In mathematics, an exact differential equation or total differential equation is a certain kind of ordinary differential equation which is widely used in Physics and engineering.
Definition
Given a simply connected and open subset D of and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form
is called an exact differential equation if there exists a continuously differentiable function F, called the potential function, so that
and
An exact equation may also be presented in the following form:
where the same constraints on I and J apply for the differential equation to be exact.
The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function , the exact or total derivative with respect to is given by
Example
The function given by
is a potential function for the differential equation
First order exact differential equations
Identifying first order exact differential equations
Let the functions , , , and , where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region . Then the differential equation
is exact if and only if
That is, there exists a function , called a potential function, such that
So, in general:
Proof
The proof has two parts.
First, suppose there is a function such that
It then follows that
Since and are continuous, then and are also continuous which guarantees their equality.
The second part of the proof involves the construction of and can also be used as a procedure for solving first order exact differential equations. Suppose that
and let there be a function for which
Begin by integrating the first equation with respect to . In practice, it doesn't matter if you integrate the first or the second equation, so long as the integration is done with respect to the appropriate variable.
where is any differentiable function such that . The function plays the role of a constant of integration, but instead of just a constant, it is function of , since we is a function of both and and we are only integrating with respect to .
Now to show that it is always possible to find an such that .
Differentiate both sides with respect to .
Set the result equal to and solve for .
In order to determine from this equation, the right-hand side must depend only on . This can be proven by showing that its derivative with respect to is always zero, so differentiate the right-hand side with respect to .
Since ,
Now, this is zero based on our initial supposition that
Therefore,
And this completes the proof.
Solutions to first order exact differential equations
First order exact differential equations of the form
can be written in terms of the potential function
where
This is equivalent to taking the exact differential of .
The solutions to an exact differential equation are then given by
and the problem reduces to finding .
This can be done by integrating the two expressions and and then writing down each term in the resulting expressions only once and summing them up in order to get .
The reasoning behind this is the following. Since
it follows, by integrating both sides, that
Therefore,
where and are differentiable functions such that and .
In order for this to be true and for both sides to result in the exact same expression, namely , then must be contained within the expression for because it cannot be contained within , since it is entirely a function of and not and is therefore not allowed to have anything to do with . By analogy, must be contained within the expression .
Ergo,
for some expressions and . Plugging in into the above equation, we find that
and so and turn out to be the same function. Therefore,
Since we already showed that
it follows that
So, we can construct by doing and and then taking the common terms we find within the two resulting expressions (that would be ) and then adding the terms which are uniquely found in either one of them - and .
Second order exact differential equations
The concept of exact differential equations can be extended to second order equations. Consider starting with the first-order exact equation:
Since both functions are functions of two variables, implicitly differentiating the multivariate function yields