Exact equations — exactness condition, finding potential function: Worked examples

Intuition What this page is for

The parent note gave you the machinery: test $\partial M / \partial y = \partial N / \partial x$ , then rebuild the hill $F (x, y)$ and write $F = c$ . Here we stress-test that machinery on every kind of equation you can meet: nice ones, sign-flipped ones, ones with a zero coefficient, ones that fail the test, ones that only become exact after a trick, and a real word problem.

Before we start, two reminders in plain words so no symbol is unearned:

$M$ is the coefficient sitting in front of $d x$ ; $N$ is the coefficient in front of $d y$ . Think of them as the two slopes of a hill $F$ : how steep it climbs as you step east ( $M = F_{x}$ ) and as you step north ( $N = F_{y}$ ).
$\partial M / \partial y$ (read "the partial of $M$ with respect to $y$ ", written $M_{y}$ ) means: freeze $x$ as if it were a fixed number, then differentiate $M$ treating only $y$ as the variable. That is the partial derivative.

The scenario matrix

Every exact-equation problem falls into one of these cells. The examples below are labelled with the cell they cover, so together they leave no gap.

#	Cell (scenario class)	What's tricky about it	Example
A	Polynomial, cleanly exact	baseline mechanics	Ex 1
B	Mixed transcendental (exp, trig, log)	integrating unusual functions	Ex 2
C	Negative / sign-flipped coefficients	signs must survive the test	Ex 3
D	One coefficient is a pure function of a single variable	a partial derivative is zero	Ex 4
E	Implicit-only solution (can't isolate $y$ )	answer stays as $F = c$	Ex 5
F	Not exact — test fails	must refuse the method	Ex 6
G	Not exact but fixable by an [[Integrating Factors for ODEs	integrating factor]]	rescue then solve
H	Real-world word problem + initial condition	pin the constant $c$	Ex 8

We also cover the degenerate limit (what happens when $M$ or $N$ vanishes entirely) inside Ex 4, and the geometry (the solution as level curves) in the figures for Ex 1 and Ex 8.

Cell A — polynomial, cleanly exact

Worked example Example 1

Solve $(3 x^{2} y + 2 x y) d x + (x^{3} + x^{2} + 2 y) d y = 0.$

Forecast: Guess before reading — does this pass the test? And will the answer contain an $x^{3} y$ term? Jot your guess.

Identify: $M = 3 x^{2} y + 2 x y$ , $N = x^{3} + x^{2} + 2 y .$

Step 1 — Test exactness. Why this step? There is no point rebuilding a hill that does not exist. Freeze $x$ , differentiate $M$ in $y$ : each term loses its $y$ , giving $M_{y} = 3 x^{2} + 2 x .$ Freeze $y$ , differentiate $N$ in $x$ : $N_{x} = 3 x^{2} + 2 x .$ They match ⟹ exact.

Step 2 — Integrate $M$ in $x$ . Why? $F_{x} = M$ , so undoing the $x$ -derivative recovers $F$ up to a $y$ -only piece. $F = \int (3 x^{2} y + 2 x y) d x + g (y) = x^{3} y + x^{2} y + g (y) .$

Step 3 — Differentiate in $y$ and match $N$ . Why? The leftover $g (y)$ is invisible to $M$ ; only $F_{y} = N$ can reveal it. $F_{y} = x^{3} + x^{2} + g^{'} (y) =! x^{3} + x^{2} + 2 y \Rightarrow g^{'} (y) = 2 y .$ Notice the $x^{3} + x^{2}$ part of $N$ was already inside $F$ — do not re-integrate it (that is the double-counting trap).

Step 4 — Integrate $g^{'}$ . $g (y) = y^{2} .$

Answer: $x^{3} y + x^{2} y + y^{2} = c .$

Verify: Take $F = x^{3} y + x^{2} y + y^{2}$ . Then $F_{x} = 3 x^{2} y + 2 x y = M$ ✓ and $F_{y} = x^{3} + x^{2} + 2 y = N$ ✓. The differential $d F$ rebuilds the original equation exactly.

Figure — Exact equations — exactness condition, finding potential function

The figure shows the contour lines $F = c$ : the ODE says "move so height never changes," so every solution curve is one of these level curves.

Cell B — mixed transcendental

Cell C — negative / sign-flipped coefficients

Worked example Example 3

Solve $(2 x - y sin (x y)) d x - x sin (x y) d y = 0.$

Forecast: $N$ is negative. Does the negative sign break exactness? Guess.

Identify: $M = 2 x - y sin (x y)$ , $N = - x sin (x y) .$

Step 1 — Test. Why? Signs matter; verify carefully. $M_{y}$ : differentiate $- y sin (x y)$ in $y$ by the product rule (freeze $x$ ): $M_{y} = - [sin (x y) + y cos (x y) \cdot x] = - sin (x y) - x y cos (x y) .$ $N_{x}$ : differentiate $- x sin (x y)$ in $x$ (freeze $y$ ): $N_{x} = - [sin (x y) + x cos (x y) \cdot y] = - sin (x y) - x y cos (x y) .$ They match ⟹ exact — the minus sign survives on both sides.

