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 exact or total derivative with respect to
, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region
: α < x < β , γ < y < δ
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.
This can be proven by showing that its derivative with respect to
First-order exact differential equations of the form
can be written in terms of the potential function
This is equivalent to taking the total derivative of
and then writing down each term in the resulting expressions only once and summing them up in order to get
In order for this to be true and for both sides to result in the exact same expression, namely
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 –
[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
is equal to its implicit ordinary derivative
could be positive, it is more intuitive to think of the integral's result as
that is missing some original extra function
is equivalent to the implicit ordinary derivative
is calculated with its arbitrary constant, it is added to
, two arbitrary constants as expected from a second-order equation.
Given the differential equation one can always easily check for exactness by examining the
In this case, both the partial and total derivative of
With one of the conditions for exactness met, one can calculate that Letting
Reduction to a first-order exact equation yields Integrating
and the full implicit solution becomes Solving explicitly for
yields The concepts of exact differential equations can be extended to any order.
th-order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced.
gives Thus, the three conditions for exactness for a third-order differential equation are: the