Math 104
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.
For example:
dy
2
 3xy
dx
2
d y
dy
 4  3y  1
2
dx
dx
ORDER = highest derivative: first order, second order...
To solve a differential equation:
…means to find a function y(x) that makes it true.
2
y 2
3x
ye 
x
solves
dy
2
 3xy
dx
2
1
3
solves
d y
dy

4

3
y

1
2
dx
dx
In Applications
Differential equations arise when we can
relate the rate of change of some quantity
back to the quantity itself.
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
g
2
dt
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
dv
2
 g  kv
dt
velocity squared
d x
 dx 
 g  k 
or
2
dt
 dt 
2
2
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
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
kP
dt
..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
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
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 savings).
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?
A)
B)
dB
 rB  W
dt
dB
 rB  W
dt
C) dB  rB  WB
dt
D)
dB
 rB  WB
dt
E)
dB
 r B  W 
dt
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.
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) ?
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...
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.
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
 3x dx
2
y
Equation of differentials...
dy
 3x dx
2
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
1 3 2
  2 x C
y
(You only need one constant of
integration).
This is called the "general solution" of the differential
equation.
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
 3xy2 ,
dx
1 3 2
is   2 x  2 ,
y
y (1)  2
or (better):
2
y
2
4  3x
DiffEq Problem:
If the function y = f (x) satisfies the initial-value problem
dy
2
2
x yx ,
dx
f (0)  5
then f (1) =
A) e3  1
B)
3e
1 e
C)
5
e
D)
2
1
E) 5 
e
4
F) 1  3
e
G) 10
H)
e
“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
minutes?
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
B. A' = A - 40
D. A' = - A/40
E. A' = - 40/A
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
C. 50e
D. 50e
1
4
 14
1
 40
E. 50e
1
 10
F. 25e
G. 50 ln(2)
H. 25
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.
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
Geometry of Differential Equations
A differential equation of the form
dy
 f ( x, y )
dx
gives geometric information about the graph of y(x).
It tells us:
If the graph of y(x)
goes through the
point (x,y), then the
slope of the graph at
that point is equal to
f(x,y).
dy
 f ( x, y )
dx
An example:
We can draw a picture of this as follows. For the
differential equation
dy
 yx
dx
, we have:
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.
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 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.
Here are a couple for you to try...
y'=2(y-
2
y)
y ' = 3 x sin(2y)
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)
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).
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
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
0
2
0.1
2.2
0.2
2.41
0.3
2.631
0.4
2.8641
0.5
3.11051
0.6
3.371561
0.7
3.6487171
0.8 3.94358881
0.9 4.257947691
1 4.59374246
f (x,y ) = y - x
2
2.1
2.21
2.331
2.4641
2.61051
2.771561
2.9487171
3.14358881
3.357947691
3.59374246
Maple...
Maple tells us that the exact solution of the equation
y ' = y - x that has y(0) = 2 is y(x) = x + 1 + e
x
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.