The purpose of this lab is to give you practice with applying some techniques of integration to solving separable differential equations, including applications to chemical kinetics.
By a differential equation involving an unknown function 
, we
mean an equation of the form
A solution to this differential equation is a function 
which
satisfies the equation. For example, 
is a solution to
the differential equation
because substituting 
into both sides of the equation
leads to the identity
A differential equation is separable if 
has the special
form
Separable differential equations can often be solved analytically by a technique known as separation of variables. That is, we rewrite the differential equation
in the form
and then integrate both sides. For example, to use this technique on
equation 
, we first separate variables to obtain
Integrating both sides gives
where C is a constant of integration. Next, we can solve for
by exponentiating both sides to get
Finally, we can get rid of the absolute value sign by replacing 
,
which is always positive, by a constant C, which can be positive or
negative. So our solution can be written as
The solution 
satisfies equation
for any value of
C. Plugging t=0 into our solution tells us that 
. That
is, the value of C is equal to the value of 
. For example, if
we wanted the solution of equation that satisfied
, the answer would be 
.
An equation of
the form 
, where K is a constant is called an initial
condition for a differential equation. For separable differential
equations, it should be clear that the solution process always
introduces a constant of integration. An initial condition is needed
to determine the value of this constant.
Separable differential equations often appear in chemical kinetics, which is the study of rates of chemical reactions. Determining rates of chemical reactions is a crucial first step in the design of any commercial chemical process, so this is an area of particular interest to chemists and chemical engineers.
As an introduction to using differential equations to model chemical
kinetics, consider the following situation. A stirred vessel contains
a solution of a chemical substance W, which decays spontaneously to
form products. If we let 
represent the concentration of species
W in the vessel as a function of time, then it has
been found experimentally that the rate at which W decays is
proportional to its concentration. In mathematical terms, this means
that 
satisfies a differential equation of the form
which can be solved by the method of separation of variables to yield
the solution 
.
As a more interesting example, suppose that two substances, Y and
Z, combine in a chemical reaction to form a substance X. We
will assume that the rate at which the molecules of X are formed is
proportional to the product of the concentrations of Y and Z at
time t. Let
, 
, and z(t) be,respectively, the concentrations of
chemicals X, Y, and Z present at time t. Then we have the
rate law
The next step is to express 
and 
in terms of 
and
then substitute the resulting expressions into the rate law to get a
differential equation involving only 
. This is done as follows.
We are assuming that one molecule of Y and one molecule of Z combine
to form one molecule of X and that there is no X present
initially. That is, for every molecule of X
formed, one molecule of Y and one molecule of Z must be consumed.
This means that 
and 
, where 
and 
are the initial concentrations of Y
and Z. Thus we have
This is a separable differential equation that can be solved using the technique described above. An example is given below.
Suppose that a solution initially containing 2 moles/liter of Y and 1 mole/liter of Z is allowed to react. Find an expression for the amount of X at time t.
We have to solve the initial value problem
(The constants 2 and 1 come from the initial concentrations.) Separating variable we get
Using the technique described above, we integrate both sides with respect to t.
(Without Maple, the integral on the left can be evaluated using partial fractions.) The integral on the right is easy and using Maple for the integral on the left we get
> int(1/((2-x)*(1-x)),x);
Thus the general solution, in implicit form, is
We now obtain an explicit solution for 
and use the initial
condition to determine c.
> a1 := solve(ln(2-x)-ln(1-x) = k*t+c,x);
> a2 := solve(subs(t=0,a1)=0,c);
> a3 := simplify(subs(c=a2,a1));
So the explicit solution to the initial value problem is
as t goes to infinity? Can you explain why?
and 
with 
,
can you find an an expression for the concentration of species X
that depends on t, A, and B? How is this solution related to
your answers to the first two questions?
