Antiderivatives and Differential Equations
Definition: A function F(x) is called an antiderivative of f(x) on an interval I if
dF(x)
 F'(x)  f(x) for
dx
all x in I. (The antiderivative of f(x) is a function you differentiate to get f(x).)
Let f(x) = cos(x) then an antiderivative of f(x) is F(x) = sin(x). But there are many antiderivatives of
f(x); for example G1(x) = sin(x) + 23 and G2(x) = sin(x) – π. In fact any expression of the form sin(x)
+ C, where C is an arbitrary constant, is an antiderivative of f(x). It can be shown that if two functions
have identical derivatives on an interval, then they must differ by a constant. Thus if F and G are
antiderivatives of f(x), then F'(x)  f(x)  G'(x) , so G(x) – F(x) = C, where C is a constant. So we can
write this as G(x) = F(x) + C. This gives us the following result.
If F(x) is an antiderivative of f(x) on the interval I, then the most general antiderivative of f(x) on I is
F(x) + C, where C is an arbitrary constant.
(Source: Calculus with Early Transcendentals, Stewart)
Comment: The independent variable need not be x; a common alternate choice is t.
A short table of common functions, derivatives, and antiderivatives in (basic) engineering.
Function: f(x)
Derivative: f '(x) 
df(x)
dx
1
x
n
n 1
nx
2
eax
aeax
sin(bx)
bcos(bx)
cos(bx)
1
a  bx
-bsin(bx)
b
k
0
k f(x)
k f '(x)
f1(x)  f2 (x)
df1(x) df2 (x)

dx
dx
3
4
5
 a  bx 2
Antiderivative F(x) ; C an arbitrary constant
xn  1
 C,n  1,for n  1, F(x)  ln( x )  C
n 1
1
F(x)  e ax  C
a
1
F(x)  cos(bx)  C
b
1
F(x)  sin(bx)  C
b
1
F(x)  ln(a  bx)  C
b
F(x) 
6
7
8
F(x) = kx + C
Antiderivative of kf(x) = k F(x)  C
Antiderivative of f1(x)  f2 (x) = F1(x)  F2 (x)  C
Copyright by David R. Hill 2015 Mathematics Department Temple University
Observe that the antiderivative of a function is a “family” or set of functions since C is an arbitrary
constant. Let f(x) = 2x, then the most general antiderivative is F(x) = x2 + C. The next figure shows
several members of the family of antiderivatives of 2x.
Rectilinear Motion
Rectilinear motion (also called linear motion) is a motion along a straight line, and can therefore be
described mathematically using only one spatial dimension.
If we assume constant acceleration g, then we can use antiderivatives to find expressions for the
velocity, v(t), and the position y(t) (if the motion is vertical) or x(t) (if the motion is horizontal). Here t
represents time.
3.76 m/s2
Mercury:
Venus
Earth
:
:
Jupiter :
23.6 m/s2
Saturn :
10.06 m/s2
Uranus :
8.87 m/s2
Neptune:
11.23 m/s2
9.04 m/s2
9.8
m/s2,
32
ft/s2
Moon: 1.622 m/s²
Mars
:
3.77 m/s2
(Source: http://planetaryfacts.blogspot.com/2012/04/g-planets-of-solar-system.html )
Gravity decreases with altitude; as one rises above the Earth's surface because a greater
altitude means a greater distance from the Earth's center. All other things being equal, an
increase in altitude from sea level to 9,000 meters (30,000 ft) causes a weight decrease of
about 0.29%. (An additional factor affecting apparent weight is the decrease in air density at
Copyright by David R. Hill 2015 Mathematics Department Temple University
altitude, which lessens an object's buoyancy. This would increase a person's apparent weight
at an altitude of 9,000 meters by about 0.08%)
It is a common misconception that astronauts in orbit are weightless because they have flown
high enough to escape the Earth's gravity. In fact, at an altitude of 400 kilometers (250 mi),
equivalent to a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the
Earth's surface. Weightlessness actually occurs because orbiting objects are in free-fall.
The effect of ground elevation depends on the density of the ground. A person flying at
30,000 ft above sea level over mountains will feel more gravity than someone at the same
elevation but over the sea. However, a person standing on the earth's surface feels less gravity
when the elevation is higher.
The following formula approximates the Earth's gravity variation with altitude:
where

gh is the gravitational acceleration at height h above sea level.

re is the Earth's mean radius.
g0 is the standard gravitational acceleration.

