Exact Equations

Suppose a curve $y(x)$ stays on a level curve $F(x, y) = C$ of a function of two variables. Differentiating with the chain rule gives $F_x + F_y\,y' = 0$, or in differential form $$F_x\,dx + F_y\,dy = 0.$$ So if an equation $M\,dx + N\,dy = 0$ has $M = F_x$ and $N = F_y$ for some $F$, its solutions are simply the level curves $F(x, y) = C$. Such equations are called exact, and recognizing them uses the mixed partials from Higher-Order Partials.

Key idea

$M(x, y)\,dx + N(x, y)\,dy = 0$ is exact on a rectangle when $M_y = N_x$ there. Then find a potential function $F$ with $F_x = M$ and $F_y = N$; the solutions are $F(x, y) = C$.

Why it works

If $M = F_x$ and $N = F_y$, then $M_y = F_{xy}$ and $N_x = F_{yx}$, and these are equal for functions with continuous second partials (Clairaut's theorem). So $M_y = N_x$ is necessary. It is also enough: integrate $M$ in $x$ to get $F = \int M\,dx + h(y)$, then choose $h$ so that $F_y = N$. The condition $M_y = N_x$ is exactly what guarantees $h'$ depends on $y$ alone.

Worked example: Find the potential

Solve $(2xy + 3x^{2})\,dx + (x^{2} + 2y)\,dy$ $= 0$.

Check: $M_y = 2x$ and $N_x = 2x$. Exact.
Integrate $M$ in $x$ (holding $y$ fixed): $F = x^{2}y + x^{3} + h(y)$.
Then $F_y = x^{2} + h'(y)$ must equal $N = x^{2} + 2y$, so $h'(y) = 2y$ and $h = y^{2}$.
$F = x^{2}y + x^{3} + y^{2}$.

Answer: $x^{2}y + x^{3} + y^{2} = C$

Worked example: With an initial condition

Solve $(y\cos x + 2x)\,dx$ $+ (\sin x - 3y^{2})\,dy$ $= 0$ with $y(0) = 1$.

$M_y = \cos x = N_x$. Exact.
$F = \int (y\cos x + 2x)\,dx$ $= y\sin x + x^{2} + h(y)$.
$F_y = \sin x + h'(y)$ $= \sin x - 3y^{2}$, so $h = -y^{3}$.
At $(0, 1)$: $F = 0 + 0 - 1 = -1$.

Answer: $y\sin x + x^{2} - y^{3} = -1$, an implicit solution.

Worked example: Not exact, but close

Is $(3xy + y^{2})\,dx + (x^{2} + xy)\,dy$ $= 0$ exact? If not, try multiplying by $x$.

$M_y = 3x + 2y$ but $N_x = 2x + y$: not exact.
Multiply by $x$: $M = 3x^{2}y + xy^{2}$, $N = x^{3} + x^{2}y$.
Now $M_y = 3x^{2} + 2xy = N_x$: exact. Integrating gives $F = x^{3}y + \frac{1}{2}x^{2}y^{2}$.

Answer: $x^{3}y + \frac{1}{2}x^{2}y^{2} = C$. (Here $\frac{M_y - N_x}{N} = \frac{1}{x}$ depends on $x$ only, which is the signal that $\mu = e^{\int dx/x} = x$ works.)

Contours of $F = x^{2}y + x^{3} + y^{2}$ from the first example. Each contour is a solution curve. The gradient $(F_x, F_y) = (M, N)$ is perpendicular to it, which is exactly what $M\,dx + N\,dy = 0$ says.

Common mistake

Two frequent slips: checking $M_x = N_y$ instead of $M_y = N_x$, and forgetting the unknown $h(y)$ after integrating $M$ in $x$ (the “constant” of a partial integration can depend on $y$). Always finish by differentiating your $F$: you should get back both $M$ and $N$.

Exact Equations

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