Exact Equation Potential Relation: Build the Potential Function

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: after verifying that $M(x,y)\,dx+N(x,y)\,dy=0$ is exact, reconstruct a potential function $\Psi$ with $\Psi_x=M$ and $\Psi_y=N$, then write the solution as $\Psi(x,y)=C$.

The invariant: the solution curves are preserved because $d\Psi=\Psi_x\,dx+\Psi_y\,dy=M\,dx+N\,dy$, so the differential equation says $d\Psi=0$ along a solution curve.

Pattern:

$$M(x,y)\,dx+N(x,y)\,dy=0,\quad M=\Psi_x,\quad N=\Psi_y \quad \Longrightarrow \quad \Psi(x,y)=C$$

Legal route	Illegal route
$(y+2x)\,dx+(x+6y)\,dy=0,\quad M_y=N_x=1 \Longrightarrow \Psi(x,y)=C$	$(y+x)\,dx+(2x+y)\,dy=0,\quad M_y=1,\ N_x=2 \not\Longrightarrow \Psi(x,y)=C$

The invalid route is tempting because both equations have the same surface form: two coefficient fields multiplying $dx$ and $dy$. Exactness is the guard that says those fields come from one shared potential.

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Conditions of Applicability

Condition: $M and N are C^1 on the working region$; exactness criterion verified

Before applying, check: use the exactness criterion for the working region. In the standard simply connected textbook setting, this means comparing $M_y$ and $N_x$ on that same region before using the potential relation.

If the condition is violated: integrating $M$ with respect to $x$ and $N$ with respect to $y$ may produce incompatible pieces, so $\Psi(x,y)=C$ would solve a different problem or no problem at all.

• The form must be written as $M(x,y)\,dx+N(x,y)\,dy=0$ before checking exactness.
• The same working region matters. A form can behave differently across singularities or disconnected regions.
• Exactness is not the same as being separable or linear; use the route that the coefficient fields license.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: integrate $M$ and $N$ separately without checking $M_y=N_x$ -> the two antiderivatives may not come from one shared potential.

Debug: before reconstructing $\Psi$, compare $M_y$ and $N_x$ on the working region.

Failure mode: treat the integration “constant” as constant in both variables -> the missing term should be a function of the other variable.

Debug: after integrating $M$ with respect to $x$, write $\Psi=\int M\,dx+h(y)$, then use $\Psi_y=N$ to find $h$.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does $M=\Psi_x$ and $N=\Psi_y$ make $M\,dx+N\,dy$ equal to $d\Psi$?
• What information is carried by the arbitrary constant $C$ in $\Psi(x,y)=C$?

For the Principle

• What fast check tells you whether the potential-function route is legal?
• Why does the missing term after integrating $M$ have to be a function of $y$ rather than a plain constant?

Between Principles

• How does this exact-equation move differ from Separable Variable Separation Rewrite, even though both produce integrable forms?

Generate an Example

• Create one exact form and one near miss by choosing $M$ and $N$ so that $M_y=N_x$ in the first case and $M_y\neq N_x$ in the second.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Verify exactness, find one potential function with partial derivatives M and N, then write the implicit solution as potential equals C.

Write the canonical pattern: _____$M(x,y)\,dx+N(x,y)\,dy=0,\ M=\Psi_x,\ N=\Psi_y \Rightarrow \Psi(x,y)=C$

State the canonical condition: _____$M and N are C^1 on the working region;\, \text{exactness criterion verified}$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from $(2xy+3)\,dx+(x^2+4y)\,dy=0$, verify exactness and reach potential level-set form.

Step	Expression	Operation
0	$M=2xy+3,\quad N=x^2+4y$	Identify the two coefficient fields in $M\,dx+N\,dy=0$.
1	$M_y=2x,\quad N_x=2x$	Verify the exactness criterion.
2	$\Psi=\int (2xy+3)\,dx=x^2y+3x+h(y)$	Integrate $M$ with respect to $x$, keeping the missing part as $h(y)$.
3	$\Psi_y=x^2+h^{\prime}(y)=x^2+4y$	Match the $y$-partial derivative to $N$.
4	$h^{\prime}(y)=4y,\quad h(y)=2y^2$	Recover the missing function of $y$.
5	$x^2y+3x+2y^2=C$	Write the potential level set.

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Drills

Action Label

What was done between these two steps? Assume $M$ and $N$ are $C^1$ and exactness has been verified.

$$(2xy+3)\,dx+(x^2+4y)\,dy=0 \quad \Longrightarrow \quad \Psi_x=2xy+3,\quad \Psi_y=x^2+4y$$

Reveal

The exact-equation potential relation was selected. The differential form is being treated as $d\Psi=0$, so the coefficients become the partial derivatives of one potential function.

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What condition licenses this transition?

$$(y+2x)\,dx+(x+6y)\,dy=0 \quad \Longrightarrow \quad \Psi(x,y)=C$$

Reveal

The condition is that $M$ and $N$ are $C^1$ on the working region and exactness is verified. Here $M_y=1$ and $N_x=1$, so the potential-level-set route is legal.

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A student proposes a potential-function route. Should it be accepted?

$$(y+x)\,dx+(2x+y)\,dy=0$$

Reveal

Reject it as an exact-equation route. Here $M_y=1$ and $N_x=2$, so exactness is not verified. The form may need another method or an integrating factor, but it is not licensed for $\Psi(x,y)=C$ as written.

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Name the move in this chain.

$$\Psi=x^2y+h(y),\quad \Psi_y=x^2+h^{\prime}(y)=x^2+e^y \quad \Longrightarrow \quad h^{\prime}(y)=e^y$$

Reveal

This is the potential-reconstruction match. After integrating $M$ with respect to $x$, the unknown $h(y)$ is determined by enforcing $\Psi_y=N$.

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What was the illegal move?

$$(2xy)\,dx+(x^2+1)\,dy=0 \quad \Longrightarrow \quad x^2y+x^2y+y=C$$

Reveal

The student integrated both coefficient fields separately and added the results. For an exact equation, the goal is one potential $\Psi$ whose partial derivatives match $M$ and $N$, not a sum of two independent antiderivatives.

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Micro-Chain

Use the potential relation. Assume the coefficient fields are $C^1$ on the working region.

$$(3x^2+y)\,dx+(x+2y)\,dy=0$$

Reveal

First check exactness:

$$M_y=1,\quad N_x=1.$$

Then integrate $M$ with respect to $x$:

$$\Psi=x^3+xy+h(y).$$

Match $\Psi_y=N$:

$$\Psi_y=x+h^{\prime}(y)=x+2y, \quad h(y)=y^2.$$

So the solution is

$$x^3+xy+y^2=C.$$

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Use the potential relation. Assume the coefficient fields are $C^1$ on the working region.

$$(e^x+y^2)\,dx+(2xy+\cos y)\,dy=0$$

Reveal

Check exactness:

$$M_y=2y,\quad N_x=2y.$$

Integrate $M$ with respect to $x$:

$$\Psi=e^x+xy^2+h(y).$$

Match $\Psi_y=N$:

$$\Psi_y=2xy+h^{\prime}(y)=2xy+\cos y, \quad h(y)=\sin y.$$

Thus

$$e^x+xy^2+\sin y=C.$$

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Reject or complete the route choice. Assume $M$ and $N$ are $C^1$.

$$(3x^2+y)\,dx+(2x+2y)\,dy=0$$

Reveal

Reject the exact-equation route as written:

$$M_y=1,\quad N_x=2.$$

Since the exactness criterion is not verified, the potential relation is not legal for this form.

