Exact equation

An exact equation is a first-order ODE in the form

$M(x,y)dx+N(x,y)dy=0$

where there exists a potential function $F(x,y)$ such that

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

If such an $F$ exists, the ODE is exact, and the implicit solution is simply $F(x,y)=c$.

The level curves of $F$ are exactly the solution curves of the ODE.

Why $F=c$ is the solution

Along any solution curve $y(x)$, compute the total derivative of $F(x,y(x))$:

$\frac{d}{dx}F(x,y(x))=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\cdot \frac{dy}{dx}=M+N\frac{dy}{dx}$

But the ODE says $M+N\frac{dy}{dx}=0$. So $\frac{d}{dx}F(x,y(x))=0$, meaning $F$ is constant along solution curves.

That’s the level curve idea: $F(x,y)=c$ defines a family of curves, and each is a solution of the ODE.

Test for exactness: Clairaut’s theorem

How do you know whether such an $F$ exists, given $M$ and $N$?

Clairaut’s theorem (mixed partials): if $F(x,y)$ has continuous second partial derivatives, then

$\frac{{\partial }^{2}F}{\partial x\partial y}=\frac{{\partial }^{2}F}{\partial y\partial x}$

If we want $F_x=M$ and $F_y=N$, then $F_{xy}=M_y$ and $F_{yx}=N_x$. So:

$\boxed{\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}$

This is the exactness condition. If it holds (in some region $R$), then $F$ exists and the ODE is exact. If it fails, no such $F$ exists, and the ODE is not exact.

How to find $F$

Once you’ve verified exactness, find $F$ by integration:

1. Integrate $M$ with respect to $x$:

$F(x,y)=\int M(x,y)dx+g(y)$

The “constant of integration” is actually a function of $y$ only, since we integrated with respect to $x$.

2. Differentiate with respect to $y$ and equate to $N$:

$\frac{\partial F}{\partial y}=\frac{\partial }{\partial y}\left[\int Mdx\right]+g^{\prime }(y)=N(x,y)$

3. Solve for $g^{\prime }(y)$, integrate to find $g(y)$.

4. Implicit solution: $F(x,y)=c$.

Worked example 1

$y^{\prime }=-\frac{2x{y}^2-1}{2x^{2}y}$.

Rearrange into standard form $Mdx+Ndy=0$: multiply through by $2x^{2}ydx$ to get $(2x{y}^2-1)dx+2x^{2}ydy=0$. So $M(x,y)=2x{y}^2-1$, $N(x,y)=2x^{2}y$.

Check exactness: $M_y=4xy$, $N_x=4xy$. Equal ✓.

Integrate $M$ wrt $x$: $F(x,y)=\int (2x{y}^2-1)dx=x^2{y}^2-x+g(y)$.

Differentiate wrt $y$, equate to $N$: $F_y=2x^{2}y+g^{\prime }(y)=2x^{2}y$, so $g^{\prime }(y)=0$, hence $g(y)=c$ (constant).

Implicit solution: $x^2{y}^2-x=c$.

Worked example 2 (not exact)

$3xy+y^2+(x^2+xy)y^{\prime }=0$. So $M=3xy+y^2$, $N=x^2+xy$.

Check exactness: $M_y=3x+2y$, $N_x=2x+y$. Not equal — the equation is not exact.

You’d need to either find an integrating factor that makes it exact, or use a different method.

Integrating factors for non-exact equations

If $M_y\ne N_x$ but the ratio $\frac{M_y-N_x}{N}$ is a function of $x$ only (or $\frac{N_x-M_y}{M}$ is a function of $y$ only), there’s an Integrating factor $\mu (x)$ (or $\mu (y)$) that makes the equation exact. Multiply through and proceed.

For the broader theory of integrating factors, see Integrating factor. For the specific test, see Clairaut’s theorem.