This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Exact differential equation" – news · newspapers · books · scholar · JSTOR (December 2009) (Learn how and when to remove this template message)

Differential equations |
---|

Scope |

Classification |

Solution |

People |

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.

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**,^{[1]}^{[2]} 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

The function given by

is a potential function for the differential equation

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:

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 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.

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 .

The concept of exact differential equations can be extended to second order equations.^{[3]} Consider starting with the first-order exact equation:

Since both functions , are functions of two variables, implicitly differentiating the multivariate function yields

Expanding the total derivatives gives that

and that

Combining the terms gives

If the equation is exact, then . Additionally, the total derivative of is equal to its implicit ordinary derivative . This leads to the rewritten equation

Now, let there be some second-order differential equation

If for exact differential equations, then

and

where is some arbitrary function only of that was differentiated away to zero upon taking the partial derivative of with respect to . Although the sign on could be positive, it is more intuitive to think of the integral's result as that is missing some original extra function that was partially differentiated to zero.

Next, if

then the term should be a function only of and , since partial differentiation with respect to will hold constant and not produce any derivatives of . In the second order equation

only the term is a term purely of and . Let . If , then

Since the total derivative of with respect to is equivalent to the implicit ordinary derivative , then

So,

and

Thus, the second order differential equation

is exact only if and only if the below expression

is a function solely of . Once is calculated with its arbitrary constant, it is added to to make . If the equation is exact, then we can reduce to the first order exact form which is solvable by the usual method for first-order exact equations.

Now, however, in the final implicit solution there will be a term from integration of with respect to twice as well as a , two arbitrary constants as expected from a second-order equation.

Given the differential equation

one can always easily check for exactness by examining the term. In this case, both the partial and total derivative of with respect to are , so their sum is , which is exactly the term in front of . With one of the conditions for exactness met, one can calculate that

Letting , then

So, is indeed a function only of and the second order differential equation is exact. Therefore, and . Reduction to a first-order exact equation yields

Integrating with respect to yields

where is some arbitrary function of . Differentiating with respect to gives an equation correlating the derivative and the term.

So, and the full implicit solution becomes

Solving explicitly for yields

The concepts of exact differential equations can be extended to any order. Starting with the exact second order equation

it was previously shown that equation is defined such that

Implicit differentiation of the exact second-order equation times will yield an th order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced. For example, differentiating the above second-order differential equation once to yield a third-order exact equation gives the following form

where

and where is a function only of and . Combining all and terms not coming from gives

Thus, the three conditions for exactness for a third-order differential equation are: the term must be , the term must be and

must be a function solely of .

Consider the nonlinear third-order differential equation

If , then is and which together sum to . Fortunately, this appears in our equation. For the last condition of exactness,

which is indeed a function only of . So, the differential equation is exact. Integrating twice yields that . Rewriting the equation as a first-order exact differential equation yields

Integrating with respect to gives that . Differentiating with respect to and equating that to the term in front of in the first-order equation gives that and that . The full implicit solution becomes

The explicit solution, then, is