us the general solution to the differential equation:
CtB 2002.0ln
02.0
1
Now we would like to solve for B. Start by multiplying by 0.02.
CtB 02.002.02002.0ln
We can rename D = 0.02C for simplicity
DtB 02.02002.0ln
Exponentiate both sides: e
left
= e
right
Dt
B
ee
02.0
2002.0ln
Use the log rule:
Ae
A
ln
Dt
eB
02.0
2002.0
Since the right side is always positive, we can drop
the absolute value sign.
Dt
eB
02.0
2002.0
Using the rule
BABA
eee
Dt
eeB
02.0
2002.0
Rename k = e
D
t
keB
02.0
2002.0
Add 20 and divide by 0.02
1000
02.002.0
20
02.002.0
tt
keke
B
Rename A = k/0.02
1000
02.0
t
AeB
Finally, we can substitute our initial value of B = 3000 when t = 0 to solve for the constant A:
10003000
)0(02.0
Ae
2000A
This gives us the equation for the account balance after t years:
10002000
02.0
t
eB
To find the balance after 10 years, we can evaluate this equation at t = 10.
81.3442$10002000)10(
)10(02.0
eB
It's worth noting that this answer isn't exactly right. Differential equations assume continuous
changes, and it is unlikely interest is compounded continuously or the fee is extracted
continuously. However, the answer is likely close to the actual answer, and differential
equations provide a relatively simple model of a complicated situation.
Chapter 3 The Integral Applied Calculus 232
Models of Growth
The bank account example demonstrated one basic model of growth: growth proportional to the
existing quantity. Bank accounts and populations both tend to grow this way if not constrained.
This type of growth is called unlimited growth.
Unlimited Growth
If a quantity or population y grows at a rate proportional that quantity's size, it can be
modeled with unlimited growth, which has the differential equation:
ryy
, where r is a constant
Example 6
A population grows by 8% each year. If the current population is 5,000, find an equation for
the population after t years.
The population is growing by a percent of the current population, so this is unlimited growth.
y
dt
dy
08.0
Separate the variables
dtdy
y
08.0
1
Integrate both sides
Cty 08.0ln
Exponentiate both sides
Ct
y
ee
08.0
ln
Simplify both sides, using the tricks we used in the bank example
t
Aey
08.0
Now substitute in the initial condition
)0(08.0
5000 Ae
, so A = 5000.
The population will grow following the equation
t
ey
08.0
5000
.
Notice that the solution to the unlimited growth equation is an exponential equation.
When a product is advertised heavily, sales will tend to grow very quickly, but eventually the
market will reach saturation, and sales will slow. In this type of growth, called limited growth,
the population grows at a rate proportional to the distance from the maximum value.
Limited Growth
If a quantity grows at a rate proportional to the distance from the maximum value, M, it can
be modeled with limited growth, which has the differential equation:
)( yMky
, where k is a constant, and M is the maximum size of y.
Chapter 3 The Integral Applied Calculus 233
Example 7
A new cell phone is introduced. The company estimates they will sell 200 thousand phones.
After 1 month they have sold 20 thousand. How many will they have sold after 9 months?
In this case there is a maximum amount of phones they expect to sell, so M = 200 thousand.
Modeling the sales, y, in thousands of phones, we can write the differential equation
)200( yky
Since it was a new phone,
0)0( y
. We also know the sales after one month,
20)1( y
.
Solving the differential equation,
)200( yk
dt
dy
Separate the variables
kdt
y
dy
200
Integrate both sides. On the left use the substitution
yu 200
Ckty 200ln
Multiply both sides by -1, and exponentiate both sides
Ckt
y
ee
200ln
Simplify
kt
Bey
200
Subtract 200, divide by -1, and simplify
200
kt
Aey
Using the initial condition
0)0( y
,
2000
)0(
k
Ae
, so
2000 A
, giving
200A
Using the value
20)1( y
,
20020020
)1(
k
e
Subtract 200 and divide by -200
Take the ln of both sides
ke
k
ln9.0ln
Divide by -1
105.09.0ln k
As a quick sanity check, this value is positive as we would expect, indicating that the sales are
growing over time. We now have the equation for the sales of phones over time:
200200
105.0
t
eA
Finally, we can evaluate this at t = 9 to find the sales after 9 months.
26.122200200
)9(105.0
eA
thousand phones.
Limited growth is also commonly used for learning models, since when learning a new skill,
people typically learn quickly at first, then their rate of improvement slows down as they
approach mastery.
Chapter 3 The Integral Applied Calculus 234
Example 7 ** FIX**
Jim is learning a new set of product codes for work. Each day he studies, and tests his recall.
Suppose that after 4 days, Jim has mastered 70% of the new codes. How long will it take for
him to master 95% of the codes, if the limited growth model applies.
In this case there is a maximum amount of mastery possible, so M = 100%. Modeling his
learning, L, as percent of mastery, we can write the differential equation
(100 )L k L

Solving as we did in the previous example, we obtain
100
kt
L Ae

.
Since he started not knowing any of the codes
(0) 0L
, so
0
0 100
k
Ae


, so
100A 
.
Using that Jim mastered 70 of the codes in 4 days,
(4) 70L
, so
4
70 100 100
k
e

