Transcript Chapter Four Review Some Practice Problems

Section 6.3
Separation of Variables and
the Logistic Equation
Section 6.3
Separation of Variables and the Logistic
Equation

We can extend the idea of separation of
variables to solve some more difficult
differential equations. Let’s look at one
such example:
ln x
dy 1

 ln x
x
dx x
1
1
1
2
dy  ln xdx   1dy   ln xdx  y   ln x   C
x
x
2
2 xy  ln x 2  0  2 xy  2 ln x  y 
Section 6.3
Separation of Variables and the Logistic
Equation

A final type of differential equation that we
will examine in this course is a class of
equations known as logistic functions.
These are functions that describe
populations that are bounded above and
below by horizontal asymptotes. The most
common application of the logistic function
is the study of populations in closed spaces.
Section 6.3
Separation of Variables and the Logistic
Equation
Below is an example of such a graph
Section 6.3
Separation of Variables and the Logistic
Equation

These functions take a distinctive form.
Their equations always look like:
L
y
 kt
1  be
Section 6.3
Separation of Variables and the Logistic
Equation

Using your calculators, and conferring with
your neighbors, experiment with different
values of L, b, and k to see if you can reach
some general conclusions about where the
horizontal asymptotes are and where the yintercept is. I’ll be quiet for a few minutes.
L
y
 kt
1  be
Section 6.3
Separation of Variables and the Logistic
Equation

Finding logistic equations in differential form
is a tricky business. Let’s look at one such
example together. Using the point (0,5) and
the differential equation below, find the
logistic function.
dy
y

 1.2 y 1  
dt
 8
Section 6.3
Separation of Variables and the Logistic
Equation

To solve differential equations like this one,
we need to be able to solve what is called a
partial fraction problem. We are faced with
the following differential to solve:
dy
dy
 y
 1.2 y 1   
 1.2dt
y
dt


 8
y 1  
 8
 We cannot integrate the left side as it is
written.
Section 6.3
Separation of Variables and the Logistic
Equation

We need to rewrite the fraction on the left
hand side as a sum of fractions. We know
that there must exist constants A and B
that satisfy the following equation:
 y
A 1    By
1
A
B
8
 y
 
 
 A 1    By  1
 y  y 1 y
 y
 8
y 1  
y 1  
8
 8
 8
Ay
A
1
 A

A
 By  1  A  y    B   1  A  1,   B  0  B 
8
8
8
 8

Section 6.3
Separation of Variables and the Logistic
Equation

So, now we can rewrite that integral and
evaluate as follows:
dy
dy
 y
 1.2 y 1   
 1.2dt  
y
dt

 8
y 1  
 8
dy
y

y 1  
 8
  1.2dt
1
1 
  y  8  y  dy   1.2dt  ln y  ln 8  y  1.2t  C
ln 8  y  ln y  1.2t  C
Section 6.3
Separation of Variables and the Logistic
Equation

Finally, we need to remember that we have
the y-intercept of the function to finally find
the curve as follows:
8 y
3
ln
 1.2t  C  ln  C
y
5
3
ln
8 y
8  y 3 1.2t
8  y 3 1.2t
 e 1.2t  e 5 
 e

 e
y
y
5
y
5
8
3
8
3
y
1
8
 1  e 1.2t   1  e 1.2t  
y
3
y
5
y
5
8 1  3 e 1.2t
1  e 1.2t
5
5