Step 2 — Integrate $N$ in $y$ this time (it's simpler; the symmetric route is always allowed). Freeze $x$ : $F = \int - x sin (x y) d y + h (x) = cos (x y) + h (x),$ because $\partial_{y} cos (x y) = - x sin (x y) .$

Step 3 — Match $M$ : Why? Now use $F_{x} = M$ to find $h (x)$ . $F_{x} = - y sin (x y) + h^{'} (x) =! 2 x - y sin (x y) \Rightarrow h^{'} (x) = 2 x .$

Step 4 — Integrate: $h (x) = x^{2} .$

Answer: $cos (x y) + x^{2} = c .$

Verify: $F_{x} = - y sin (x y) + 2 x = M$ ✓. $F_{y} = - x sin (x y) = N$ ✓.

Cell D — a coefficient depends on one variable only (partial = 0)

Cell E — implicit-only solution

Cell F — NOT exact (must refuse the method)

Cell G — not exact, but fixable

Worked example Example 7

Solve $(y^{2} + 2 x y) d x - x^{2} d y = 0.$

Forecast: Which variable will the integrating factor depend on? Guess $x$ or $y$ .

Identify: $M = y^{2} + 2 x y, N = - x^{2} .$

Step 1 — Test. $M_{y} = 2 y + 2 x, N_{x} = - 2 x .$ Not equal ⟹ not exact.

Step 2 — Try $μ (x)$ . Why? Check if the shortcut ratio is $x$ -only. $\frac{M _{y} - N _{x}}{N} = \frac{( 2 y + 2 x ) - ( - 2 x )}{- x ^{2}} = \frac{2 y + 4 x}{- x ^{2}},$ which still has $y$ — not $x$ -only. So $μ (x)$ fails; try $μ (y)$ .

Step 3 — Try $μ (y)$ . The mirror shortcut: $\frac{N _{x} - M _{y}}{M}$ should be $y$ -only. $\frac{N _{x} - M _{y}}{M} = \frac{- 2 x - ( 2 y + 2 x )}{y ^{2} + 2 x y} = \frac{- 4 x - 2 y}{y ( y + 2 x )} .$ Hmm, still mixed. Instead spot the factor directly: multiply by $μ = 1/ y^{2}$ : $(1 + \frac{2 x}{y}) d x - \frac{x ^{2}}{y ^{2}} d y = 0.$ Re-test: new $M^{*} = 1 + 2 x / y$ , new $N^{*} = - x^{2} / y^{2}$ . $M_{y}^{*} = - 2 x / y^{2}$ , $N_{x}^{*} = - 2 x / y^{2}$ . Match ⟹ now exact. Why $1/ y^{2}$ ? Because $- x^{2} / y^{2}$ is $\partial_{y}$ of $x^{2} / y$ , hinting the potential.

Step 4 — Integrate $M^{*}$ in $x$ : $F = x + \frac{x ^{2}}{y} + g (y) .$

Step 5 — Match $N^{*}$ : $F_{y} = - \frac{x ^{2}}{y ^{2}} + g^{'} (y) =! - \frac{x ^{2}}{y ^{2}} \Rightarrow g^{'} (y) = 0 \Rightarrow g = const .$

Answer: $x + \frac{x ^{2}}{y} = c .$

Verify: $F_{x} = 1 + 2 x / y = M^{*}$ ✓, $F_{y} = - x^{2} / y^{2} = N^{*}$ ✓. Multiplying back by $y^{2}$ returns the original equation, so solution curves are unchanged.

Cell H — real-world word problem with an initial condition

Worked example Example 8 — a conservative force field

A particle moves in the plane so that its motion always stays on a level curve of a hidden energy $F (x, y)$ . Measurements give the differential relation $(2 x + y) d x + (x + 2 y) d y = 0,$ and the particle passes through the point $(x, y) = (1, 2)$ (units: metres). Find the energy contour it rides on.

Forecast: This is a symmetric field ( $M$ and $N$ swap roles under $x \leftrightarrow y$ ). Guess whether $F$ has an $x y$ term.

Identify: $M = 2 x + y, N = x + 2 y .$

Step 1 — Test. Why? Confirm a potential/energy exists (a conservative field). $M_{y} = 1, N_{x} = 1.$ Match ⟹ exact, so an energy $F$ exists.

Step 2 — Integrate $M$ in $x$ : $F = x^{2} + x y + g (y) .$

Step 3 — Match $N$ : $F_{y} = x + g^{'} (y) =! x + 2 y \Rightarrow g^{'} (y) = 2 y \Rightarrow g (y) = y^{2} .$

Step 4 — General energy: $F = x^{2} + x y + y^{2} .$ General solution $x^{2} + x y + y^{2} = c .$

Step 5 — Apply the initial condition $(1, 2)$ : Why? To pin the specific contour, not the whole family (units: m²). $c = 1^{2} + (1) (2) + 2^{2} = 1 + 2 + 4 = 7.$

Answer: $x^{2} + x y + y^{2} = 7.$

Verify: $F_{x} = 2 x + y = M$ ✓, $F_{y} = x + 2 y = N$ ✓. Plug $(1, 2)$ : $1 + 2 + 4 = 7$ ✓ (units m² consistent as an energy contour).

The red dot is $(1, 2)$ ; the highlighted ellipse is the single contour $x^{2} + x y + y^{2} = 7$ the particle stays on. Nearby thin curves are other energy levels $c$ .

Recall Which cell does each example cover?

Ex1 = A (poly) ::: Ex2 = B (transcendental) Ex3 = C (signs) ::: Ex4 = D (degenerate / separable) Ex5 = E (implicit-only) ::: Ex6 = F (non-exact refusal) Ex7 = G (integrating factor rescue) ::: Ex8 = H (word problem + IC)