The formula treats the Earth as a perfect sphere with a radially symmetric distribution of mass.
(Source: https://en.wikipedia.org/wiki/Gravity_of_Earth#Altitude)
Basic equations of rectilinear motion
Velocity v(t) = the instantaneous rate of change of distance y(t) with respect to time
v( t ) 
dy( t )
dt
Acceleration a(t) = the instantaneous rate of change of distance y(t) with respect to time
a( t ) 
dv( t )
dt
Suppose that a(t) = g, a constant. Then the antiderivative of a(t) is v(t), so v(t) = gt + C1,
where C1 is an arbitrary constant. Similarly, Then the antiderivative of v(t) is y(t), so
g 2
yt 
t  C1t  C2 , where C2 is an arbitrary constant. Summarizing we have the basic
2
equations
a( t )  g
v( t )  gt  C1
g 2
t  C1t  C2
2
Copyright by David R. Hill 2015 Mathematics Department Temple University
yt 
Example: An object is dropped from a height of 2 meters at time t = 0. Determine the formulas for
velocity and the distance of the object above the ground.
Let a(t) = -9.8 m/sec2, then since the object was dropped v(0) = 0. Using v(t) = gt + C1 we have
that C1 = 0. Then v(t) = -9.8 t. Then y(t), the height of the object above the ground is
9.8 2
9.8 2
yt 
t + C2 . Since y(0) = 2 m we have that C2 = 2. Thus y  t  
t + 2.
2
2
Now we can determine the time at which the object hits the ground and its impact velocity. To
9.8 2
determine the time of impact set y(t) = 0 and solve for t; 0 
t + 2  t ≈ 0.6389 sec. and
2
the impact velocity is v ≈ 6.2610 m/sec.
Example: An object is thrown up from ground level with initial velocity 20 ft/sec. Determine the
formulas for velocity and the distance of the object above the ground.
Let a(t) = -32 ft/sec2, then since the object was tossed starting at ground level at 20 ft/sec we
have v(0) = 20. Using v(t) = -32t + C1 we have that C1 = 20. Thus v(t) = -32t + 20. Then y(t), the
height of the object above the ground is y  t   16t 2 + 20t  C2 . Since y(0) = 0 ft. we have that
C2 = 0, and so y  t   16t 2 + 20t .
Now we can determine the time at which the object returns to the ground, its impact velocity, and
the maximum height of the object. To determine the time the object returns to earth set y(t) = 0
and solve for t; 0  16t 2 + 15t  t = 0 or 20/16 = 1.25 sec. and the impact velocity is v = -20
ft/sec. The maximum height of the object occurs when v(t) = 0. The time when the velocity is zero
is t= 5/8 sec. then y(5/8) = 6.25 ft. is the maximum height.
Example: Two workmen are repairing things on a shed 15 feet high. One workman is atop the
shed and the other is on the ground near its base. The workman on the ground tosses a tool to
his partner on the roof using an initial velocity of 40 ft/sec. At what times can the man on the roof
“try” to catch the tool as it passes the top of the shed?
We know v(0) = 40ft/sec and y(0) = 0. So we have v(t) = -32t + 40 and y  t   16t 2 + 20t . To
find the times the man on the roof should try to catch the tool, set y(t) = 15 and solve for t; solving
15  16t 2 + 20t we get that t is approximately 0.4594 sec. on the way up and 2.0406 sec. on
the way down.
Some (simple) applications from differential equations
Population growth
Falling object with air resistance
Radioactive decay
Measuring collisions for belted vs. unbelted
passengers
Compound interest
Newton’s Law of heating/cooling
Building designs for snow loading
Mixing problems
Design of energy absorbing aviation seats
Copyright by David R. Hill 2015 Mathematics Department Temple University
A first order differential equation (DE) involving independent variable t and dependent variable y
has the form
dy
 f(t, y)
dt
where f is a function of t and y. It is called first order since only first derivatives are involved. If an
auxiliary condition, like y(a) = b, accompanies the differential equation then the pair
dy
 f(t, y) , y(a)  b
dt
is called an initial value problem (IVP). (In the examples above we had auxiliary conditions
specifies, so we were solving IVPs.) The condition y(a) = b specifies a point t = a, y = b that lies on
the solution curve. It would be nice if we could solve every first order differential equation.
Unfortunately, that is not always possible.
dy( t )
dv( t )
and a( t ) 
.
dt
dt
We solved these using antiderivatives since v(t) and a(t) were functions strictly in terms of time t. Not
all first order DEs are of this form, but there is a certain type of first order DE that can be solved by
antiderivatives. If we can determine the antiderivatives we can complete a solution of the DE.
Previously in our discussion of rectilinear motion we had first order DEs v( t ) 
dy
 f(t, y) can be factored into a
dt
function of t times a function of y. That is, the DE can be “rearranged” into the form
Definition: A first order DE is called separable provided f(t, y) in
dy
 g(t) h(y) ,
dt
then if h(y) ≠ 0 we have
1
dy  g(t) dt .
h(y)
If we can find the antiderivative of each side we have an expression which, in general, gives an
implicit expression for the solution y(t) for the DE.
Example: (a) Solve the DE
dy t 3
 .
dt
y
1 2 1 2
y  t C.
2
4
(We could have put a constant C1 on the left side and a constant C2 or the right side. But then we can
rearrange to have C2 – C1 on the right side and rename C2 – C1 = C.) Of course we can simplify the
