Contributed by:

Differential equations arise when we can relate the rate of change of some quantity back to the quantity itself. Here we will discuss some applications of differential equations.

1.
Math 104

2.
Differential Equations

•The most important application of integrals is to the

solution of differential equations.

•From a mathematical point of view, a differential

equation is an equation that describes a relationship

among a function, its independent variable, and the

derivative(s) of the function.

•The most important application of integrals is to the

solution of differential equations.

•From a mathematical point of view, a differential

equation is an equation that describes a relationship

among a function, its independent variable, and the

derivative(s) of the function.

3.
For example:

dy 2

3xy

dx

2

d y dy

2

4 3 y 1

dx dx

ORDER = highest derivative: first order, second order...

dy 2

3xy

dx

2

d y dy

2

4 3 y 1

dx dx

ORDER = highest derivative: first order, second order...

4.
To solve a differential equation:

…means to find a function y(x) that makes it true.

2 dy 2

y 2 solves 3xy

3x dx

2

x d y dy

y e 1

3

solves

2

4 3 y 1

dx dx

…means to find a function y(x) that makes it true.

2 dy 2

y 2 solves 3xy

3x dx

2

x d y dy

y e 1

3

solves

2

4 3 y 1

dx dx

5.
In Applications

Differential equations

arise when we can relate

the rate of change of some

quantity back to the

quantity itself.

Differential equations

arise when we can relate

the rate of change of some

quantity back to the

quantity itself.

6.
Example (#1)

The acceleration of gravity is constant (near the

surface of the earth). So, for falling objects:

the rate of change of velocity is constant

dv

g

dt

Since velocity is the rate of change of position, we could

write a second order equation:

2

d x

2

g

dt

The acceleration of gravity is constant (near the

surface of the earth). So, for falling objects:

the rate of change of velocity is constant

dv

g

dt

Since velocity is the rate of change of position, we could

write a second order equation:

2

d x

2

g

dt

7.
Example (#2)

Here's a better one -- with air resistance, the acceleration of a

falling object is the acceleration of gravity minus the

acceleration due to air resistance, which for some objects is

proportional to the square of the velocity. For such an object we

have the differential equation:

rate of change of velocity is

gravity minus

something proportional to velocity squared

2

dv 2 d x 2

dx

g kv or 2

g k

dt dt dt

Here's a better one -- with air resistance, the acceleration of a

falling object is the acceleration of gravity minus the

acceleration due to air resistance, which for some objects is

proportional to the square of the velocity. For such an object we

have the differential equation:

rate of change of velocity is

gravity minus

something proportional to velocity squared

2

dv 2 d x 2

dx

g kv or 2

g k

dt dt dt

8.
Example (#3)

In a different field:

Radioactive substances decompose at a rate

proportional to the amount present.

Suppose y(t) is the amount present at

time t.

rate of change of amount is

proportional to the amount (and decreasing)

dy

k y

dt

In a different field:

Radioactive substances decompose at a rate

proportional to the amount present.

Suppose y(t) is the amount present at

time t.

rate of change of amount is

proportional to the amount (and decreasing)

dy

k y

dt

9.
Other problems that

yield the same

equation:

In the presence of abundant resources (food and

space), the organisms in a population will reproduce

as fast as they can --- this means that

the rate of increase of the population will be

proportional to the population itself:

dP

k P

dt

yield the same

equation:

In the presence of abundant resources (food and

space), the organisms in a population will reproduce

as fast as they can --- this means that

the rate of increase of the population will be

proportional to the population itself:

dP

k P

dt

10.
..and another

The balance in an interest-paying bank

account increases at a rate (called the interest

rate) that is proportional to the current

balance. So

dB

kB

dt

The balance in an interest-paying bank

account increases at a rate (called the interest

rate) that is proportional to the current

balance. So

dB

kB

dt

11.
More realistic situations for the

last couple of problems

For populations: An ecosystem may have a maximum capacity to

support a certain kind of organism (we're worried about this

very thing for people on the planet!).

In this case, the rate of change of population is proportional both

to the number of organisms present and to the amount of excess

capacity in the environment (overcrowding will cause the

population growth to decrease).

