Integrating factors for non-exact equations: Worked examples

This page is the "shooting range" for Integrating factors for non-exact equations. We fire every kind of problem at you — clean $x$ -factors, clean $y$ -factors, the trap where you must switch mid-solve, an already-exact equation (the degenerate case), a limiting/undefined case, a word problem, and an exam twist — and we work each to the finish, then verify.

Before we start: everything here rests on one test and two formulas from the parent note. Let me restate them in plain words so no symbol appears un-earned.

Recall The three tools we lean on

The exactness test. Write the equation as $M d x + N d y = 0$ . Here $M$ is "the stuff multiplying a tiny step $d x$ to the right" and $N$ is "the stuff multiplying a tiny step $d y$ upward". The equation is exact when $M_{y} = N_{x}$ — read "the $y$ -slope of $M$ equals the $x$ -slope of $N$ ". ( $M_{y}$ means: freeze $x$ , ask how fast $M$ changes as $y$ grows.) $x$ -factor test: if $\frac{M _{y} - N _{x}}{N}$ has no $y$ left in it, call it $g (x)$ and set $μ = exp \int g d x$ . $y$ -factor test: if $\frac{N _{x} - M _{y}}{M}$ has no $x$ left in it, call it $h (y)$ and set $μ = exp \int h d y$ . A picture of what " $μ$ fixes" is coming in the first figure.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are tagged with the cell they hit, and together they cover all of them.

#	Cell (scenario class)	What makes it special	Example
A	Already exact ( $M_{y} = N_{x}$ )	degenerate case: $μ = 1$ , no fixing needed	Ex 1
B	Clean $x$ -factor	ratio loses its $y$	Ex 2
C	Clean $y$ -factor	first ratio still has $x$ ; switch and win	Ex 3
D	Sign trap in the $y$ -numerator	$N_{x} - M_{y}$ vs $M_{y} - N_{x}$ — must not carry old sign	Ex 4
E	Neither pure- $x$ nor pure- $y$ (limiting/failure)	both ratios keep both letters — what to do	Ex 5
F	Word problem (real units)	model → identify $M, N$ → solve → check units	Ex 6
G	Linear ODE in disguise	$μ = e^{\int P d x}$ is a special $x$ -factor	Ex 7
H	Exam twist: negative/zero regions	sign of $μ$ , domain where $μ$ undefined	Ex 8

Figure — Integrating factors for non-exact equations

Cell A — already exact (the degenerate case)

Worked example Ex 1. Solve

$(2 x y + 1) d x + (x^{2}) d y = 0.$ Forecast: guess before reading — will you need an integrating factor at all? Compute the two slopes in your head first.

Step 1 — read off $M, N$ . $M = 2 x y + 1$ , $N = x^{2}$ . Why this step? Everything downstream needs $M$ and $N$ named.

Step 2 — test exactness. $M_{y} = 2 x$ (freeze $x$ , differentiate in $y$ ), $N_{x} = 2 x$ (freeze $y$ , differentiate in $x$ ). Equal! Why this step? If $M_{y} = N_{x}$ we are in Cell A: $μ = 1$ , skip all the factor machinery.

Step 3 — build $F$ . $F = \int M d x = \int (2 x y + 1) d x = x^{2} y + x + ϕ (y) .$ Why the $ϕ (y)$ ? Integrating in $x$ freezes $y$ , so the "constant" may secretly depend on $y$ .

Step 4 — match $N$ . $F_{y} = x^{2} + ϕ^{'} (y)$ must equal $N = x^{2}$ , so $ϕ^{'} (y) = 0$ , $ϕ = const$ .

$x^{2} y + x = C .$

Verify: differentiate the answer implicitly: $d (x^{2} y + x) = (2 x y + 1) d x + x^{2} d y = 0$ . Exactly the original left side. ✓

Cell B — clean $x$ -factor

Worked example Ex 2. Solve

$(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0.$ Forecast: which single-variable ratio do you bet loses its extra letter — the $x$ -one or the $y$ -one?

