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

Name: Exact Equation Solver
Availability: InStock
Author: Nham Vu

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

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.

How it works

A first-order ODE M(x,y)dx + N(x,y)dy = 0 is exact when dM/dy equals dN/dx everywhere. This tool evaluates both partial derivatives using central finite differences: df/dy = (f(x,y+h) - f(x,y-h)) / (2h). It samples several (x,y) pairs, compares the two partials numerically, and reports the maximum absolute difference. If that difference is below the tolerance, the equation is exact and a numerical estimate of F(x,y) is computed by integrating M dx at a reference point.

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: 2026-05-23 · Reviewed by Nham Vu