PERTEMUAN VII
A p l i k a si PD Linear & PD Bernoulli
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Example : 2
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Example : 3
Example 4 : A radioactive isotope has a half-life of 16 days. You wish
to have 30 g at the end of 30 days. How much radioisotope should you start
with?
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Solution: Since the half-life is given in days we will measure time in days.
Let Q(t) be the amount present at time t and
looking for (the initial
the amount we are
amount). We know that
,
where r is a constant. We use the half-life T to determine r. Indeed, we have
Hence, since
,
we get
Newton's Law of Cooling
Example 5
From experimental observations it is known that (up to a ``satisfactory''
approximation) the surface temperature of an object changes at a rate
proportional to its relative temperature. That is, the difference between its
temperature and the temperature of the surrounding environment. This is
what is known as Newton's law of cooling. Thus, if
is the temperature
of the object at time t, then we have
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where S is the temperature of the surrounding environment. A qualitative study
of this phenomena will show that k >0. This is a first order linear differential
equation. The solution, under the initial condition
, is given by
Hence,
,
which implies
This equation makes it possible to find k if the interval of time
and vice-versa.
is known
Example 6 : Time of Death Suppose that a corpse was discovered in a
motel room at midnight and its temperature was
room is kept constant at
dropped to
. The temperature of the
. Two hours later the temperature of the corpse
. Find the time of death.
Solution: First we use the observed temperatures of the corpse to find the
constant k. We have
.
In order to find the time of death we need to remember that the temperature of a
corpse at time of death is
(assuming the dead person was not sick!).
Then we have
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which means that the death happened around 7:26 P.M.
One of our interested readers, E.P. Esterle, wrote a program that helps find the
time of death based on the above notes. Click HERE to download it. Have fun
with it.
Population Dynamics
Here are some natural questions related to population problems:

What will the population of a certain country be in ten years?

How are we protecting the resources from extinction?
More can be said about the problem but, in this little review we will not discuss
them in detail. In order to illustrate the use of differential equations with regard
to this problem we consider the easiest mathematical model offered to govern
the population dynamics of a certain species. It is commonly called the
exponential model, that is, the rate of change of the population is proportional
to the existing population. In other words, if P(t) measures the population, we
have
,
where the rate k is constant. It is fairly easy to see that if k > 0, we have growth,
and if k <0, we have decay. This is a linear equation which solves into
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,
where
is the initial population, i.e.
. Therefore, we conclude the
following:

if k>0, then the population grows and continues to expand to infinity,
that is,

if k < 0, then the population will shrink and tend to 0. In other words we
are facing extinction.
Clearly, the first case, k > 0, is not adequate and the model can be dropped. The
main argument for this has to do with environmental limitations. The
complication is that population growth is eventually limited by some factor,
usually one from among many essential resources. When a population is far
from its limits of growth it can grow exponentially. However, when nearing its
limits the population size can fluctuate, even chaotically. Another model was
proposed to remedy this flaw in the exponential model. It is called the logistic
model (also called Verhulst-Pearl model). The differential equation for this
model is
,
where M is a limiting size for the population (also called the carrying
capacity). Clearly, when P is small compared to M, the equation reduces to the
exponential one. In order to solve this equation we recognize a nonlinear
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equation which is separable. The constant solutions are P=0 and P=M. The
non-constant solutions may obtained by separating the variables
,
and integration
The partial fraction techniques gives
,
which gives
Easy algebraic manipulations give
where C is a constant. Solving for P, we get
If we consider the initial condition
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(assuming that P0 is not equal to both 0 or M), we get
,
which, once substituted into the expression for P(t) and simplified, we find
It is easy to see that
However, this is still not satisfactory because this model does not tell us when a
population is facing extinction since it never implies that. Even starting with a
small population it will always tend to the carrying capacity M.
Population Dynamics :
Example: 7 Let P(t) be the population of a certain animal species. Assume
that P(t) satisfies the logistic growth equation
1. Is the above differential equation separable?
2. Is the differential equation autonomous?
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3. Is the differential equation linear?
4. Without solving the differential equation, give a sketch of the graph of
P(t).
5. What is the long-term behavior of the population P(t)?
6. Show that the solution is of the form
Find A and B.
Hint: Use the initial condition and the result of 5.
7. Where is the solution's inflection point?
Hint: This can be done without using the answer of 6.
8. What is special about the growth rate of the population P(t) at the
inflection point (found in 7)?
Answer:
1. This differential equation is autonomous, i.e. the variable t is missing.
Therefore, this equation is indeed separable.
2. The answer is Yes! (see 1.) Every autonomous differential equation is
separable.
3. The equation is not linear because of the presence of
.
4. The graph of the solution P(t) is as follows:
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5. Clearly, because of the initial condition (see the graph of the solution below),
we have
,
200 being the carrying capacity.
6. Let us solve this equation (use the technique for solving separable
equations). First, we
look for the constant solutions (equilibrium points or critical points). We
have
Then, the non-constant solutions can be generated by separating the
variables
,
and the integration
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.
Next, the left hand-side can be handled by using the technique of
integration of rational functions. We get
,
which gives
Hence, we have
.
Easy algebraic manipulations give
,
where
. Therefore, all the solutions are
where C is a constant parameter.
Remark: We may rewrite the non-constant solutions as
,
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where a and B are two parameters. If we use the conditions
we will be able to get the desired solution. Indeed, we have
Thus,
and consequently
.
7. From the graph (see 4.), the graph of P(t) has an inflection point between 0
and 200. Let us find it. Since we must have P''(t) = 0, we get
,
where we used the chain rule and the fact that
.
Since our solution is not one of the two constant solutions we are only
left with the equation
This simplifies to
,
which gives P=100 (half-way between 0 and 200).
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Remark: You still need to convince yourself that t=100 is indeed the
inflection point, that is the second derivative changes sign at that point.
8.
The growth rate at t=100 is maximal.
Population Dynamics:
Example: 8 The fox squirrel is a small mammal native to the Rocky
Mountains. These squirrels are very territorial. Note the following
observations:

if the population is large, their rate of growth decreases or even becomes
negative;

if the population is too small, fertile adults run the risk of not being able
to find suitable mates, so again the rate of growth is negative
The carrying capacity N indicates when the population is too big, and the
sparsity parameter M indicates when the population is too small. A
mathematical model which agrees with the above assumptions is the modified
logistic model:
.
1. Find the equilibrium (critical) points. Classify them as : source, sink
or node. Justify your answers.
2. Sketch the slope-field
.
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3. Assume N=100 and M=1 and k = 1. Sketch the graph of the solution
which
satisfies the initial condition y(0)=20.
4. Assume that squirrels are emigrating (from a certain region) with a
fixed rate E. Write down the new differential equation.
Also, discuss the equilibrium (critical) points under the parameter E.
When do you observe a bifurcation?
Answer:
1. The equilibrium (or critical) points are the roots of the equation
Clearly, we have P=0, P=N, and P=M. Using the graph of
,
we get the phase-line of the equilibrium points,
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We conclude that P=0 and P=N are sinks, while P=M is a source.
2. The Slope-Field is given by the following graph:
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3. From the Slope-Field, we get the graph of the particular solution
satisfying the condition P(0) = 20.
4. First, the new equation is
,
where E > 0. Clearly, the graph of the function
can be obtained by shifting the graph of
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down E-unit along the vertical axis. Clearly we have three cases
according to the value of E and the value of f at the local maximum
:

If 0 < E < f(h), then we have similar behavior as for E=0:

If E=f(h), we have two critical points:
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
If E > f(h), we have only one critical point:
Note that
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Clearly, the bifurcation is happening when E = f (h ).
Perkiraan penyebaran AID
The HIV Virus invades a white blood cell...
Populations usually grow in an exponential fashion at first:
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However, populations do not continue to grow forever, because food, water and
other resources get used up over time. Differential equations are used to predict
populations of people, animals, bacteria and viruses that are being affected by
external events.
The logistic equation (developed in the mid-19th century) allows for a growth
term AND an inhibition term and it is predicted that the AIDS epidemic will
follow the pattern of the logistic equation.
If A = number of people affected by the virus at time t,
P = the total population (a constant), and c is a constant,
then the rate of growth of the virus at time t is given by the differential
equation:
The term cPA is the growth term and - cA2 is the inhibition term.
AIDS
Example 9
AIDS is spreading through a city of 50.000 people who take no precautions.
The virus was brought to the town by 100 people and it was found that 1000
people were infected after 10 weeks. How long will it take for half of the
population to be infected?
Solution.
Here, P = 50.000 . So our differential equation becomes:
Solving using SN gives:
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Exact solution is:
Now, we are told that A(10) = 1000 . So now we solve for c:
, Solution is: c = 4. 6416 × 10 -6 . So
We see in the graph the S-shaped curve which the logistic equation can take.
Now to find how long it takes for half the population (
) to be infected:
, Solution is:
So we conclude that half of the population will be infected after about 27
weeks.
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Newly Reported AIDS Cases in Australia
Year Cumulative Totals
1982 1
1985 50
1988 413
1991 1100
1993 2426
1995 4179
1996 4836
1998 5525
1999 5721
Men: 5439 Women: 282 Total: 5721
We see that the data roughly follows the expected S-shaped logistic equation
curve. The decrease in growth in the late 1990s is largely due to suppressant
drugs and education.
Reported AIDS Cases - Australia
NOTE: This data is from several sources and some of it is conflicting. Some
sources include "HIV/AIDS" while others "AIDS".
TERIMA KASIH
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