Exact differential equations

Some first-order ODEs are secretly a total differential of a single function in disguise. Write the equation as $M(x,y)\,dx + N(x,y)\,dy = 0$ and ask whether the left side is exactly $du$ for some potential function $u(x,y)$ — if so, the level curves $u(x,y) = C$ are the solution, with no separation of variables and no integrating trick needed beyond one test. Within the MA10001 Ordinary Differential Equations unit, exact equations sit right after separable equations as the second core technique: many first-order, first-degree ODEs are not separable but are exact, and a huge further class becomes exact once multiplied by the right integrating factor. This article builds the exactness test, the two-step construction of $u(x,y)$ , and the standard integrating-factor shortcuts, then drills both on fully worked examples.

The core insight

An equation $M\,dx + N\,dy = 0$ is exact exactly when $M_y = N_x$ , and then its solution is the potential function $u(x,y) = C$ built by partially integrating $M$ or $N$ and matching the leftover unknown function to the other side.

The core idea

Differential form. Any first-order, first-degree ODE can be rearranged into

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

where: $M$ and $N$ are functions of $x$ and $y$ with continuous first partial derivatives on the region of interest. This is the standard form every exactness discussion starts from — even an equation given as $dy/dx = f(x,y)$ must first be moved to this shape by careful regrouping.

What "exact" means. The equation is exact if some potential function $u(x,y)$ has $M$ and $N$ as its own partial derivatives:

u_x = M, \qquad u_y = N,

where: $u_x$ and $u_y$ denote $\partial u/\partial x$ and $\partial u/\partial y$ . When this holds, the total differential $du = u_x\,dx + u_y\,dy$ is literally $M\,dx + N\,dy$ , so the original ODE reads $du = 0$ , and integrating once gives the implicit solution $u(x,y) = C$ .

The exactness test. Because mixed partial derivatives of a well-behaved function agree, $u_{xy} = u_{yx}$ , exactness is equivalent to a single checkable condition:

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.

where: both sides are ordinary partial derivatives of the given $M$ and $N$ — no potential function needs to be known in advance to run this test. It is a necessary and sufficient condition (given continuity of the first partials), so a single computation settles the question before any integration is attempted.

Geometric picture. The solution curves $u(x,y) = C$ are the level curves (contours) of the potential $u$ . Moving along one of these curves keeps $u$ constant, so $du = 0$ there — which is precisely the differential equation. Exactness is the statement that the ODE is the "contour equation" of some hidden surface $z = u(x,y)$ .

Figure

Level curves of a potential function u equal to C shown as nested contour rings, each labeled with an increasing constant, with the exact equation M dx plus N dy equals zero holding along every contour

Sometimes it isn't exact — yet. If $M_y \neq N_x$ , multiplying every term by a suitable integrating factor $\mu(x,y)$ can restore exactness: the modified equation $\mu M\,dx + \mu N\,dy = 0$ passes the test even though the original did not. Finding a general $\mu$ is hard, but two special cases are routine to check: if $(M_y - N_x)/N$ depends on $x$ only, then $\mu(x) = e^{\int (M_y-N_x)/N\,dx}$ ; if $(N_x - M_y)/M$ depends on $y$ only, then $\mu(y) = e^{\int (N_x-M_y)/M\,dy}$ .

Derivation AConstructing the solution u(x,y) from an exact equation

The goal is to turn the bare exactness condition $M_y = N_x$ into an explicit, mechanical recipe that produces the potential function $u(x,y)$ and hence the implicit general solution $u(x,y) = C$ , using nothing beyond partial integration and matching one unknown function of a single variable.

Step 1. Start from the exactness condition $u_x = M$ . Treat $y$ as a fixed constant and integrate both sides with respect to $x$ . Because the "constant" of integration with respect to $x$ can still depend on $y$ , write

u = \int M\,dx + \phi(y),

where $\phi(y)$ is an unknown function of $y$ alone, not yet determined. This is the one place students most often go wrong: dropping $\phi(y)$ and replacing it by a plain numeric constant throws away part of the solution, because any function of $y$ alone differentiates to zero with respect to $x$ and is therefore invisible to the integration in this step.

Step 2. Use the second condition, $u_y = N$ , to pin down $\phi(y)$ . Differentiate the expression for $u$ from Step 1 with respect to $y$ :

u_y = \frac{\partial}{\partial y}\left(\int M\,dx\right) + \phi'(y).

Set this equal to $N$ and solve for $\phi'(y)$ . Because $M_y = N_x$ was already guaranteed by the exactness test, every $x$ -dependent term is guaranteed to cancel out in this equation, leaving $\phi'(y)$ as a function of $y$ only — if an $x$ -term survives, an arithmetic slip has occurred somewhere upstream, since consistency is not optional here.

Step 3. Integrate $\phi'(y)$ with respect to $y$ to recover $\phi(y)$ itself (the numeric constant of this last integration is absorbed into the overall constant $C$ later, so it can be dropped). Substitute $\phi(y)$ back into the expression from Step 1 to obtain the complete potential function $u(x,y)$ .