1
1
expression y2  t 2  C by multiplying both sides by 4 and renaming 4C to C; we then have
2
4
Separating the variables we have ydy = t3dt. Finding the antiderivatives we have
2y2  t 2  C .
dy t 3
 , y(1)  5 .
dt
y
Copyright by David R. Hill 2015 Mathematics Department Temple University
(b) Find the solution of the IVP
Using 2y2  t 2  C we set t = 1 and y = 5 giving us 2(5)2 = 1 + C  C = 49. Hence the solution of the
initial value problem 2y2  t 2  49 . (See the next figure.)
Note: The solution of the DE is a family of curves because of arbitrary constant C, while the
solution of the IVP is a particular member of the family. Here the particular member is the one
that goes through the point t = 1 and y = 5.
Example: Population Growth
Let P(t) represent the population where there is no immigration or emigration. (Emigration is the act of
leaving one's native country with the intent to settle elsewhere. Conversely, immigration describes the
movement of persons into one country from another.) One model for the rate at which the population
is growing says the rate of growth is proportional to the population. If the population has a continuous
growth rate of r% per unit of time, then we have the DE
dP
 rP
dt
dP
dP
 r dt or equivalently
 r dt . Determining the antiderivatives
P
P
using our table we have from row 5 that the antiderivative of the left side is ln(P) (for a = 0 and b = 1)
and the antiderivative for the right side rt uses row 6. Then the solution of the DE is ln(P) = rt + C.
Solve for P by using the exponential function we obtain eln(P)  ert  C  ert eC and simplifying we get P
= Cert where we have renamed eC to be C. This equation provides a model for unlimited population
Copyright by David R. Hill 2015 Mathematics Department Temple University
Separating the variable we have
growth and is only applicable for short periods of time or in cases where the population is quite
isolated.
Determining the growth rate r is often time consuming. We can estimate the growth rate using
population data as follows. (The used data is hypothetical.)
Suppose we know that at t = 0 that P= 450 and at t = 3 that P= 510. Then in the model
equation P = Cert we have 450 = Ce0r hence C = 450. Now the model equation is P = 450 ert.
Using the second set of data we have 510 = 450 e3r so solving for r we get
ln(510 / 450)
r
 0.0417 .
3
Thus the growth rate is approximately 4% over the time interval [0 ,3].
Example: Mixing Problem
Typical mixing problems employ a tank or other receptacle that holds a fluid. For illustration consider
a tank that contains water that has salt dissolved in it, and assume the mixture is kept well stirred.
The simple example we use below should not be underestimated since tanks can be replaced by the
heart, stomach, or gastrointestinal systems.
Suppose we have an input to the tank which is water containing a certain concentration of salt. In
addition we have an output, or drain, from the bottom of the tank. Let y(t) represent the number of
dy
pounds of salt in the tank at time t. Then
represents the rate of change of the amount of salt in the
dt
tanks, hence the units are pounds per unit of time (often we use lbs/min). The differential equation
that models this situation is given by
dy
 rate in of salt  rate out of salt
dt
Both the in rate and out rate of salt are composed of a product of the fluid flow rate times the
concentration of the salt in the fluid:
Rate in = fluid flow in rate × input concentration of salt
Out rate = fluid flow out rate × output concentration of salt
The fluid flow rates must be specified as well as the input concentration. The output concentration is
trickier; it depends on the volume V(t) of the fluid in the tank at time t and the y(t), the amount of salt
in the tank at time t. Consider the following situation.
A 500 gallon tank contains 300 gallons of salt solution containing 25 lbs of salt. The input flow rate
is 4 gal/min and the input concentration 2lb/gal while the output flow rate is also 4 gal/min. Note
that the in and out flow rates are equal so that there is always 300 gallons of solution in the tank.
The IVP modelling this situation is
Copyright by David R. Hill 2015 Mathematics Department Temple University
ylb
dy
 4 gal / min  2 lb / gal  4 gal/ min 
, y(0)  25
dt
300 gal
dy
4y
 8
, y(0)  25
dt
300
To solve the DE we use separation of variables. We simplify the equation before separating we
dy 2400  4 y

get
. The separation of variables yields
dt
300
dy
dt

2400  4 y 300
Applying rows 5 and 6 from our antiderivative table we have
1
1
ln 2400  4 y 
tC
4
300
Applying the initial condition y(0) = 25 we get C 
1
ln(2400  100)  1.9351 hence we an
4
implicit solution of the IVP given by
1
1
ln 2400  4 y 
t  1.9351
4
300
Using exponential function we solve for y:
4
t  7.7407
1
1
4
ln 2400  4 y
 e 300
ln 2400  4 y 
t  1.9351  ln 2400  4 y 
t  7.7407  e
4
300
300
 2400  4y 
4
t
7.7407 300
e
e
 y
4
t
7.7407 300
2400  e
e
4

4
t
300
600  575e
(Sources: Calculus Early Transcentdentals by Stewart, Differential Equations by Polking,
Boggess, and Arnold, Introductory Mathematics for Engineers by Rattan and Klingbeil.
Applied Calculus By Hughes-Hallet et. al.)
Copyright by David R. Hill 2015 Mathematics Department Temple University