4
30 100
k
e

4
0.3
k
e

4
ln 0.3 ln 4
k
ek

ln 0.3
0.301
4
k 
Now we have the equation
0.301
100 100
t
Le
, we can solve for when L = 95.
0.301
95 100 100
t
e
0.301
5 100
t
e
0.301
0.05
t
e
0.301
ln 0.05 ln 0.301
t
et
ln 0.05
9.95
0.301
t 
Jim will have reached 95% mastery after about 10 days.
Earlier we used unlimited growth to model a population, but often a population will be
constrained by food, space, and other resources. When a population grows both proportional to
its size, and relative to the distance from some maximum, that is called logistic growth. This
leads to the differential equation
)( yMkyy
, which is accurate but not always convenient to
use. We will use a slight modification. Since solving this differential equation requires
integration techniques we haven't learned, the solution form is given.
Chapter 3 The Integral Applied Calculus 235
Logistic Growth
If a quantity grows at a rate proportional to its size and to the distance from the maximum
value, M, it can be modeled with logistic growth, which has the differential equation:
M
y
ryy 1
r can be interpreted as "the growth rate absent constraints" - how the population would
grow if there wasn't a maximum value.
This differential equation has solutions of the form
rt
Ae
M
y
1
Example 8
A colony of 100 rabbits is introduced to a reclaimed forest. After 1 year, the population has
grown to 300. It is estimated the forest can sustain 5000 rabbits. The forest service plans to
reintroduce wolves to the forest when the rabbit population reaches 3000 rabbits. When will
that occur?
The maximum sustainable population was given as M = 5000. Using the solution form,
rt
Ae
y
1
5000
Using the
100)0( y
, we can solve for A
)0(
1
5000
100
r
Ae
Simplify
A
1
5000
100
Multiply both sides by 1+A
Divide by 100
501 A
49A
Now, using
300)1( y
, we can solve for r.
)1(
491
5000
300
r
e
5000491300
r
e
300
5000
491
r
e
3197.0
49
1
3
50
r
e
Chapter 3 The Integral Applied Calculus 236
3197.0ln r
1404.13197.0ln r
We now have the equation for the population after t years.
t
e
y
1404.1
491
5000
To answer the original equation, of when the rabbit population will reach 3000, we need to
solve for t when y = 3000.
t
e
1404.1
491
5000
3000
50004913000
1404.1
t
e
01361.0
49
1
3
5
1404.1
t
e
01361.0ln1404.1 t
77.3
1404.1
01361.0ln
t
years.
Logistic growth is also a good model for unadvertised sales. A new product that is not
advertised will have sales increase slowly at first, then grow as word of mouth spreads and
people become familiar with the product. Sales will level off as they approach market saturation.
3.8 Exercises
In problems 1 – 4, check that the function y is a solution of the given differential equation.
1. y' + 3y = 6. y = e
–3x
+ 2. 2. y' – 2y = 8. y = e
2x
– 4.
3. y' = – x/y. y = 7 – x
2
. 4. y' = x – y. y = x – 1 + 2e
–x
.
In problems 5 – 8 check that the function y is a solution of the given initial value problem.
5. y' = 6x
2
– 3 and y(1) = 2 . y = 2x
3
– 3x + 3.
6. y' = 6x + 4 and y(2) = 3. y = 3x
2
+ 4x – 17.
7. y' = 5y and y(0) = 7. y = 7e
5x
.
8. y' = –2y and y(0) = 3. y = 3e
–2x
.
Chapter 3 The Integral Applied Calculus 237
In problems 9 – 12, a family of solutions of a differential equation is given. Find the value of the
constant C so the solution satisfies the initial value condition.
9. y' = 2x and y(3) = 7. y = x
2
+ C. 10. y' = 3x
2
– 5 and y(1) = 2. y = x
3
– 5x + C.
11. y' = 3y and y(0) = 5. y = Ce
3x
. 11. y' = –2y and y(0) = 3. y = Ce
–2x
.
In problems 13 – 18, solve the differential equation. (Assume that x and y are restricted so
that division by zero does not occur.)
13. y' = 2xy 14. y' = x/y 15. xy' = y + 3
16. y' = x
2
y + 3y 17. y' = 4y 18. y' = 5(2 – y)
In problems 19 – 22, solve the initial value separable differential equations.
19. y' = 2xy for y(0) = 3, y(0) = 5, and y(1) = 2.
20. y' = x/y for y(0) = 3, y(0) = 5, and y(1) = 2.
21. y' = 3y for y(0) = 4, y(0) = 7, and y(1) = 3.
22. y' = –2y for y(0) = 4, y(0) = 7, and y(1) = 3.
23. The rate of growth of a population P(t) which starts with 3,000 people and increases by 4% per year
is P '(t) = 0.0392
.
P(t). Solve the differential equation and use the solution to estimate the population
in 20 years.
24. The rate of growth of a population P(t) which starts with 5,000 people and increases by 3% per
year is P '(t) = 0.0296
.
P(t). Solve the differential equation and use the solution to estimate the
population in 20 years.
25. A manufacturer estimates that she can sell a maximum of 130 thousand cell phones in a city.
By advertising heavily, her total sales grow at a rate proportional to the distance below this
upper limit. If she enters a new market, and after 6 months her total sales are 59 thousand
phones, find a formula for the total sales (in thousands) t months after entering the market,
and use this to estimate the total sales at the end of the first year.
26. The temperature of a turkey in the oven will grow like limited growth. The turkey starts out
at 40 degrees Fahrenheit, and is placed into a 350 degree oven. After 30 minutes, the
Chapter 3 The Integral Applied Calculus 238
turkey's temperature has risen to 55 degrees. How long will it take until the turkey's
temperature reaches 165 degrees?
27. A new cell phone is introduced into the market. It is predicted that sales will grow
logistically. The manufacturer estimates that they can sell a maximum of 100 thousand cell
phones. After 44 thousand cell phones have been sold, sales are increasing by 4 thousand
phones per month. Use this to estimate the total sales at the end of the first year.
28. Biologists stocked a lake with 400 fish and estimated the carrying capacity of the lake to
be 8000 fish. The number of fish tripled in the first year. How long will it take the
population to increase to 4000?