Step 4. State the implicit general solution as $u(x,y) = C$ , where $C$ is an arbitrary constant fixed later by any given initial condition. As a check, differentiate the found $u$ partially with respect to $x$ and $y$ and confirm the results reproduce the original $M$ and $N$ exactly — this catches nearly every algebra slip made in Steps 1 through 3. The mirror-image route — integrating $N$ with respect to $y$ first and matching against $M$ — works identically and is preferable whenever $N$ is easier to integrate than $M$ .

Worked example

Solve $(2xy + 3)\,dx + (x^2 - 1)\,dy = 0$ , and find the particular solution satisfying $y(0) = 2$ .

Step 1. Identify $M$ and $N$ in the standard differential form: $M = 2xy + 3$ and $N = x^2 - 1$ .

Step 2. Test exactness. $M_y = 2x$ and $N_x = 2x$ . Since $M_y = N_x$ , the equation is exact, so the potential-function method applies directly.

Step 3. Integrate $M$ with respect to $x$ , holding $y$ fixed:

u = \int (2xy + 3)\,dx = x^2 y + 3x + \phi(y).

Step 4. Differentiate with respect to $y$ and match to $N$ : $u_y = x^2 + \phi'(y)$ , and this must equal $N = x^2 - 1$ , so $\phi'(y) = -1$ , giving $\phi(y) = -y$ .

Step 5. Assemble the general solution: $u = x^2y + 3x - y = C$ . Check: $u_x = 2xy + 3 = M$ and $u_y = x^2 - 1 = N$ , as required.

Step 6. Apply the initial condition $x = 0, y = 2$ : $(0)^2(2) + 3(0) - 2 = C$ , so $C = -2$ .

Final answer: the particular solution is $x^2y + 3x - y = -2$ .

Competitive stretch

The same family $(3x^2 + 6xy^2)\,dx + (6x^2y + 4y^3)\,dy = 0$ with $y(1) = 1$ is worth knowing by heart: here it is faster to integrate $N$ with respect to $y$ first, giving $u = 3x^2y^2 + y^4 + \psi(x)$ , then matching $u_x = 6xy^2 + \psi'(x)$ to $M = 3x^2 + 6xy^2$ gives $\psi'(x) = 3x^2$ , so $\psi(x) = x^3$ . The general solution is $x^3 + 3x^2y^2 + y^4 = C$ , and the initial condition forces $C = 1 + 3 + 1 = 5$ , so the particular solution is $x^3 + 3x^2y^2 + y^4 = 5$ . Recognising which side, $M$ or $N$ , is quicker to integrate first is a genuine time-saver under exam pressure.

Common traps

Skipping the exactness test. Running the potential-function procedure on a non-exact equation produces an inconsistent or meaningless result — always confirm $M_y = N_x$ before integrating anything.
Treating $\phi(y)$ as a numeric constant. It is an unknown function of the other variable; dropping it at the first integration step silently discards part of the solution.
Getting an $x$ -term when solving for $\phi'(y)$ . If matching $u_y$ to $N$ leaves an $x$ -dependent piece, that signals an earlier arithmetic error, since $M_y = N_x$ already guarantees clean cancellation.
Sign or regrouping errors converting $dy/dx = f(x,y)$ into $M\,dx + N\,dy = 0$ . Move every term to one side deliberately before reading off $M$ and $N$ ; a stray sign flips the whole exactness test.

Ask the tutor: if the two-step construction of $u(x,y)$ feels shaky, or you want more drill on spotting exact equations and integrating factors quickly, ask the AI tutor to walk through the reasoning step by step.

Why does M sub y equal N sub x guarantee exactness instead of just being a convenient coincidence?

It follows from a genuine theorem, not a lucky pattern. If $u_x = M$ and $u_y = N$ for some smooth potential $u$ , then differentiating the first equation with respect to $y$ and the second with respect to $x$ gives $u_{xy} = M_y$ and $u_{yx} = N_x$ . Since mixed partial derivatives of a function with continuous derivatives are always equal, $u_{xy} = u_{yx}$ , it must be that $M_y = N_x$ . The converse — that $M_y = N_x$ is enough to guarantee such a $u$ exists — is a standard result proved by actually constructing $u$ via the two-step integration, which is exactly the procedure used in the worked examples.

What do I do if M sub y does not equal N sub x?

The equation is not exact as written, but it may become exact after multiplying through by an integrating factor $\mu$ . First check whether $(M_y - N_x)/N$ depends on $x$ alone; if so, $\mu(x) = e^{\int (M_y-N_x)/N\,dx}$ works. If instead $(N_x - M_y)/M$ depends on $y$ alone, use $\mu(y) = e^{\int (N_x-M_y)/M\,dy}$ . Multiply the whole equation by whichever $\mu$ applies, recheck exactness (it will now hold), and then solve exactly as usual. If neither special case applies, finding $\mu$ in general is a harder problem outside this unit's standard toolkit.

Is checking u by differentiating back really necessary, or is it just extra work?

It is worth the thirty seconds. The construction in Steps 1 to 3 involves two separate integrations and one differentiation, any of which can carry a sign or algebra slip that would otherwise go unnoticed until marks are lost. Differentiating the final $u$ partially with respect to $x$ and $y$ and comparing against the original $M$ and $N$ catches essentially every such error immediately, because the check must reproduce the two given functions exactly — any mismatch points straight back to where the mistake happened.

Exact differential equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads