Exact Equations and Potential Curves

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The idea

An equation $M(x,y)dx+N(x,y)dy=0$ is exact when it is $d\Phi=0$ for a potential $\Phi$ . Solutions then lie on level curves $\Phi=C$ . This connects first-order equations to conservative fields in multivariable calculus.

Visual guide

VISUAL GUIDESolutions follow potential level curves

xy = 0.5
xy = 1
xy = 2

For y dx + x dy = 0, the conserved potential is xy. Each hyperbola xy = C is a solution curve where it can be written as a graph. Moving along it leaves the potential unchanged.

Method and assumptions

On a suitable simply connected domain with continuous first partial derivatives, test $M_y=N_x$ . Integrate $M$ with respect to $x$ , add an unknown $g(y)$ , and match the resulting $\Phi_y$ with $N$ . An implicit curve can be locally solved for $y(x)$ where $N=\Phi_y\ne0$ .

Worked example: recover a potential

For $(2xy+1)dx+(x^2+2y)dy=0$ , the cross partials both equal $2x$ . Integrating $M$ gives $\Phi=x^2y+x+g(y)$ ; matching $N$ gives $g=y^2$ . Thus $x^2y+x+y^2=C$ . Through $(0,1)$ , $C=1$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For M(x,y)dx+N(x,y)dy=0, which derivative equality tests exactness on a suitable simply connected domain?

M=N

Mᵧ=Nₓ

Mₓ=Nᵧ

Hint 1 · Find a starting point

If M=ψₓ and N=ψᵧ, compare mixed partials.

Hint 2 · Take the next step

Differentiate M with respect to y and N with respect to x.

Show the reasoning

Answer: Mᵧ=Nₓ

Mᵧ=Nₓ is the compatibility condition for a potential under the stated domain and regularity assumptions.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a multiplying factor

$2y\,dx+x\,dy=0$ is not exact: $M_y=2$ , $N_x=1$ . Multiplying by $x$ gives $2xy\,dx+x^2dy=d(x^2y)$ . On an interval excluding $x=0$ , solutions satisfy $x^2y=C$ .

Interpreting the result

Multiplying by a factor that vanishes or blows up can change which points are admissible. State the working domain and check excluded points separately. A level curve may have a vertical tangent and fail to be one global graph.

Practice

Is $y\,dx+x\,dy=0$ exact?
Find its level curves through $(1,2)$ .
Is $y\,dx+2x\,dy=0$ exact as written?

Show worked solutions

Yes: both cross partials equal $1$ .
The potential is $xy$ , so $xy=2$ and locally $y=2/x$ .
No: the cross partials are $1$ and $2$ . A different method or factor is needed.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

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