Exact Differential Equations

Definition Let $M$ and $N$ be continuous real functions on an open set $D$ in the plane. The equation

is a first-order differential equation in differential form. A solution is a differentiable function $y = \varphi(x)$ on an open interval $I$ whose graph lies in $D$ and which satisfies

Solutions of the form $x = \psi(y)$, with $M\,\psi'(y) + N = 0$, are allowed in the same way, so the form treats $x$ and $y$ symmetrically.

Definition A potential function for the form $M\,dx + N\,dy$ on an open set $D$ is a function $F$ with continuous first partial derivatives on $D$ such that

Then $dF = M\,dx + N\,dy$, and the vector field $\langle M, N \rangle$ is the gradient of $F$. When $D$ is connected, any two potentials differ by a constant.

Definition The equation $M\,dx + N\,dy = 0$ is exact on the open set $D$ if the form has a potential $F$ on $D$. In that case a differentiable function $\varphi$ on an interval $I$, with graph in $D$, is a solution exactly when $F(x, \varphi(x))$ is constant on $I$, because by the chain rule

So the solution curves lie on the level sets $F(x, y) = C$.

Definition Suppose $M$ and $N$ have continuous first partial derivatives on an open set $D$. If the equation is exact on $D$, then

The reason is that a potential $F$ then has continuous second partials $F_{xy} = M_y$ and $F_{yx} = N_x$, and Clairaut’s theorem says these mixed partials are equal. This condition is necessary on every open set. Whether it is also sufficient depends on the shape of $D$, which is the next section.

Definition An open set $D$ in the plane is simply connected if it is connected and every simple closed curve in $D$ encloses only points of $D$, so $D$ has no holes. The theorem on exact equations states the following. Let $D$ be an open, simply connected set, and let $M$ and $N$ have continuous first partial derivatives on $D$. Then $M\,dx + N\,dy = 0$ is exact on $D$ if and only if $M_y = N_x$ at every point of $D$. In the simplest case $D$ is an open rectangle $a < x < b$, $c < y < d$. Fix a point $(x_0, y_0)$ in it. Integrating the form from $(x_0, y_0)$ horizontally to $(x, y_0)$ and then vertically to $(x, y)$, two segments that stay inside the rectangle, gives a potential on the whole rectangle,

Definition A function $\mu(x, y)$ with continuous first partials and $\mu \ne 0$ on $D$ is an integrating factor for $M\,dx + N\,dy = 0$ if $\mu M\,dx + \mu N\,dy = 0$ is exact on $D$. Two cases reduce to one variable. Suppose $D$ is open and simply connected and $M$ and $N$ have continuous first partials on $D$. In each case below the new coefficients have equal cross partials, so the previous theorem makes the new equation exact.

• If $M_y - N_x = h(x)\,N$ on $D$ for a function $h$ continuous on an interval containing every $x$ coordinate of $D$, then $\mu(x) = e^{\int h(x)\,dx}$ is an integrating factor.
• If $N_x - M_y = k(y)\,M$ on $D$ for a continuous $k$ in the same sense, then $\mu(y) = e^{\int k(y)\,dy}$ is an integrating factor.

Because $\mu$ never vanishes on $D$, the old and new equations have the same solutions in $D$. Multiplying by a factor that is zero or undefined somewhere can gain or lose solutions there, so those points must be checked by hand.

Definition An equation $F(x, y) = C$ is an implicit solution of $M\,dx + N\,dy = 0$ on an interval $I$ if some differentiable $\varphi$ on $I$ has $F(x, \varphi(x)) = C$ there and solves the differential equation. For an exact equation with potential $F$ on $D$ and a point $(x_0, y_0)$ of $D$ with $N(x_0, y_0) \ne 0$, the implicit function theorem gives an open interval around $x_0$ on which the level set $F(x, y) = F(x_0, y_0)$ is, near $(x_0, y_0)$, the graph of a unique differentiable $\varphi$ with $\varphi(x_0) = y_0$. That $\varphi$ is the solution of the initial value problem. It extends as long as its graph stays in $D$ and $N \ne 0$ along it.