Step 1 — name and test. $M = 3 x y + y^{2}$ , $N = x^{2} + x y$ . $M_{y} = 3 x + 2 y$ , $N_{x} = 2 x + y$ . Not equal → not exact. Why this step? Confirms we need a factor before wasting effort building $F$ .

Step 2 — try the $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{( 3 x + 2 y ) - ( 2 x + y )}{x ^{2} + x y} = \frac{x + y}{x ( x + y )} = \frac{1}{x} .$ The $(x + y)$ cancels and we are left with a pure- $x$ function — so the $x$ -factor exists. This is a clean Cell B. Why this step? The whole point of Cell B: the $x$ -ratio genuinely loses its $y$ .

Step 3 — build $μ$ . $μ = exp \int \frac{1}{x} d x = e^{l n x} = x .$

Step 4 — multiply and re-test. Using $M_{⋆} = μ M$ , $N_{⋆} = μ N$ : $M_{⋆} = x (3 x y + y^{2}) = 3 x^{2} y + x y^{2}$ , $N_{⋆} = x (x^{2} + x y) = x^{3} + x^{2} y$ . $(M_{⋆})_{y} = 3 x^{2} + 2 x y$ , $(N_{⋆})_{x} = 3 x^{2} + 2 x y$ . Equal ✓ Why re-test? The parent note's #1 mistake: never trust the factor without re-checking.

Step 5 — build $F$ . $F = \int M_{⋆} d x = \int (3 x^{2} y + x y^{2}) d x = x^{3} y + \frac{1}{2} x^{2} y^{2} + ϕ (y) .$ $F_{y} = x^{3} + x^{2} y + ϕ^{'} (y) = N_{⋆} = x^{3} + x^{2} y \Rightarrow ϕ^{'} = 0.$

$x^{3} y + \frac{1}{2} x^{2} y^{2} = C .$

Verify: $d (x^{3} y + \frac{1}{2} x^{2} y^{2}) = (3 x^{2} y + x y^{2}) d x + (x^{3} + x^{2} y) d y = μ (M d x + N d y)$ with $μ = x$ . ✓

Cell C — first ratio has $x$ , switch and win

Worked example Ex 3. Solve

$y d x + (2 x - y e^{y}) d y = 0.$ (the parent's Ex 2, re-checked here) Forecast: which test wins, and what is $ϕ (y)$ going to require (a plain integral or by-parts)?

Step 1 — name and test. $M = y$ , $N = 2 x - y e^{y}$ . $M_{y} = 1$ , $N_{x} = 2$ . Not exact.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{- 1}{2 x - y e ^{y}}$ — still has $x$ . ✗ $y$ -test. $\frac{N _{x} - M _{y}}{M} = \frac{2 - 1}{y} = \frac{1}{y}$ — pure $y$ . ✓ (Cell C.)

Step 3 — $μ$ . $μ = exp \int \frac{1}{y} d y = y .$

Step 4 — multiply, re-test. $M_{⋆} = y^{2}$ , $N_{⋆} = 2 x y - y^{2} e^{y}$ . $(M_{⋆})_{y} = 2 y = (N_{⋆})_{x}$ . ✓

Step 5 — $F$ . $F = \int y^{2} d x = x y^{2} + ϕ (y)$ . Then $F_{y} = 2 x y + ϕ^{'} (y) = 2 x y - y^{2} e^{y} \Rightarrow ϕ^{'} = - y^{2} e^{y}$ . Integrate by parts twice: $\int y^{2} e^{y} d y = e^{y} (y^{2} - 2 y + 2)$ , so $ϕ = - e^{y} (y^{2} - 2 y + 2)$ .

$x y^{2} - e^{y} (y^{2} - 2 y + 2) = C .$

Verify: $ϕ^{'} (y) = \frac{d}{d y} [- e^{y} (y^{2} - 2 y + 2)] = - e^{y} (y^{2} - 2 y + 2) - e^{y} (2 y - 2) = - e^{y} y^{2}$ . ✓

Cell D — the sign trap

Worked example Ex 4. Solve

$(x y - 1) d x + (x^{2} - x y) d y = 0.$ Forecast: the danger here is copying $M_{y} - N_{x}$ into the $y$ -formula. What sign belongs on top?

