What is an exact equation?

A first-order ODE M(x,y)dx + N(x,y)dy = 0 is called exact if there exists a function F(x,y) such that dF/dx = M and dF/dy = N. The necessary and sufficient condition (for functions with continuous partial derivatives) is dM/dy = dN/dx.

Why numerical differentiation instead of symbolic?

Symbolic differentiation requires a full computer-algebra system, which is impractical to load in a lightweight browser tool. Numerical central differences give excellent accuracy (error ~ h^2) and work for any expression math.js can evaluate.

What syntax can I use?

Expressions are evaluated by math.js. Use x and y as variables. Supported: +, -, *, /, ^ (power), sin, cos, tan, exp, log, sqrt, abs, pi, e. Example: x^2 + y^2, sin(x)*cos(y).

How is the potential function F estimated?

When the equation is exact, F(x,y) = integral of M dx + g(y). The tool numerically integrates M with respect to x using Simpson's rule from x=0 to a sample x, holding y fixed at several values, to approximate F at a grid of points.

What does the tolerance setting control?

Because finite-difference derivatives have small numerical errors, the tool uses a tolerance (default 1e-6). If max|dM/dy - dN/dx| is below this value across all sample points, the equation is considered exact.

Exact Equation Solver

Check whether M(x,y)dx + N(x,y)dy = 0 is an exact ODE by verifying dM/dy = dN/dx using numerical differentiation, with step-by-step output.

Presets:

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

M(x,y)

N(x,y)

Tolerance

Sample points

Result

Enter M and N, then press "Check Exactness".

Step-by-Step: Partial Derivative Evaluation

Central finite differences with h = 1e-5. Columns show ∂M/∂y and ∂N/∂x at each sample (x,y) pair.

#	x	y	∂M/∂y	∂N/∂x	\|diff\|	Match?

Potential Function F(x,y) — Numerical Estimate

Since dF/dx = M, integrate M with respect to x (Simpson's rule, 200 sub-intervals) from x = 0, at several y values. The implicit solution is F(x,y) = C.

x	y = 0.5	y = 1.0	y = 1.5	y = 2.0

Summary

Enter expressions for M(x,y) and N(x,y), then verify exactness by comparing the numerical partial derivatives dM/dy and dN/dx at several sample points. If the equation is exact, the tool also numerically estimates the potential function F(x,y) by integrating M with respect to x.

How it works

Enter expressions for M(x,y) and N(x,y) using x and y — e.g. 2*x*y and x^2 - 1.
The tool evaluates dM/dy and dN/dx at several sample (x,y) points using central finite differences.
It reports the maximum absolute difference between the two partials and compares it to the tolerance.
If the difference is below tolerance, the equation is declared exact.
When exact, the tool numerically integrates M with respect to x to estimate the potential function F(x,y).

Use cases

Quickly verify whether a given ODE is exact before attempting an analytic solution.
Check textbook problems on exact equations step-by-step.
Explore how M and N must relate for an equation to be exact.
Estimate the potential function F(x,y) numerically when the equation is exact.
Understand partial derivatives of two-variable expressions interactively.

Frequently Asked Questions

Last updated: 2026-05-25 · Reviewed by Nham Vu