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Use the potential relation after checking exactness.

$$(2x\cos y)\,dx+(-x^2\sin y+3y^2)\,dy=0$$

Reveal

Exactness check:

$$M_y=-2x\sin y,\quad N_x=-2x\sin y.$$

Integrate $M$ with respect to $x$:

$$\Psi=x^2\cos y+h(y).$$

Match $\Psi_y=N$:

$$\Psi_y=-x^2\sin y+h^{\prime}(y)=-x^2\sin y+3y^2, \quad h(y)=y^3.$$

So

$$x^2\cos y+y^3=C.$$

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Transition Identification

Where does exactness enter this chain?

$$(y+2x)\,dx+(x+6y)\,dy=0 \quad \Longrightarrow \quad M_y=1,\ N_x=1 \quad \Longrightarrow \quad \Psi=x^2+xy+3y^2 \quad \Longrightarrow \quad \Psi=C$$

Reveal

Exactness enters in the second state, where $M_y$ and $N_x$ are compared. The later potential reconstruction is legal because that condition check passed.

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What is missing from this worked chain?

$$(2xy)\,dx+(x^2+4y)\,dy=0 \quad \Longrightarrow \quad \Psi=x^2y+2y^2 \quad \Longrightarrow \quad x^2y+2y^2=C$$

Reveal

The chain skipped the exactness check. It should verify $M_y=2x$ and $N_x=2x$ before using one potential function.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $(4x^3+2xy)\,dx+(x^2+3y^2)\,dy=0$, verify exactness and reach potential level-set form.

Full solution

Step	Expression	Move
0	$M=4x^3+2xy,\quad N=x^2+3y^2$	Identify coefficient fields.
1	$M_y=2x,\quad N_x=2x$	Verify exactness on the working region.
2	$\Psi=\int (4x^3+2xy)\,dx=x^4+x^2y+h(y)$	Integrate $M$ with respect to $x$.
3	$\Psi_y=x^2+h^{\prime}(y)=x^2+3y^2$	Enforce $\Psi_y=N$.
4	$h^{\prime}(y)=3y^2,\quad h(y)=y^3$	Recover the missing function.
5	$x^4+x^2y+y^3=C$	Write the implicit solution.

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Related Guides

• Differential Equations Subdomain - See where exact equations sit in the first-order ODE sequence.
• Separable Equation Product Form - Compare exactness with the route that separates variable factors.
• Separable Variable Separation Rewrite - Practice another legal transformation from ODE form to integrable form.
• Differential Equation Solution Condition - Check a candidate implicit solution against the original differential equation.
• Principle Structures - Treat the name, equation, and condition as separate recall targets.

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FAQ

What is Exact Equation Potential Relation?

Exact Equation Potential Relation is the move from an exact differential form $M(x,y)\,dx+N(x,y)\,dy=0$ to a potential level set $\Psi(x,y)=C$. It works when one potential function has $\Psi_x=M$ and $\Psi_y=N$.

How do I know an equation is exact?

Use the exactness criterion for the working region. In the standard simply connected textbook setting with $C^1$ coefficient fields, this means checking whether $M_y=N_x$; when that criterion is verified, the potential-function route is licensed.

Why is the solution written as potential equals C?

If $M\,dx+N\,dy=d\Psi$, then the differential equation says $d\Psi=0$ along a solution curve. That means $\Psi$ stays constant on the curve, so the implicit solution is $\Psi(x,y)=C$.

What is the common mistake in solving exact equations?

The common mistake is to integrate both coefficient fields separately and add the answers. Instead, integrate one coefficient, include the missing function of the other variable, and use the second coefficient to determine that missing function.

How is this different from separation of variables?

Separation rewrites an ODE so each variable’s differential appears with its own variable-dependent factor. Exact-equation solving instead recognizes a two-variable differential form as the total differential of one potential.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats exact equations as a route-selection move: check the coefficient fields, reconstruct the potential only when exactness is verified, then use the level set as the solution form. The Unisium Study System pairs that condition check with retrieval practice, self-explanation, and compact problem-solving chains so the potential-function route becomes fluent without hiding the legality check.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the complete framework in Masterful Learning.

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