Exact Equations and the Integrating Factor

This chapter completes the toolkit of elementary first-order techniques with the most structurally interesting one: exactness. An exact equation is secretly the statement that a function $F(x,y)$ is constant along solution curves, and the entire solution method amounts to reconstructing $F$ from its own partial derivatives. When an equation is not exact, an integrating factor — a multiplier chosen to force exactness — often rescues it, generalising the linear-equation integrating factor of Chapter 2 to a genuinely two-variable setting.

Have to hand: partial derivatives and the equality of mixed partials (Clairaut's theorem) from multivariable calculus, and the differential-form notation $M\,dx+N\,dy=0$ introduced in Chapter 2.

1. Exact Equations

Write a first-order ODE in differential form

Definition. The equation is exact on a region $R$ if there exists a function $F(x,y)$ (a potential function) with continuous partial derivatives on $R$ such that

$$\frac{\partial F}{\partial x} = M, \qquad \frac{\partial F}{\partial y} = N.$$

In that case $M\,dx+N\,dy = dF$ (the total differential of $F$), so the ODE reads $dF=0$, and every solution satisfies $F(x,y)=C$ for a constant $C$.

The entire solving method rests on recognising when such an $F$ exists and constructing it explicitly.

2. The Exactness Test

Theorem. Suppose $M,N$ have continuous partial derivatives on a simply connected region $R$. Then $M\,dx+N\,dy=0$ is exact on $R$ if and only if

Proof. ($\Rightarrow$) If the equation is exact, $M=F_x$ and $N=F_y$ for some $F$. Since $M,N$ have continuous partials, $F$ has continuous second partials, so by Clairaut's theorem (equality of mixed partial derivatives), $M_y = F_{xy} = F_{yx} = N_x$.

($\Leftarrow$) Suppose $M_y=N_x$ on the simply connected region $R$. Define

for a fixed $x_0$ and an unknown function $h(y)$ to be determined. By construction $F_x = M$. Differentiating with respect to $y$,

using $M_y=N_x$ and the fundamental theorem of calculus. Requiring $F_y=N$ forces $h'(y) = N(x_0,y)$, which determines $h$ (up to an additive constant) by a single integration in $y$ alone. The simple connectedness of $R$ is exactly what is needed to guarantee this $F$, built from a path integral from a fixed basepoint, is well defined and single-valued throughout $R$. $\blacksquare$

Why simple connectedness matters. On a region with a hole, $M_y=N_x$ everywhere is not sufficient for a single-valued potential to exist. The classic counterexample is $M=\dfrac{-y}{x^2+y^2},\,N=\dfrac{x}{x^2+y^2}$ on $\mathbb{R}^2\setminus\{(0,0)\}$: direct computation gives $M_y=N_x$ everywhere on the domain, yet $\oint M\,dx+N\,dy = 2\pi\ne 0$ around a loop encircling the origin, so no single-valued $F$ exists there. This is the exact same obstruction noted for the curl of a vector field in vector calculus, and it foreshadows the far-reaching role simple connectedness plays in complex integration (Chapter 11).

3. Method: Constructing the Potential Function

In practice, the proof above is carried out as an explicit algorithm:

1. Verify $M_y=N_x$.
2. Integrate $M$ with respect to $x$ (holding $y$ fixed): $F(x,y) = \int M\,dx + h(y)$, where the "constant" of integration may depend on $y$.
3. Differentiate this $F$ with respect to $y$, set it equal to $N$, and solve for $h'(y)$ — the $x$-dependence must cancel out of this equation; if it does not, $M_y\ne N_x$ and a computational error has been made.
4. Integrate $h'(y)$ to find $h(y)$ (no new arbitrary function is needed here, only a constant, which is absorbed into the final answer).
5. State the implicit solution $F(x,y) = C$.

Worked example. Solve $(2xy+3)\,dx + (x^2-1)\,dy = 0$.

Here $M=2xy+3,\ N=x^2-1$. Check exactness: $M_y = 2x$ and $N_x=2x$, so $M_y=N_x$ and the equation is exact.

Integrate $M$ with respect to $x$: $F = \displaystyle\int (2xy+3)\,dx = x^2y+3x+h(y)$.

Differentiate with respect to $y$ and match to $N$: $F_y = x^2+h'(y) = x^2-1$, so $h'(y)=-1$, giving $h(y)=-y$ (dropping the redundant constant).

Hence $F(x,y) = x^2y+3x-y$, and the general solution is

As a check, implicit differentiation of $x^2y+3x-y=C$ gives $2xy\,dx+x^2\,dy+3\,dx-dy=0$, i.e. $(2xy+3)dx+(x^2-1)dy=0$ — exactly the original equation.

4. Integrating Factors for Non-Exact Equations

If $M\,dx+N\,dy=0$ is not exact, it may become exact after multiplying through by a suitable function $\mu(x,y)$, called an integrating factor: $\mu M\,dx + \mu N\,dy=0$ is exact precisely when

In general this is itself a (harder) partial differential equation for $\mu$, but two special cases reduce to an ordinary differential equation and are the standard syllabus cases.

