4.1
Applications of Separation of Variables
Many of the differential equations that arise in the natural and social sciences are
separable. In this section we describe three applications of separable equations to
physics, chemistry, and biology.
Newton’s Law of Cooling
This law was first formulated by Isaac
Newton in 1701 in an anonymous article
entitled Scala Graduum Caloris (“Scale of
the Degrees of Heat”).
Newton’s law of cooling is a differential equation that predicts the cooling of a warm
body placed in a cold environment. According to the law, the rate at which the
temperature of the body decreases is proportional to the difference in temperatures
between the body and its environment. In symbols,
dT
−r (T − Te )
dt
where T is the temperature of the object, Te is the (constant) temperature of the
environment, and r is a constant of proportionality.
We can solve this differential equation using separation of variables. Dividing
through by T − Te gives
1
dT
−r
T − Te dt
so
Z
Z
1
dT −r dt.
T − Te
Integrating gives
ln |T − Te | −rt + C,
and hence
T − Te A e −rt ,
a Newton illustrated his law by describing
the cooling of hot iron.1
where A ±e C . Thus the solutions are
T Te + A e −rt
where A is an arbitrary constant.
Figure 1 shows the cooling of a warm body over time as predicted by Newton’s
law of cooling. The behavior is very similar to exponential decay, except that the
temperature T approaches Te instead of 0 as t → ∞. Indeed, Newton’s law of cooling
can be interpreted as saying that the temperature difference T − Te decays exponentially
over time.
a Figure 1: Cooling of a warm body.
EXAMPLE 1
An apple pie with an initial temperature of 170 ◦ C is removed from the oven and left to cool
in a room with an air temperature of 20 ◦ C. Given that the temperature of the pie initially
decreases at a rate of 3.0 ◦ C/min, how long will it take for the pie to cool to a temperature
of 30 ◦ C?
SOLUTION
The pie should obey Newton’s law of cooling with Te 20 ◦ C. Thus
T ( t ) 20 + A e −rt
1 Photo
and
T 0 ( t ) −rAe −rt
by Jeff Kubina, licensed under CC BY-SA 2.0, cropped from original.
APPLICATIONS OF SEPARATION OF VARIABLES
2
for some constants A and r. We also have the initial conditions
T (0) 170 ◦ C
T 0 (0) −3.0 ◦ C/min.
and
Plugging these in gives the equations
170 20 + A
and
−3 −rA,
so A 150 ◦ C and r 0.02. Thus
T 20 + 150 e −0.02t .
a
Figure 2: The temperature function T ( t )
in Example 1. The diagonal dashed line
shows the initial 3.0 ◦ C/min rate of
temperature decrease.
To find how long it will take for the temperature to reach 30 ◦ C, we plug in 30 for T and solve
for t. The result is that t 135 min, as shown in Figure 2.
Reaction Rates
The study of the rates at which chemical
reactions occur is a branch of chemistry
known as chemical kinetics. For more
complicated reactions, the chemical
kinetics can involve a system of
differential equations, with one equation
for each reactant.
In chemistry, the rate at which a chemical reaction occurs is determined by a differential
equation called a rate equation. For a reaction with a single reactant, the concentration C
of the reactant obeys a rate equation of the form
dC
−rC n ,
dt
where r is a constant called the rate constant, and n is a constant called the order of
the reaction.
Typically the order of the reaction is the same as its molecularity, i.e. the number
of molecules of the reactant that must come together to react. For example, the
decomposition of sulfuryl chloride
SO2 Cl2 −→ SO2 + Cl2 .
is a first-order reaction (n 1), since it involves the decomposition of a single molecule
of SO2 Cl2 . On the other hand, the decomposition of nitrogen dioxide
2 NO2 −→ 2 NO + O2 .
a A sample of nitrogen dioxide gas.
2
is a second order reaction (n 2), since it only occurs when two molecules of NO2
come together.
First and second order reactions behave somewhat differently. The rate equation
for a first-order reaction is
dC
−rC,
dt
which is an instance of the exponential decay equation. Thus the concentration of the
reactant decays exponentially. A second-order reaction, on the other hand, is governed
by the equation
dC
−rC 2 ,
dt
We can solve this equation using separation of variables. Dividing through by C 2 gives
C −2
2 Photo
dC
−r,
dt
by W. Oelen, licensed under CC BY-SA 3.0, via Wikimedia Commons
APPLICATIONS OF SEPARATION OF VARIABLES
3
so the solutions are given by
Z
Z
C
−2
dC −r dt.
