Many differential equations have solutions that can be written in implicit form:
F(x,y) = C
Now instead of starting with the differential equation and finding the solution, suppose we look at this ``backwards'': start with the solution F(x,y) = C and differentiate it, obtaining a differential equation. For example, the solutions
y2 + x y = Cwould lead to the differential equation
In general, we have (with y = y(x))
An equation of the form
is said to be an exact differential equation. Thus is exact, and we could make as many examples as we want by taking an arbitrary (differentiable) F and differentiating.
Now let's look ``forwards'' again: we have a differential equation which we suspect might be exact. We need to be able to
Thus we want to be able to look at the differential equation and see that and where F = x y + y2, and therefore that the general solution is x y + y2 = C.
The test for exactness comes from the equality of mixed partial derivatives:
So if is exact, we must have
In our example that is indeed the case:
It turns out that (under suitable conditions) this test is all you need. The details are as follows:
Suppose M, N, and are continuous in a rectangular region R: a < x < b, c < y < d. Then the differential equation is exact in R (i.e. there is a function F defined in R with and there) if and only if everywhere in R.
Now how do we find F when the equation is exact? By integration. In our example, we want and . Look at the first equation, and integrate with respect to x:
A few points to notice:
This time the ``constant'' c really is constant, and we can ignore it (since at the end the solution F(x,y) = C will have a constant in it anyway). We conclude that F(x,y) = x y + y2 = C is the general solution of our differential equation.
Thus you can use the following procedure for solving an exact equation: