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
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