Integrating gives
−C −1 −rt + A
a Figure 3: Concentration of a
for some constant A, so
second-order reactant.
C 1
rt + B
where B −A.
Figure 3 shows the decrease in the concentration of a second-order reactant according
to this equation. This decrease behaves somewhat differently than exponential decay,
with the concentration decreasing very quickly at first but having a very long tail. In
particular, the half-life of the reactant is very small at first but increases over the course
of the reaction, as shown in Figure 4.
a
Figure 4: Halving times for the reactant
in a second-order reaction. Note the
increasing half-life.
EXAMPLE 2
The decomposition of nitrogen dioxide (NO2 ) is a second-order reaction. During a chemistry
experiment, nitrogen dioxide with an initial concentration of 0.20 M is decomposed at a high
temperature. If 90% of the NO2 decomposes during the first ten seconds, what is the rate
constant for the reaction?
SOLUTION
Figure 5 shows the concentration [NO2 ] of nitrogen dioxide in this example. We
know that
1
,
rt + B
and we are given that [NO2 ] 0.20 M at t 0, and [NO2 ] 0.020 M (10% of 0.20 M) at
t 10 sec. This information yields two equations:
[NO2 ] 0.20 M a
Figure 5: The concentration of NO2 in
Example 2.
1
B
and
0.020 M 1
.
r (10 sec) + B
Then B 5.0/M, so the rate constant r is 4.5/ (M · sec) .
Logistic Population Growth
a A colony of Paenibacillus dendritiformis
bacteria.3
Consider a colony of bacteria growing in an environment with limited resources.
For example, there may be a scarcity of food, or space constraints on the size of the
colony. In this case, it is not reasonable to expect the population of the colony to grow
exponentially—indeed, the colony will unable to grow larger than some maximum
population Pmax .
In the study of population ecology, the maximum number of organisms of a certain
type that a given environment can support is known as the carrying capacity of the
population. Exponential population growth is a reasonable model in situations where
the carrying capacity Pmax is very large, but for environments with a limited carrying
capacity other models must be used to predict population growth.
The most common model for population growth in an environment with limited
carrying capacity is the logistic equation:
3 Photo from the laboratory of Dr. Eshel Ben Jacob, licensed under CC BY-SA 3.0, via Wikimedia Commons
APPLICATIONS OF SEPARATION OF VARIABLES
4
P
dP
kP 1 −
dt
Pmax
The idea of this equation is that the factor on the right approaches zero as P gets close
to Pmax , causing the population growth to taper off.
It is possible to solve the logistic equation using separation of variables, though the
calculation is a bit difficult. (See Solving the Logistic Equation on the next page.) The
solutions have the form
Pmax
1 + A e −kt
P a Figure 6: Logistic population growth.
where A is an arbitrary constant.
Figure 6 shows the graph of a typical solution to the logistic equation. Note that
the population grows quickly at first, but the rate of increase slows as the population
reaches the maximum. As t → ∞, the population asymptotically approaches the
carrying capacity, i.e.
lim P ( t ) Pmax .
t→∞
Logistic functions are sometimes called
sigmoid functions because of the S-like
shape of their graphs, though this term is
also used more broadly to refer to any
function whose graph has an S-like shape.
The solutions to the logistic equation are known as logistic functions, and can be used
to model any situation where a variable is growing but the growth is bounded above.
EXAMPLE 3
A population of bacteria is undergoing logistic growth with a maximum population of 100,000.
Initially the population is 16,000. By t 1.0 hour, the population has grown to 64,000. What
will the population be at t 2.0 hours?
SOLUTION
We know that
100,000
1 + A e −kt
for some constants A and k. Plugging in the two data points gives
P (t ) 16,000 100,000
1+A
and
64,000 100,000
.
1 + A e −k
Solving for A in the first equation gives A 5.25. Plugging this into the second equation and
solving for k gives k 2.2336. Then
a
P (2) Figure 7: The bacteria population in
Example 3.