Case 1: $\mu$ a function of $x$ alone. Expanding the exactness condition for $\mu=\mu(x)$:

This has a solution $\mu(x)$ (via separation) if and only if the right side $\dfrac{M_y-N_x}{N}$ is a function of $x$ alone; when it is, call it $g(x)$, and then

Case 2: $\mu$ a function of $y$ alone. By the symmetric computation,

which is solvable for $\mu(y)$ if and only if $\dfrac{N_x-M_y}{M}$ is a function of $y$ alone, say $k(y)$, giving $\mu(y)=e^{\int k(y)\,dy}$.

Relation to Chapter 2. For a linear equation $y'+p(x)y=q(x)$, written in differential form as $\bigl(p(x)y-q(x)\bigr)dx + dy = 0$, one finds $M_y=p(x)$, $N_x=0$, so $(M_y-N_x)/N = p(x)$, a function of $x$ alone — recovering exactly Chapter 2's integrating factor $\mu(x)=e^{\int p\,dx}$ as the special Case 1 of this more general theory.

Worked example. Solve $(3xy+y^2)\,dx + (x^2+xy)\,dy=0$.

Here $M=3xy+y^2,\,N=x^2+xy$. Check exactness: $M_y=3x+2y$, $N_x=2x+y$; these are unequal, so the equation is not exact.

Try Case 1: $\dfrac{M_y-N_x}{N} = \dfrac{(3x+2y)-(2x+y)}{x^2+xy} = \dfrac{x+y}{x(x+y)} = \dfrac{1}{x}$, a function of $x$ alone. So $\mu(x) = e^{\int (1/x)\,dx} = x$ (for $x>0$).

Multiplying through by $\mu=x$: $(3x^2y+xy^2)\,dx + (x^3+x^2y)\,dy=0$. Checking: new $M_y = 3x^2+2xy$, new $N_x=3x^2+2xy$ — equal, confirming exactness. Integrating the new $M$ with respect to $x$: $F = x^3y + \dfrac{x^2y^2}{2} + h(y)$; differentiating in $y$ and matching to the new $N=x^3+x^2y$ gives $h'(y)=0$. The general solution is

5. Common Exam Pitfalls

• Checking exactness on the original equation but forgetting to recheck it after multiplying by an integrating factor. The whole point of $\mu$ is to restore exactness; always verify $(\mu M)_y=(\mu N)_x$ on the new equation before proceeding to construct $F$.
• Conflating this chapter's integrating factor with Chapter 2's. Chapter 2's $\mu(x)=e^{\int p\,dx}$ is the special case of Case 1 applied to a linear equation written in differential form; the general $(M_y-N_x)/N$ test must still be checked for other equations — it is not automatically $p(x)$.
• Forgetting that $h(y)$ (or $h(x)$) is found by differentiating the partial potential function, not by guessing. The $x$-terms must cancel identically when $F_y$ is set equal to $N$; if they do not cancel, exactness has failed (or an algebra error was made) — it is not something to force by hand.
• Assuming an integrating factor of the simple $\mu(x)$ or $\mu(y)$ form always exists. The ratio $(M_y-N_x)/N$ (or $(N_x-M_y)/M$) must genuinely simplify to a function of one variable alone; if neither does, a more general $\mu(x,y)$ (beyond this chapter's scope) or an entirely different method is required.
• Dropping the requirement of simple connectedness. $M_y=N_x$ guarantees a local potential function; on a region with a hole (as in the Section 2 counterexample), the global solution curves $F=C$ may not close up into a genuine single-valued function.
• Sign errors when converting $y'=f(x,y)$ to $M\,dx+N\,dy=0$. Writing $y'=-M/N$ means a given $y'=f(x,y)$ corresponds to $M=f(x,y)$, $N=-1$ (one valid choice among infinitely many equivalent scalings), and mixing up this sign convention corrupts the exactness test.

Summary

• Exact equation: $M\,dx+N\,dy=0$ with $M=F_x,\,N=F_y$ for some potential $F$; solutions are the level curves $F(x,y)=C$.
• Exactness test: $M_y=N_x$ on a simply connected region is necessary and sufficient, proved via Clairaut's theorem (necessity) and an explicit path-integral construction of $F$ (sufficiency).
• Method: integrate $M$ in $x$ to get $F$ up to $h(y)$; differentiate in $y$, match to $N$, solve for $h'(y)$, integrate, assemble $F(x,y)=C$.
• Integrating factors: if $(M_y-N_x)/N$ depends only on $x$, use $\mu(x)=e^{\int (M_y-N_x)/N\,dx}$; if $(N_x-M_y)/M$ depends only on $y$, use $\mu(y)=e^{\int (N_x-M_y)/M\,dy}$; Chapter 2's linear integrating factor is the special case of the first applied to $(py-q)dx+dy=0$.
• Simple connectedness of the domain is essential for a global potential function to exist, even when $M_y=N_x$ holds pointwise everywhere it is defined.