If the carrying capacity of the environment is the constant Pmax ,

then we get the equation:

dP

kP Pmax P

dt

last couple of problems

For populations: An ecosystem may have a maximum capacity to

support a certain kind of organism (we're worried about this

very thing for people on the planet!).

In this case, the rate of change of population is proportional both

to the number of organisms present and to the amount of excess

capacity in the environment (overcrowding will cause the

population growth to decrease).

If the carrying capacity of the environment is the constant Pmax ,

then we get the equation:

dP

kP Pmax P

dt

12.
and for the Interest

Problem...

For annuities: Some accounts pay interest but

at the same time the owner intends to withdraw

money at a constant rate (think of a retired

person who has saved and is now living on the

Problem...

For annuities: Some accounts pay interest but

at the same time the owner intends to withdraw

money at a constant rate (think of a retired

person who has saved and is now living on the

13.
Question:

If the interest rate is r , and the retiree wants to

withdraw W dollars per year, which is the correct

differential equation for the balance B in the account

at time t?

dB dB

A) rB W

D) rB WB

dt dt

dB dB

B) rB W E) r B W

dt dt

dB

C) rB WB

dt

If the interest rate is r , and the retiree wants to

withdraw W dollars per year, which is the correct

differential equation for the balance B in the account

at time t?

dB dB

A) rB W

D) rB WB

dt dt

dB dB

B) rB W E) r B W

dt dt

dB

C) rB WB

dt

14.
Another application:

According to Newton's law of cooling , the

temperature of a hot or cold object will

change at a rate proportional to the difference

between the object's temperature and the

ambient temperature.

temperature

If the ambient temperature is kept constant at

A, and the object's temperature is u(t), what is

the differential equation for u(t) ?

According to Newton's law of cooling , the

temperature of a hot or cold object will

change at a rate proportional to the difference

between the object's temperature and the

ambient temperature.

temperature

If the ambient temperature is kept constant at

A, and the object's temperature is u(t), what is

the differential equation for u(t) ?

15.
Solving Differential

Equations

Since the the process of solving of a differential equation

recovers a function from knowing something about its

derivative, it's not too surprising that we have to use integrals to

solve differential equations.

And since we’re using integrals, we should also expect to see

some "arbitrary" constants in the solutions of differential

equations. In general, there will be one constant in the solution

of a first-order equation, two in a second-order one, etc...

Equations

Since the the process of solving of a differential equation

recovers a function from knowing something about its

derivative, it's not too surprising that we have to use integrals to

solve differential equations.

And since we’re using integrals, we should also expect to see

some "arbitrary" constants in the solutions of differential

equations. In general, there will be one constant in the solution

of a first-order equation, two in a second-order one, etc...

16.
In practice...

In practice, we can solve for the constants

by having some information about the value

of the unknown function (and/or the value

of its derivative(s)) at some point. From an

applications point of view, such initial

conditions are clearly needed, since you

can't determine the value of something just

from information about how it is changing.

You also need to know its value at some

("initial") time.

In practice, we can solve for the constants

by having some information about the value

of the unknown function (and/or the value

of its derivative(s)) at some point. From an

applications point of view, such initial

conditions are clearly needed, since you

can't determine the value of something just

from information about how it is changing.

You also need to know its value at some

("initial") time.

17.
Some examples will make this clear

Let's go back to the very first example,

dy 2

3xy

dx

This is an example of a separable first-order equation

(the only kind we'll worry about today).

If you view dy and dx as variables (so you can multiply both

sides by dx), you can get all the x's on one side and all the y's on

the other by algebraic manipulation. Here, you can write:

dy

2

3 x dx

y

Let's go back to the very first example,

dy 2

3xy

dx

This is an example of a separable first-order equation

(the only kind we'll worry about today).

