Exact Differential Equation Solver

Math Calculus • Differential Equations

Written by STEM Calculators Team Published December 31, 2025 Updated February 24, 2026

6. Exact DE Solver

Solve $M(x,y)\,dx + N(x,y)\,dy = 0$ when $\partial M/\partial y = \partial N/\partial x$ by finding a potential $\Phi(x,y)$ such that $\Phi_x=M,\ \Phi_y=N$ , hence $\Phi(x,y)=C$ .

Input supports: pi, e, sqrt( ), sin, cos, exp, log. Use * for multiplication.

$M(x,y)$: $N(x,y)$:

Initial condition (optional)

$x_0$: $y_0$: Use to set $C=\Phi(x_0,y_0)$: Manual $C$:

The graph marks $(x_0,y_0)$ with a coordinate label. If “Use IC” is enabled, the plotted main level is $\Phi(x,y)=\Phi(x_0,y_0)$.

Graph settings (optional)

$x_{\min}$: $x_{\max}$: $y_{\min}$: $y_{\max}$: Extra levels (comma): Grid resolution:

Drag to pan, mouse wheel to zoom, double-click to reset view. Axes units are your $x,y$ input units.

Ready

Steps

Enter $M(x,y)$ and $N(x,y)$, then click Solve.

Graph

Implicit solution: $\Phi(x,y)=C$ (level curves)

Hover to probe $(x,y)$ and $\Phi(x,y)$.

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Theory

Exact differential equations

An equation of the form \[ M(x,y)\,dx + N(x,y)\,dy = 0 \] is called exact on a region if there exists a potential function $\Phi(x,y)$ such that \[ \Phi_x = \frac{\partial \Phi}{\partial x} = M(x,y), \qquad \Phi_y = \frac{\partial \Phi}{\partial y} = N(x,y). \] In that case the differential form equals $d\Phi$, and the solution is the level set \[ \Phi(x,y)=C. \]

Exactness test

If $M$ and $N$ have continuous first partial derivatives on a simply connected region, then exactness is equivalent to \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \]

Finding the potential $\Phi$

A common method is:

Integrate $M(x,y)$ with respect to $x$: \[ \Phi(x,y)=\int M(x,y)\,dx + g(y). \]
Differentiate with respect to $y$ and match $N$: \[ \Phi_y = \frac{\partial}{\partial y}\!\left(\int M\,dx\right) + g'(y) = N(x,y). \] Hence \[ g'(y)=N(x,y) - \frac{\partial}{\partial y}\!\left(\int M\,dx\right), \] and then $g(y)=\int g'(y)\,dy$.

Initial condition

If the curve passes through a point $(x_0,y_0)$, substitute into $\Phi(x,y)=C$ to get \[ C=\Phi(x_0,y_0). \] This selects the specific level curve through $(x_0,y_0)$.

If it’s not exact: integrating factors

Sometimes multiplying by $\mu(x)$ or $\mu(y)$ makes the equation exact. Useful quick checks:

If $\displaystyle \frac{M_y - N_x}{N}$ depends only on $x$, then \[ \mu(x)=e^{\int \frac{M_y - N_x}{N}\,dx}. \]
If $\displaystyle \frac{N_x - M_y}{M}$ depends only on $y$, then \[ \mu(y)=e^{\int \frac{N_x - M_y}{M}\,dy}. \]

Geometric view: when exact, $\Phi(x,y)=C$ is an implicit family of curves (level curves) and $\nabla\Phi=\langle M,N\rangle$ is normal to those curves.

Frequently Asked Questions

What is an exact differential equation?

An equation M(x,y) dx + N(x,y) dy = 0 is exact if there exists a potential function Phi(x,y) such that Phi_x = M and Phi_y = N. The solution is given implicitly by Phi(x,y) = C.

How do you check whether M(x,y) dx + N(x,y) dy = 0 is exact?

A common test is comparing the mixed partials: dM/dy and dN/dx. If they are equal on a region where M and N have continuous first partial derivatives, the equation is exact.

How is the potential function Phi(x,y) found for an exact equation?

Integrate M(x,y) with respect to x to get Phi(x,y) = integral M dx + g(y), then differentiate with respect to y and match N(x,y) to determine g(y).

How does an initial condition determine the constant C?

After finding Phi(x,y), substitute the point (x0,y0) into Phi(x,y) = C to get C = Phi(x0,y0). This selects the specific level curve passing through the given point.

Why might the solver report that the equation is not exact?

If dM/dy and dN/dx are not equal, the differential form is not exact under the standard test. In some cases an integrating factor depending only on x or only on y can make it exact, but that requires additional analysis beyond the basic exactness condition.

Steps

Graph

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Frequently Asked Questions

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