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 \(y.\))

E 2.4 Exercises