If you view dy and dx as variables (so you can multiply both

sides by dx), you can get all the x's on one side and all the y's on

the other by algebraic manipulation. Here, you can write:

dy

2

3 x dx

y

18.
Equation of differentials...

dy

2

3 x dx

y

This is an actual "equation of differentials". Then,

simply integrate both sides:

dy

y 2 3x dx

1 3 2

2 x C

y

dy

2

3 x dx

y

This is an actual "equation of differentials". Then,

simply integrate both sides:

dy

y 2 3x dx

1 3 2

2 x C

y

19.
1 3 2 (You only need one constant of

2 x C integration).

y

This is called the "general solution" of the differential

We can determine C if we were given one point on the graph

of the function y(x).

For instance, if you were given that y(1)=2 , then you could

substitute 2 for y and 1 for x and get: 12 32 C

and so you would conclude that C = -2, so the solution of the

initial-value problem:

dy

3 xy 2 , y (1) 2

dx

1 3 2 2

is 2 x 2 , or (better): y

y 4 3x 2

2 x C integration).

y

This is called the "general solution" of the differential

We can determine C if we were given one point on the graph

of the function y(x).

For instance, if you were given that y(1)=2 , then you could

substitute 2 for y and 1 for x and get: 12 32 C

and so you would conclude that C = -2, so the solution of the

initial-value problem:

dy

3 xy 2 , y (1) 2

dx

1 3 2 2

is 2 x 2 , or (better): y

y 4 3x 2

20.
DiffEq Problem:

If the function y = f (x) satisfies the initial-value problem

dy

x 2 y x 2 , f (0) 5

dx

then f (1) =

A) e 3 1 1

E) 5

3e e

B) 4

1 e F) 1 3

5 e

C)

e G) 10

D) 2 H) e

If the function y = f (x) satisfies the initial-value problem

dy

x 2 y x 2 , f (0) 5

dx

then f (1) =

A) e 3 1 1

E) 5

3e e

B) 4

1 e F) 1 3

5 e

C)

e G) 10

D) 2 H) e

21.
“DiffEq Greatest Hits”

A tank contains 1000 liters of brine (salty water) with 50 kg of

dissolved salt. Pure water enters the tank at the rate of 25 liters

per minute, The solution is kept thoroughly mixed and drains at

an equal rate. How many kg of salt remain in the tank after 10

A tank contains 1000 liters of brine (salty water) with 50 kg of

dissolved salt. Pure water enters the tank at the rate of 25 liters

per minute, The solution is kept thoroughly mixed and drains at

an equal rate. How many kg of salt remain in the tank after 10

22.
The setup….

The first step in most DiffEq problems is to identify the

unknown function.

Since we want to know the amount of salt at different

times, use A(t) for the amount of salt (in kg) in the tank

at time t (minutes). We are given that A(0)=50. The rate

of change of A could come from salt being added to the

tank (but there is none), or from salt flowing out of the

tank (the solution flows out at 25 liters per minute, and

there are A(t) kg in 1000 liters, so there are A(t)/40 kg in

25 liters. So, which of the following is the differential

equation for this problem?

A. A' = A/40 C. A' = 40 - A E. A' = - 40/A

B. A' = A - 40 D. A' = - A/40

The first step in most DiffEq problems is to identify the

unknown function.

Since we want to know the amount of salt at different

times, use A(t) for the amount of salt (in kg) in the tank

at time t (minutes). We are given that A(0)=50. The rate

of change of A could come from salt being added to the

tank (but there is none), or from salt flowing out of the

tank (the solution flows out at 25 liters per minute, and

there are A(t) kg in 1000 liters, so there are A(t)/40 kg in

25 liters. So, which of the following is the differential

equation for this problem?

A. A' = A/40 C. A' = 40 - A E. A' = - 40/A

B. A' = A - 40 D. A' = - A/40

23.
Answer this...

Now, what is the answer to the problem?

(i.e., what is A(10) if A’= -A/40 and A(0) = 50 ?)

A. 0

B. 40

1

C. 50e 4

1

D. 50e 4

1

E. 50e 40

1

F. 25e 10

G. 50 ln(2)

H. 25

Now, what is the answer to the problem?

(i.e., what is A(10) if A’= -A/40 and A(0) = 50 ?)

A. 0

B. 40

1

C. 50e 4

1

D. 50e 4

1

E. 50e 40

1

F. 25e 10

G. 50 ln(2)

H. 25

24.
Growth and decay:

Connect

What is the solution of the differential equation

y ’ = ky ?

How about the initial value problem

y ’ = ky , y(0) = y0 ?

As noted previously, this differential equation is useful

for talking about radioactive decay, compound interest

and unrestricted population growth.

Connect

What is the solution of the differential equation

y ’ = ky ?

How about the initial value problem

y ’ = ky , y(0) = y0 ?

As noted previously, this differential equation is useful

for talking about radioactive decay, compound interest

and unrestricted population growth.

25.
One more greatest hits problem:

For obvious reasons, the dissecting room of a medical examiner

is kept very cool, at a constant temperature of 5 degrees C.

While doing an autopsy early one morning, the medical

examiner himself is killed. At 10 am, the examiner's assistant

discovers the body and finds its temperature to be 23 degrees C,

and at noon the body's temperature is down to 18.5 degrees C.

Assuming that the medical examiner had a normal temperature

of 37 degrees C when he was alive, when was he murdered?

A. 3 am E. 7 am

B. 4 am F. 8 am

C. 5 am G. 9 am

D. 6 am

For obvious reasons, the dissecting room of a medical examiner

is kept very cool, at a constant temperature of 5 degrees C.

While doing an autopsy early one morning, the medical

examiner himself is killed. At 10 am, the examiner's assistant

discovers the body and finds its temperature to be 23 degrees C,

and at noon the body's temperature is down to 18.5 degrees C.

Assuming that the medical examiner had a normal temperature

of 37 degrees C when he was alive, when was he murdered?

A. 3 am E. 7 am

B. 4 am F. 8 am

C. 5 am G. 9 am

D. 6 am

26.
Geometry of Differential Equations

A differential equation of the form

dy

f ( x, y )

dx

gives geometric information about the graph of y(x).

If the graph of y(x)

It tells us:

goes through the

point (x,y), then the

slope of the graph at dy

that point is equal to f ( x, y )

f(x,y). dx

A differential equation of the form

dy

f ( x, y )

dx

gives geometric information about the graph of y(x).

If the graph of y(x)

It tells us:

goes through the

point (x,y), then the

slope of the graph at dy

that point is equal to f ( x, y )

f(x,y). dx

27.
An example:

We can draw a picture of this as follows. For the

dy

differential equation y x , we have:

dx

If the graph goes through (2,3), the slope must be 1 there.

If the graph goes through (0,0), the slope must be 0 there.

If the graph goes through (-1,-2), the slope must be -1 there.

We can draw a picture of this as follows. For the

dy

differential equation y x , we have:

dx

If the graph goes through (2,3), the slope must be 1 there.

If the graph goes through (0,0), the slope must be 0 there.

If the graph goes through (-1,-2), the slope must be -1 there.

28.
Put it on a graph...

The slope of the arrow at any point is equal to y - x at that point.

This kind of picture is called a "direction field" for the differential

equation dy/dx = y - x .

We can use this to solve the differential equation geometrically and

recover the graph of the function.

The slope of the arrow at any point is equal to y - x at that point.

This kind of picture is called a "direction field" for the differential

equation dy/dx = y - x .

We can use this to solve the differential equation geometrically and

recover the graph of the function.

29.
The idea is to start somewhere on the

direction field and simply follow the arrows:

The idea is to start

This graphical technique is useful for getting qualitative

information about solutions of differential equations, especially

when they cannot be integrated.

direction field and simply follow the arrows:

The idea is to start

This graphical technique is useful for getting qualitative

information about solutions of differential equations, especially

when they cannot be integrated.

30.
Here are a couple for you to try...

y ' = 2 ( y - y 2)

y ' = 2 ( y - y 2)

31.
y ' = 3 x sin(2y)

32.
Numerical methods

Another way to gain insight into solutions of differential equations

is to use numerical methods for their solution. The simplest

numerical method is called Euler's method.

Euler's method is easy to understand if you relate it to two things

you already know:

1. The left endpoint (rectangle) method for estimating integrals,

and

2. The fundamental theorem of calculus.

Or, you can think of Euler's method in terms of

differentials:

y x x y ( x) x y ' ( x)

Another way to gain insight into solutions of differential equations

is to use numerical methods for their solution. The simplest

numerical method is called Euler's method.

