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 \(R.\)
$$
M(x,y) \; dx + N(x,y) \; dy = 0
$$
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
- 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
$$
F(x,y) = \int M(x,y) \; dx + g(y).
$$ - 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 \(g'(y).\)
- 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 \(y.\))