Exact Equations

Main idea

Some equations $M + N\,y' = 0$ are secretly saying that a function $\Psi(x, y)$ isn't changing along the solution. Then the solutions are just its level curves $\Psi(x, y) = C$ , and the quick test $M_y = N_x$ tells you when such a $\Psi$ exists.

M + N\,y' = 0 \text{ is exact} \iff M_y = N_x, \qquad {\htmlData{sym=0}{\Psi}}_x = M,\;\; {\htmlData{sym=0}{\Psi}}_y = N

What the symbols mean

$\Psi$
Potential function
Solutions stay on its level curves, because dΨ/dx = M + N y′ = 0.

Explanation

An exact equation is simply one that's secretly "the derivative of one function $\Psi(x, y)$ equals zero" in disguise. "Exact" because the left side is exactly a total derivative (an exact differential, in the older language), with nothing left over. When that happens the solutions are just the level curves of $\Psi$ , like contour lines on a map, which is how energy conservation shows up in physics. In $M + N\,y' = 0$ , the pieces $M$ (the part without $y'$ ) and $N$ (the part multiplying $y'$ ) are the two partial derivatives of $\Psi$ : $\Psi_x = M$ (how $\Psi$ changes as $x$ moves) and $\Psi_y = N$ (how it changes as $y$ moves), and the test $M_y = N_x$ just checks that such a $\Psi$ can exist. It looks like a lot of letters, but it's just a few simple parts, so bear with it.

Separable equations worked because the left side turned out to be the derivative of something. Exact equations push that idea into two variables. Suppose you have a function $\Psi(x, y)$ , and you walk along a solution curve $y(x)$ . The multivariable chain rule says

\frac{d}{dx}\Psi(x, y(x)) = \Psi_x + \Psi_y\,y'.

Now look at the equation $M + N\,y' = 0$ . It has exactly that shape.

Why it works. If $\Psi_x = M$ and $\Psi_y = N$ , the chain rule gives $\frac{d}{dx}\Psi(x, y(x)) = M + Ny' = 0$ . So the equation is really saying " $\Psi$ doesn't change as you move along a solution", which means each solution stays on one level curve $\Psi = C$ . When does such a $\Psi$ exist? If it does, Clairaut's theorem says $\Psi_{xy} = \Psi_{yx}$ , which is $M_y = N_x$ . So that test is necessary, and on nice regions it's also sufficient.

Finding $\Psi$ (the same process as finding a potential for a conservative vector field):

$\Psi = \int M\,dx + h(y)$ .
Differentiate in $y$ and set equal to $N$ to find $h'(y)$ .
Integrate for $h(y)$ . The solution is $\Psi(x, y) = C$ , with $C$ from the initial condition.

A common mistake. It's tempting to stop at $\Psi = \int M\,dx + C$ , because that's how integrating normally ends. It isn't enough. You integrated in $x$ while treating $y$ as a constant, so the "constant" of integration can be any function of $y$ . That's the $h(y)$ , and without it $\Psi_y$ usually won't match $N$ .

See it

These are solutions of (2xy + 3) + (x² − 1)y′ = 0: the level curves x²y + 3x − y = C, which here can be solved for y. As C changes, the curve sweeps through the whole family. Notice that every curve breaks at x = ±1, where the coefficient x² − 1 in front of y′ is zero. The solution with y(0) = 2 is the level curve C = −2.

Worked examples

Example 1

Solve $2xy - 9x^2 + \left(2y + x^2 + 1\right)y' = 0$ .

1
Check the test first: $M_y = 2x = N_x$ , so it's exact.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Is $(2xy + 3) + (x^2 - 1)y' = 0$ exact?

Problem 2+5 XP

For that equation $\Psi = x^2y + 3x - y$ . If $y(0) = 2$ , what is $C$ in $\Psi = C$ ?

Problem 3+5 XP

Is $y + (-x)y' = 0$ exact?

Problem 4+5 XP

Find $k$ so that $(kxy + 1) + (x^2 + y)y' = 0$ is exact.

Problem 5+5 XP

The solutions of $2x + y + (x + 2y)y' = 0$ are $x^2 + xy + y^2 = C$ . Find $C$ for the solution through $(1, 1)$ .

Problem 6+5 XP

A drone flies so that the signal strength $\Psi(x, y) = x^2 + xy + y^2$ stays constant, which means its path satisfies $(2x + y) + (x + 2y)y' = 0$ . It starts at $(1, 1)$ . Where does its path cross the positive $x$ -axis?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $(2xy - 9x^2) + (2y + x^2 + 1)y' = 0$ with $y(0) = -3$ , and give $y$ explicitly as a function of $x$ .

Where it's used

Exact equations are energy conservation in disguise: for a frictionless system the energy function is Ψ, and trajectories stay on its level curves.

Air Flowing Around a CylinderDiff Eq · Physics

Separable Equations Bernoulli Equations