Euler's method is easy to understand if you relate it to two things

you already know:

1. The left endpoint (rectangle) method for estimating integrals,

and

2. The fundamental theorem of calculus.

Or, you can think of Euler's method in terms of

differentials:

y x x y ( x) x y ' ( x)

33.
Euler’s method

You can algebraically manipulate most first-order equations,

until they are in the form:

y'(x) = f(x,y)

Euler's method then combines the differential formula with the

differential equation:

y x x y ( x) x f ( x, y )

In Euler's method, we simply ignore the small errors and

repeatedly use the resulting equation with a small value of x

to construct a table of values for y(x) (that can then be graphed,

for instance).

You can algebraically manipulate most first-order equations,

until they are in the form:

y'(x) = f(x,y)

Euler's method then combines the differential formula with the

differential equation:

y x x y ( x) x f ( x, y )

In Euler's method, we simply ignore the small errors and

repeatedly use the resulting equation with a small value of x

to construct a table of values for y(x) (that can then be graphed,

for instance).

34.
An example...

y ' = y - x, y(0) = 2

(this is the example we graphed before).

We'll use x = 0.1 The choice of x is usually dictated by the

problem or the situation. The smaller x is the more accurate the

approximated solution will be, but of course you need to do more

work to cover an interval of a given length.

For the first step, we can use that x=0 and y=2, therefore

y ' = 2. Euler's method then tells us that:

y(x + x) = y(x) + f(x,y) x

y(0.1) = 1 + (2 - 0) 0.1 = 1.2

y ' = y - x, y(0) = 2

(this is the example we graphed before).

We'll use x = 0.1 The choice of x is usually dictated by the

problem or the situation. The smaller x is the more accurate the

approximated solution will be, but of course you need to do more

work to cover an interval of a given length.

For the first step, we can use that x=0 and y=2, therefore

y ' = 2. Euler's method then tells us that:

y(x + x) = y(x) + f(x,y) x

y(0.1) = 1 + (2 - 0) 0.1 = 1.2

35.
Continue...

For the second step, we have x = 0.1, y = 2.2, therefore y' = 2.1.

Euler's method then gives:

y(0.2) = 2.2 + 2.1(0.1) = 2.41

We continue in this manner and fill in the following table

x y f (x,y ) = y - x

0 2 2

0.1 2.2 2.1

0.2 2.41 2.21

0.3 2.631 2.331

0.4 2.8641 2.4641

0.5 3.11051 2.61051

0.6 3.371561 2.771561

0.7 3.6487171 2.9487171

0.8 3.94358881 3.14358881

0.9 4.257947691 3.357947691

1 4.59374246 3.59374246

For the second step, we have x = 0.1, y = 2.2, therefore y' = 2.1.

Euler's method then gives:

y(0.2) = 2.2 + 2.1(0.1) = 2.41

We continue in this manner and fill in the following table

x y f (x,y ) = y - x

0 2 2

0.1 2.2 2.1

0.2 2.41 2.21

0.3 2.631 2.331

0.4 2.8641 2.4641

0.5 3.11051 2.61051

0.6 3.371561 2.771561

0.7 3.6487171 2.9487171

0.8 3.94358881 3.14358881

0.9 4.257947691 3.357947691

1 4.59374246 3.59374246

36.
Maple...

Maple tells us that the exact solution of the equation

x

y ' = y - x that has y(0) = 2 is y(x) = x + 1 + e

and so we have y(1) = 4.718281828.

So the Euler method result is pretty close (within 10%). We

could do better by decreasing x, but of course then we'd

need more steps to reach x = 1.

You'll get to try a couple of these on this week's homework.

Maple tells us that the exact solution of the equation

x

y ' = y - x that has y(0) = 2 is y(x) = x + 1 + e

and so we have y(1) = 4.718281828.

So the Euler method result is pretty close (within 10%). We

could do better by decreasing x, but of course then we'd

need more steps to reach x = 1.

You'll get to try a couple of these on this week's homework.