Step 1 — name and test. $M = x y - 1$ , $N = x^{2} - x y$ . $M_{y} = x$ , $N_{x} = 2 x - y$ . Not exact.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{x - ( 2 x - y )}{x ^{2} - x y} = \frac{- x + y}{x ( x - y )} = \frac{- ( x - y )}{x ( x - y )} = \frac{- 1}{x}$ — pure $x$ . ✓ Why did it clean up? The numerator $y - x$ is exactly $- (x - y)$ , cancelling the $(x - y)$ hiding in $N$ .

Step 3 — $μ$ . $μ = exp \int \frac{- 1}{x} d x = e^{- l n x} = x^{- 1} .$

Step 4 — multiply, re-test. $M_{⋆} = \frac{x y - 1}{x} = y - \frac{1}{x}$ , $N_{⋆} = \frac{x ^{2} - x y}{x} = x - y$ . $(M_{⋆})_{y} = 1$ , $(N_{⋆})_{x} = 1$ . ✓

Step 5 — $F$ . $F = \int (y - \frac{1}{x}) d x = x y - ln ∣ x ∣ + ϕ (y)$ . $F_{y} = x + ϕ^{'} (y) = x - y \Rightarrow ϕ^{'} = - y \Rightarrow ϕ = - \frac{1}{2} y^{2} .$

$x y - ln ∣ x ∣ - \frac{1}{2} y^{2} = C .$

Verify: $d (answer) = (y - \frac{1}{x}) d x + (x - y) d y = \frac{1}{x} (M d x + N d y)$ . ✓

Cell E — neither pure- $x$ nor pure- $y$ (the limiting/failure case)

Worked example Ex 5. Attempt

$(y) d x + (x^{2} + y^{2}) d y = 0$ with a single-variable factor. Forecast: do you expect $μ (x)$ or $μ (y)$ to exist? Predict "neither" and see why.

Step 1 — test. $M = y$ , $N = x^{2} + y^{2}$ . $M_{y} = 1$ , $N_{x} = 2 x$ . Not exact.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{1 - 2 x}{x ^{2} + y ^{2}}$ — has BOTH letters. ✗ $y$ -test. $\frac{N _{x} - M _{y}}{M} = \frac{2 x - 1}{y}$ — the $x$ survives. ✗ Why does this happen? Neither ratio collapses to one variable, so no single-variable $μ$ exists. This is Cell E, the honest failure case.

Step 3 — what to do instead. You must either guess a factor of the form $μ (x^{a} y^{b})$ or a more general shape, or fall back on a different method entirely. Why mention it? So you never grind forever on a formula that cannot apply. Recognising failure fast is a skill.

Escape route (bonus). Rewrite as $\frac{d x}{d y} = - \frac{x ^{2} + y ^{2}}{y}$ — not our topic, but a homogeneous-type substitution $x = v y$ handles it. The lesson stays: Cell E means the parent's two formulas legitimately give nothing.

Verify (that both ratios really keep both letters): at $(x, y) = (1, 1)$ the $x$ -ratio is $\frac{1 - 2}{2} = - \frac{1}{2}$ but at $(1, 2)$ it is $\frac{1 - 2}{5} = - \frac{1}{5}$ — the value changed with $y$ at fixed $x$ , proving it is not a function of $x$ alone. ✓

Cell F — a word problem with real units

Worked example Ex 6. A conserved-quantity model gives

$2 x y d x + (x^{2} + 3 y^{2}) d y = 0$ , where $x$ (metres) is a position coordinate and $y$ (metres) is a second position coordinate of a particle tracing an equipotential. Find the family of curves and check dimensional consistency. Forecast: exact or not? Peek at the two slopes.

Step 1 — name and test. $M = 2 x y$ , $N = x^{2} + 3 y^{2}$ . $M_{y} = 2 x$ , $N_{x} = 2 x$ . Already exact (Cell A pattern inside a word problem). Why this step? Real models are often built exact on purpose; always test before reaching for $μ$ .

