Exact Equation Checker | ∂M/∂y=∂N/∂x Test

📖 Tutorial｜Exact Equations

1. Definition

If there exists u(x,y) with du = M(x,y)dx + N(x,y)dy, then M dx + N dy = 0 is an exact equation. The test is ∂M/∂y = ∂N/∂x.

2. Symbols

Symbol	Meaning
M(x,y)	coefficient of dx
N(x,y)	coefficient of dy
u(x,y)	potential with du=Mdx+Ndy
∂M/∂y	partial derivative of M w.r.t. y

3. How this tool works

Necessary condition: ∂M/∂y = ∂N/∂x (also sufficient on a simply connected region);
This tool uses central differences to compute both partials numerically;
If they agree, u(x,y)=C is the general solution.

4. Steps

Enter M(x,y) and N(x,y);
Enter a test point;
Click Calculate to see both partials and the verdict.

5. Example

Example: 2xy dx + (x²+3y²) dy = 0. ∂M/∂y=2x, ∂N/∂x=2x, equal, so it is exact. The potential is u=x²y+y³=C.

6. Pitfalls

Finite differences have error; a difference below 1e-3 is treated as equal;
The region must be simply connected for sufficiency;
Exactness helps find the potential, not a numerical solution directly.

❓ FAQ｜Exact Equations

What is an exact equation?

One that has a potential u with du=Mdx+Ndy; then the equation is u(x,y)=C.

What is the test?

On a simply connected region, ∂M/∂y = ∂N/∂x everywhere.

Why finite differences here?

To demonstrate the test quickly; by hand, differentiate symbolically.

How to solve after the test passes?

Integrate M in x: u=∫M dx + φ(y), then differentiate in y and match N to find φ(y).

What if they are not equal?

Try an integrating factor μ(x,y) to make it exact.