Growth
Authors: Bill Davis, Horacio Porta and Jerry Uhl
©19962007
Publisher: Math Everywhere, Inc.
Version 6.0
1.06 The Differential Equations of Calculus
BASICS
B.1) The most important of all differential equations:
y
£
@
x
D
=
r y
@
x
D
and why you already know how to solve it
·
B.1.a)
Given a number
r
, to solve the differential equation
y'
@
x
D
=
r y
@
x
D
,
you want a function
y
@
x
D
with the property that
y'
@
x
D
=
r y
@
x
D
for all
x
's.
This means
100
y
£
@
x
D
y
@
x
D
=
100 r
no matter what
x
is.
How does this give away a formula for
y
@
x
D
?
·
Answer:
The equation is
y'
@
x
D
=
r y
@
x
D
for all
x
's.
This means
100
y
£
@
x
D
y
@
x
D
=
100 r
no matter what
x
is.
This tells you that
y
@
x
D
grows with a steady instantaneous percentage growth rate of
100 r percent
. And this tells you that
y
@
x
D
is the exponential function
y
@
x
D
=
k e
r x
for some number
k
.
Try it out:
Clear
@
x, y, k, r
D
;
y
@
x
_
D
=
k E
r x
;
y
£
@
x
D
==
r y
@
x
D
True
No problemo.
You can get
Mathematica
to spit this out directly:
Clear
@
y, x
D
;
DSolve
@
y'
@
x
D
==
r y
@
x
D
, y
@
x
D
, x
D
88
y
@
x
D
Ø ‰
r x
C
@
1
D<<
This means
C
@
1
D
can be any constant. You can reconcile the solution
y
@
x
D
=
k e
r x
above with the
Mathematica
solution
y
@
x
D
=
C
@
1
D
e
r x
by taking
k
=
C
@
1
D
.
·
B.1.b)
Now that you know that
y
@
x
D
=
k e
3
x
solves the differential equation
y'
@
x
D
=
3
y
@
x
D
,
what data do you need to nail down what the value of
k
is?
·
Answer:
You know
y
@
x
D
=
k e
3
x
;
so you have one unknown (namely
k
) to determine.
You need one data point on
y
@
x
D
to do this.
For instance, if you know
y
@
1.71
D
is equal to
18.06
, then you determine
k
by solving
for
k
as follows:
Clear
@
x, k
D
;
Solve
A
k
I
E
3 x
ê
. x
Ø
1.71
M
==
18.06, k
E
88
k
Ø
0.106853
<<
This gives you the specific solution
y
@
x
D
=
0.106853 e
3
x
.
You can also get an equivalent solution directly from
Mathematica
:
Clear
@
y, x, starterx, startery
D
;
starterx
=
1.71;
startery
=
18.06;
sol
=
DSolve
@8
y'
@
x
D
==
3 y
@
x
D
,
y
@
starterx
D
==
startery
<
, y
@
x
D
, x
D
;
y
@
x
D ê
. sol
@@
1
DD
0.106853
‰
3 x
Good.
B.2) The logistic differential equation
y
£
@
x
D
=
r y
@
x
D I
1

y
@
x
D
b
M
and how you get a formula for its solution
·
B.2.a)
Given numbers
r
and
b
, you want a function
y
@
x
D
with the property that
y
£
@
x
D
=
r y
@
x
D J
1

y
@
x
D
b
N
.
How do you find a formula for
y
@
x
D
?
·
Answer:
This one is harder than the others. Let's see what mighty
Mathematica
thinks about this
one:
Clear
@
y, x, r, b
D
;
DSolve
B
y
£
@
x
D
==
r y
@
x
D
1

y
@
x
D
b
, y
@
x
D
, x
F
::
y
@
x
D
Ø
b
‰
r x
+
b C
@
1
D

1
+ ‰
r x
+
b C
@
1
D
>>
Here
C
@
1
D
can be any constant. Call it
k
and check
Mathematica
's solution:
Clear
@
y, x, k, r, b
D
;
y
@
x
_
D
=
b E
r x
E
r x

E
k
;
Expand
@
y
£
@
x
DD
==
Expand
B
r y
@
x
D
1

y
@
x
D
b
F
True
Good.
Later in the course, you'll learn how to pull this formula out of a hat with your own hands.