100,000
94,315.8,
1 + (5.25) e − (2.2336)(2)
so the population at t 2.0 hours will be approximately 94,000, as shown in Figure 7.
EXERCISES
1. A bottle of water with an initial temperature of 25 ◦ C is placed in a refrigerator
with an internal temperature of 5 ◦ C. Given that the temperature of the water is
21 ◦ C ten minutes after it is placed in the refrigerator, what will the temperature of
the water be after one hour?
2. For the logistic population model, find a formula for the constant A that appears in
the solution in terms of the maximum population Pmax and the initial population P0 .
APPLICATIONS OF SEPARATION OF VARIABLES
5
A Closer Look Solving the Logistic Equation
It is possible to solve the logistic equation using separation of variables. Starting with the
equation
dP
P
,
kP 1 −
dt
Pmax
we can separate the variables to get
Pmax
dP
k.
P ( Pmax − P ) dt
Thus the solutions are given by
Z
Pmax
dP P ( Pmax − P )
Z
k dt.
(1)
The integral on the left requires the integration technique known as partial fractions decomposition. The idea of partial fractions decomposition is to write a complicated fraction as a
sum of simpler fractions. In this case, observe that
1
1
Pmax
+
.
P ( Pmax − P )
P
Pmax − P
This decomposition allows us to evaluate the integral:
Z
Pmax
dP P ( Pmax − P )
Z
1
dP +
P
Z
1
dP
Pmax − P
ln |P| − ln |Pmax − P| + C
P
+ C
ln Pmax − P Plugging this into equation (1) and integrating the right side gives
P
kt + C.
Pmax − P ln Then
P
A e kt
Pmax − P
and solving for P gives
P where B 1/A.
Pmax
,
1 + B e −kt
3. In 1974, Stephen Hawking discovered that black holes emit a small amount of
radiation, causing them to slowly evaporate over time. According to Hawking, the
mass M of a black hole obeys the differential equation
dM
k
− 2
dt
M
where k 1.26 × 1023 kg3 /year.
(a) Use separation of variables to find the general solution to this equation
(b) After a supernova, the remnant of a star collapses into a black hole with an
initial mass of 6.00 × 1031 kg. How long will it take for this black hole to
evaporate completely?
APPLICATIONS OF SEPARATION OF VARIABLES
6
4. The decomposition of hydrogen iodide is a second-order reaction:
2 HI −→ H2 + I2
Initially the concentration of a sample of hydrogen iodide is 0.250 M, and the
concentration is decreasing at a rate of 1.00 × 10−4 M/sec.
(a) How long will it take for half of the hydrogen iodide to be consumed?
(b) How long will it take for three quarters of the hydrogen iodide to be consumed?
5. The velocity of an object moving through a fluid can be modeled by the drag
equation
dv
−kv 2
dt
where k is a constant.
(a) Find the general solution to this equation.
(b) An object moving through the water has an initial velocity of 16 m/sec. After
2.0 seconds, the velocity has decreased to 12 m/sec. What will the velocity be
after ten seconds?
6. A population of bacteria is undergoing logistic growth, with a maximum possible
population of 100,000. Initially, the bacteria colony has 5,000 members, and the
population is increasing at a rate of 400/minute.
(a) How large will the population be 30 minutes later?
(b) When will the population reach 80,000?
7. Water is being drained from a spout in the bottom of a cylindrical tank. According
to Torricelli’s law, the volume V of water left in the tank obeys the differential
equation
√
dV
−k V
dt
where k is a constant.
(a) Use separation of variables to find the general solution to this equation.
(b) Suppose the tank initially holds 30.0 L of water, which initially drains at a rate
of 1.80 L/min. How long will it take for tank to drain completely?
8. The Gompertz equation has been used to model the growth of malignant tumors.
The equation states that
dP
kP (ln Pmax − ln P )
dt
where P is the population of cancer cells, and k and Pmax are constants.
(a) Use separation of variables to find the general solution to this equation.
(b) A tumor with 5000 cells is initially growing at a rate of 200 cells/day. Assuming
the maximum size of the tumor is Pmax 100,000 cells, how large will the tumor
be after 100 days?