Step 2 — build $F$ . $F = \int 2 x y d x = x^{2} y + ϕ (y)$ . $F_{y} = x^{2} + ϕ^{'} (y) = x^{2} + 3 y^{2} \Rightarrow ϕ^{'} = 3 y^{2} \Rightarrow ϕ = y^{3} .$

$x^{2} y + y^{3} = C .$

Step 3 — dimensional check (done right). With both $x$ and $y$ in metres, the term $x^{2} y$ has units $m^{2} \cdot m = m^{3}$ and $y^{3}$ has units $m^{3}$ — they match, so the sum $x^{2} y + y^{3} = C$ is dimensionally sound with $C$ in $m^{3}$ . Why this matters: if the two terms of $F$ had different units you would know an algebra slip occurred — a free error-catcher. (This is why I chose both variables in metres rather than mixing seconds and degrees, which would force artificial rescaling.)

Verify: $d (x^{2} y + y^{3}) = 2 x y d x + (x^{2} + 3 y^{2}) d y = 0$ . ✓

Cell G — a linear ODE in disguise

Worked example Ex 7. Solve the linear ODE

$\frac{d y}{d x} + \frac{2}{x} y = x$ using an integrating factor, and see it as a special $x$ -factor. Forecast: a first-order linear ODE $y^{'} + P y = Q$ always has $μ = e^{\int P d x}$ . Guess $μ$ before computing.

Step 1 — put in differential form. Move everything to one side: $(\frac{2}{x} y - x) d x + d y = 0$ . So $M = \frac{2}{x} y - x$ , $N = 1$ . Why this step? To use the $M, N$ machinery we need the $d x, d y$ layout.

Step 2 — $x$ -test. $M_{y} = \frac{2}{x}$ , $N_{x} = 0$ . $\frac{M _{y} - N _{x}}{N} = \frac{2/ x}{1} = \frac{2}{x}$ — pure $x$ . ✓ Why it's guaranteed: for linear $y^{'} + P y = Q$ this ratio is always exactly $P (x)$ , so the $x$ -factor formula reproduces $μ = e^{\int P d x}$ . See Linear first-order ODEs.

Step 3 — $μ$ . $μ = exp \int \frac{2}{x} d x = e^{2 l n x} = x^{2} .$

Step 4 — multiply, re-test. $M_{⋆} = 2 x y - x^{3}$ , $N_{⋆} = x^{2}$ . $(M_{⋆})_{y} = 2 x = (N_{⋆})_{x}$ . ✓

Step 5 — $F$ . $F = \int (2 x y - x^{3}) d x = x^{2} y - \frac{1}{4} x^{4} + ϕ (y)$ . $F_{y} = x^{2} + ϕ^{'} = x^{2} \Rightarrow ϕ^{'} = 0.$

$x^{2} y - \frac{1}{4} x^{4} = C ⟺ y = \frac{C}{x ^{2}} + \frac{x ^{2}}{4} .$

Verify: with $y = \frac{C}{x ^{2}} + \frac{x ^{2}}{4}$ , $y^{'} = - \frac{2 C}{x ^{3}} + \frac{x}{2}$ ; then $y^{'} + \frac{2}{x} y = - \frac{2 C}{x ^{3}} + \frac{x}{2} + \frac{2 C}{x ^{3}} + \frac{x}{2} = x$ . ✓ (matches $Q = x$ )

Cell H — exam twist: monomial factor, signs, zeros

Worked example Ex 8. Solve

$(3 y + 4 x y^{2}) d x + (2 x + 3 x^{2} y) d y = 0$ , then state where the integrating factor is undefined or zero. Forecast: both single-variable tests will fail — so we look for a monomial $μ = x^{a} y^{b}$ . Guess $a, b$ before reading.

Step 1 — test. $M = 3 y + 4 x y^{2}$ , $N = 2 x + 3 x^{2} y$ . $M_{y} = 3 + 8 x y$ , $N_{x} = 2 + 6 x y$ . Not exact.

Step 2 — single-variable tests both fail. $\frac{M _{y} - N _{x}}{N} = \frac{1 + 2 x y}{2 x + 3 x ^{2} y}$ (has $y$ ); $\frac{N _{x} - M _{y}}{M} = \frac{- 1 - 2 x y}{3 y + 4 x y ^{2}}$ (has $x$ ). Both fail → try $μ = x^{a} y^{b}$ . Why this step? When both single-variable tests fail (Cell E symptom), a monomial guess often rescues an exam problem.

