# 2.4 | Exact Solutions

The differential form $$M(x,y)\; dx + N(x,y) \; dy$$ is said to be exact in a rectangle $$R$$ if there is a function $$F(x,y)$$ such that

$$\frac{\partial F}{\partial x}(x,y) = M(x,y)$$
and
$$\frac{\partial F}{\partial y}(x,y) = N(x,y)$$

for all $$(x,y)$$ in $$R.$$
If $$M(x,y)\; dx + N(x,y) \; dy$$ is an exact differential form, then the equation
$$M(x,y)\; dx + N(x,y) \; dy = 0$$
is called an exact equation.

Suppose the first partial derivatives of $$M(x,y)$$ and $$N(x,y)$$ are continuous in a rectangle $$R.$$ Then
$$M(x,y) \; dx + N(x,y) \; dy = 0$$
is an exact equation in $$R$$ if and only if the compatibility condition
$$\frac{\partial M}{\partial y}(x,y) = \frac{\partial N}{\partial x}(x,y)$$
holds for all $$(x,y)$$ in $$R.$$
Method for Solving Exact Equations

1. If $$M \; dx + N \; dy = 0$$ is exact, then $$\frac{\partial F}{\partial x} = M.$$ Integrate this last equation with respect to $$x$$ to get
$$F(x,y) = \int M(x,y) \; dx + g(y).$$
2. To determine $$g(y),$$ take the partial derivative with respect to $$y$$ of both sides of the equation for $$F(x,y)$$ and substitute $$N$$ for $$\partial F / \partial y.$$ We can now solve for $$g'(y).$$
3. Integrate $$g'(y)$$ to obtain $$g(y)$$ up to some constant. Substituting $$g(y)$$ into the equation in step (1), we obtain the solution $$F(x,y) = C$$ to the given exact differential equation
Alternatively, starting with $$\partial F/ \partial y = N,$$ the implicit solution can be found by first integrating with respect to $$y.$$)