Review
• Average absolute
" P ( b) ! P ( a ) #
$
%
b!a
&
'
and relative
" P(b) ! P(a) %
$$
''
# (b ! a)P(a) &
growth rates.
•Formulas for a population at time t if the population
has a constant absolute ( P(t) = A + Bt ) or relative ( P(t) = A(1+ r) ) growth rate.
•Found the relative change in the population based on its relative growth
rate, by integrating the relative growth rate to find ln(P(b)/P(a)),
and then solving for P(b)/P(a).
b P '(t )
P(b)
t
!a
P(t )
dt = ln P(b)" ln P( a)= ln
P( a)
•The difference between constant instantaneous and constant average
relative growth. Two ways to view “increase by 2% per year”:
• Instantaneous rate of 2% per year
- This leads to y' = 0.02 y, and y = Ce0.02t.
• By the end of each year, we have increased by 2%.
- This leads to A(1.02)t.
Review Problem
Relative growth rate
given by the graph.
Find the relative change
in the population from
year 0 to 15.
Hint: Use what we know about the relationship between area and
the integral.
Review Problem
Relative growth rate
given by the graph.
Find the relative change
in the population from
year 0 to 15.
15
P '( t )
P(15)
"0 P ( t ) dt
=e
= e0.1!0.025 = e0.075 # 1.078
P(0)
So P(15) is about 107.8% of P(0).
Or: The population increased by about 7.8%.
Lesson 10
Generalization of Exponential Growth:
Another Differential Equation
4
Differential Equations Review
Differential Equations: An equation with derivatives in it.
They express the relationship involving the rates of change
• Easy to confirm solutions
• Harder to find solutions
Solving the Equation
We say that y = Cekx is the general solution to y' = k y
We call it a general solution because we do not know
the value of C.
5
Exponential
Growth
Exponential
Decay
y' = k y, k > 0: Exponential
growth
y' = k y, k < 0: Exponential
decay
Ex: y' = 2y; Two solutions:
Ex: y' = –y; Two solutions:
y = 5e2x
y = –e2x
y = 2e–x
y = –3e–x
6
Growth Proportional to a
Difference
We consider the differential equation y ! = k(y " A)
Represents growth proportional to the difference between y and a
constant A.
Solution?
Equilibrium Solution
One special solution to y' = k(y – A): y = A
Works because y' = 0 and k(y – A) = k(A – A) = 0.
Called an equilibrium solution. Are there other equilibrium solutions?
7
Equilibrium Solutions
Any equilibrium solution must have y' = 0, since y is to be
constant.
For our equation, this means
y' = k(y – A) = 0
So y = A is the only equilibrium solution to our equation.
Special Case
If we let A = 0, then we get y' = k y which we have already studied.
(Note: y = 0 is the equilibrium solution for y' = k y.)
8
Solving the Equation
To solve
y' = k(y – A)
Make the substitution u = y – A. Then u' = y', and we get
u' = k u
This we can solve!
u = Cekt, or
y – A = Cekt, so
y = A + Cekt
The general solution to y' = k(y – A) is y = A + Cekt
where C is any constant.
9
Example
Find the solution to the initial value problem
y' = 2 ( y – 10 ), y(0) = 8.
General solution: y = 10 + C e 2 t.
Equilibrium
solution A
k
Have y(0) = 10 + Ce0 = 10 + C
So 10 + C = 8, or C = 8 – 10 = -2
Solution: y(t) = 10 – 2e2t
10
Example
1- Solve the initial value problem y' = –2y – 6, y(0) = 1.
First put the equation into the right form:
y' = –2y – 6 = –2(y + 3) = –2(y–(–3))
General solution:
y = –3 + Ce–2t
Initial condition:
y(0) = –3 + C = 1, so C = 4.
Solution:
y(t) = –3 + 4e–2t
2- What if we had y' = –2y – 6, y(0) = – 3?
General Solution:
y(t) = – 3 + Ce–2t
If y(0) = – 3, then – 3 + Ce0 = – 3 + C = – 3, so C = 0:
y(t) = – 3 (Which is the equilibrium solution.)
11
Newton’s Law of Cooling
The rate at which an
object cools down or
heats up…
…is
…is
proportional
proportional
to…
to…
That is,
…the difference
between the object’s
temperature and the
surrounding (ambient)
temperature.
k (y – A)
y' = k
y' Where
y(t) = temperature at time t, and
A is the ambient (surrounding) temperature.
The Temperature of the object after time t has passed is
y = A + Cekt
12
Example
Cup of coffee at 120°. Left on the table in a 70° room, it cools to
115° in one minute. How long before the temperature is 100°?
What Do We Want? Solve for t when y(t) = 100
What Do We Know?
y(0) = 120
y(1) = 115 y' = k(y – 70).
That’s Easy!
Here, y' = k(y – 70). General: y = 70 + Cekt. Need both C and k…
C: since y(0) = 120, we have 70 + C = 120 so C = 50.
k; use y(1) = 115: y(1) = 70 + 50ek = 115 so k = ln(45/50) = –.1054
y(t) = 70 + 50e–.1054t
When y(t) = 100, we get 100 = 70 + 50e –. 1054t
ln (30 / 50 )
t=
! 4.8 minutes
".1054
13
Stable and Unstable Equilibrium
Solutions
Recall: Equilibrium solution is a constant solution.
What do the other solutions do?
• If other solutions tend toward the equilibrium
solution, it is said to be a stable equilibrium.
• If other solutions tend away from the equilibrium
solution, it is said to be unstable.
14
Example: Exponentials
y' = ky, where k < 0:
y' = ky where k > 0
Equilibrium: y = 0
Equilibrium: y = 0
Others: y = Cekt
Others: y = Cekt
y = 0 is stable because the other
solutions move toward it as t
increases.
In this case, y = 0 is unstable.
15
Example
y' = k(y – A)
Equilibrium: y = A
Others: y = A + Cekt
Is the equilibrium solution stable or unstable?
1- Suppose k < 0. What happens to A + Cekt as t goes to infinity?
lim( A + Ce
kt
t !"
)= A
This goes to zero as t
goes to infinity if k < 0.
So y = A is stable.
2- Suppose k > 0. What happens to A + Cekt as t goes to infinity?
lim( A + Ce
kt
t !"
So y = A is unstable in this case.
) = ±"
This goes to infinity as t
goes to infinity if k > 0.
16
Example
Which is for k > 0 and which for k < 0?
Which is stable and which is unstable?
k > 0; unstable
k < 0; stable
17
Summary
•
•
•
•
y' = k(y – A): growth proportional to a difference.
General solution: y = A + Cekt
Newton’s law of cooling (or warming)
Stable equilibrium (other solutions move toward)
and unstable equilibrium (other solutions move
away).
18
InClass Group Work
1- Find the solution for:
dP
= 5P ! 50
dt
and
P(0) = 8
2- If an object takes 40 minutes to cool from 30
degrees to 24 degrees in a 20 degree room, how
long will it take the object to cool to 21 degrees?
What Do We Want? Solve for t when y(t) = ?
y(40) = ? y' = k(y – ?).
What Do We Know? y(0) = ?
19
dP
1- Find the solution for: = 5P ! 50 and P(0) = 8
dt
P! = 5( P " 10) P(t ) = 10 + Ce5t # P(t ) = 10 " 2e5t
2- If an object takes 40 minutes to cool from 30 degrees to 24
degrees in a 20 degree room, how long will it take the object
to cool to 21 degrees?
Solve for t when y(t) = 21
y (0) = 20 + Cek !0 = 20 + C = 30 " C = 10
ln(2 / 5)
y (40) = 24 = 20 + 10e40k " k =
= -0.023
40
#40ln(10)
y (t ) = 21 = 20 + 10exp(-0.023 ! t ) " t =
= 100.5
ln(2 / 5)
20