Step 3 — systematic matching for $a, b$ . With $μ = x^{a} y^{b}$ , write $M_{⋆} = μ M$ , $N_{⋆} = μ N$ and impose exactness $(M_{⋆})_{y} = (N_{⋆})_{x}$ . Compute: $M_{⋆} = 3 x^{a} y^{b + 1} + 4 x^{a + 1} y^{b + 2}, N_{⋆} = 2 x^{a + 1} y^{b} + 3 x^{a + 2} y^{b + 1} .$ $(M_{⋆})_{y} = 3 (b + 1) x^{a} y^{b} + 4 (b + 2) x^{a + 1} y^{b + 1},$ $(N_{⋆})_{x} = 2 (a + 1) x^{a} y^{b} + 3 (a + 2) x^{a + 1} y^{b + 1} .$ Now match like power-terms (the parser of the equation): the $x^{a} y^{b}$ terms and the $x^{a + 1} y^{b + 1}$ terms must separately agree. $3 (b + 1) = 2 (a + 1) (first term),$ $4 (b + 2) = 3 (a + 2) (second term) .$ Why "match like terms"? Two polynomials are equal only if every matching monomial's coefficient is equal — this turns one exactness equation into a small linear system.

Step 4 — solve the system. From the first: $3 b + 3 = 2 a + 2 \Rightarrow 2 a - 3 b = 1$ . From the second: $4 b + 8 = 3 a + 6 \Rightarrow 3 a - 4 b = 2$ . Solve: multiply first by $3$ , second by $2$ : $6 a - 9 b = 3$ and $6 a - 8 b = 4$ ; subtract: $- b = - 1 \Rightarrow b = 1$ , then $2 a - 3 = 1 \Rightarrow a = 2$ . So $μ = x^{2} y$ — no guessing, the equations forced it.

Step 5 — multiply, re-test. $M_{⋆} = 3 x^{2} y^{2} + 4 x^{3} y^{3}$ , $N_{⋆} = 2 x^{3} y + 3 x^{4} y^{2}$ . $(M_{⋆})_{y} = 6 x^{2} y + 12 x^{3} y^{2}$ , $(N_{⋆})_{x} = 6 x^{2} y + 12 x^{3} y^{2}$ . Equal ✓

Step 6 — build $F$ . $F = \int (3 x^{2} y^{2} + 4 x^{3} y^{3}) d x = x^{3} y^{2} + x^{4} y^{3} + ϕ (y)$ . $F_{y} = 2 x^{3} y + 3 x^{4} y^{2} + ϕ^{'} = N_{⋆} = 2 x^{3} y + 3 x^{4} y^{2} \Rightarrow ϕ^{'} = 0.$

$x^{3} y^{2} + x^{4} y^{3} = C .$

Step 7 — where $μ$ dies. $μ = x^{2} y$ is zero on the axes $x = 0$ and $y = 0$ . On those lines multiplying by $μ$ destroys information (we multiplied by zero), so any solution branch lying on an axis must be checked separately — the exam's hidden marks.

Verify: $d (x^{3} y^{2} + x^{4} y^{3}) = (3 x^{2} y^{2} + 4 x^{3} y^{3}) d x + (2 x^{3} y + 3 x^{4} y^{2}) d y = x^{2} y (M d x + N d y)$ . ✓

Coverage check

All eight cells of the scenario matrix now have a fully worked, verified example. ✓

Connections

Integrating factors for non-exact equations — the parent method these examples drill
Exact differential equations — Cell A is this, unchanged
Total differentials and potential functions — every "build $F$ " step
Linear first-order ODEs — Ex 7's $μ = e^{\int P d x}$
Separable equations — the fallback in Cell E
Mixed partial derivatives (Clairaut's theorem) — why the exactness test works
Conservative vector fields — $M d x + N d y = d F$ is the conservative-field picture

Worked examples — Integrating factors for non-exact equations