Integrating factors for non-exact equations

A "fixer" you multiply through a stubborn ODE to make it exact, so you can solve it by potential functions.

The setup: what "exact" means

We write a first-order ODE in differential form: $M(x,y)\,dx + N(x,y)\,dy = 0.$

The exactness test. Since mixed partials commute ($F_{xy} = F_{yx}$ for smooth $F$): $M_y = F_{xy} = F_{yx} = N_x.$ So: $\boxed{\text{Exact} \iff M_y = N_x.}$

The problem: most equations are NOT exact

If $M_y \neq N_x$, there is no $F$ and we're stuck... unless we multiply the whole equation by a function $\mu(x,y)$, the integrating factor: $\mu M\,dx + \mu N\,dy = 0.$ This has the SAME solution curves (we only multiplied by something nonzero), but now we demand it be exact.

Deriving the condition on $\mu$ from scratch

Exactness of the multiplied equation requires $\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N).$ Expand both sides with the product rule: $\mu_y M + \mu M_y = \mu_x N + \mu N_x.$ Rearrange: $\boxed{\;N\,\mu_x - M\,\mu_y = \mu\,(M_y - N_x)\;}$

This is a PDE for $\mu$ — generally as hard as the original ODE! The trick is to guess that $\mu$ depends on only one variable, which kills a term.

Case 1: $\mu = \mu(x)$ only

Then $\mu_y = 0$, $\mu_x = \mu'$. The PDE becomes $N\,\mu' = \mu\,(M_y - N_x) \implies \frac{\mu'}{\mu} = \frac{M_y - N_x}{N}.$

Case 2: $\mu = \mu(y)$ only

By symmetry ($\mu_x=0$): $-M\,\mu' = \mu(M_y - N_x) \implies \frac{\mu'}{\mu} = \frac{N_x - M_y}{M}.$

Full procedure (the 80/20 core)

1. Write as $M\,dx + N\,dy = 0$.
2. Test: is $M_y = N_x$? If yes, skip to step 5.
3. Compute $\frac{M_y-N_x}{N}$ → function of $x$? Use $\mu(x)$. Else compute $\frac{N_x-M_y}{M}$ → function of $y$? Use $\mu(y)$.
4. Multiply through by $\mu$. Now exact.
5. Find $F$: integrate $F_x=M^*$ in $x$, add unknown $\phi(y)$, differentiate in $y$, match $N^*$.
6. Solution: $F(x,y)=C$.

Worked Example 1 — factor in $x$

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

Here $M = 3xy + y^2$, $N = x^2 + xy$.

$M_y = 3x + 2y$, $\;N_x = 2x + y$. Why? Differentiate $M$ w.r.t. $y$, $N$ w.r.t. $x$. Not exact since $3x+2y \neq 2x+y$.

$\frac{M_y-N_x}{N} = \frac{(3x+2y)-(2x+y)}{x^2+xy} = \frac{x+y}{x(x+y)} = \frac{1}{x}.$ Why this step? The numerator factors and cancels the $(x+y)$, leaving a pure-$x$ function — so Case 1 works.

$\mu = \exp\!\left(\int \tfrac{1}{x}\,dx\right) = e^{\ln x} = x.$

Multiply: $M^* = 3x^2 y + xy^2$, $N^* = x^3 + x^2 y$. Check exact: $M^*_y = 3x^2 + 2xy$, $N^*_x = 3x^2 + 2xy$. ✓

Find $F$: $F = \int (3x^2y + xy^2)\,dx = x^3 y + \tfrac12 x^2 y^2 + \phi(y).$ Why $\phi(y)$? Integration "constant" w.r.t. $x$ may depend on $y$. $F_y = x^3 + x^2 y + \phi'(y)$ must equal $N^* = x^3 + x^2 y$, so $\phi'(y)=0$.

$\boxed{x^3 y + \tfrac12 x^2 y^2 = C.}$

Worked Example 2 — factor in $y$

Solve $y\,dx + (2x - ye^y)\,dy = 0.$

$M = y$, $N = 2x - ye^y$. $M_y = 1$, $N_x = 2$. Not exact.

Try Case 1: $\frac{M_y-N_x}{N} = \frac{-1}{2x-ye^y}$ — still has $x$. ✗ Try Case 2: $\frac{N_x-M_y}{M} = \frac{2-1}{y} = \frac{1}{y}$ — pure $y$. ✓

$\mu = \exp\!\left(\int \tfrac1y\,dy\right) = y.$

Multiply: $M^* = y^2$, $N^* = 2xy - y^2 e^y$. Check: $M^*_y = 2y$, $N^*_x = 2y$. ✓

$F = \int y^2\,dx = x y^2 + \phi(y)$. Then $F_y = 2xy + \phi'(y) = 2xy - y^2 e^y$, so $\phi'(y) = -y^2 e^y$. Integrate by parts: $\phi = -e^y(y^2 - 2y + 2)$.

$\boxed{xy^2 - e^y(y^2 - 2y + 2) = C.}$

Common mistakes

Flashcards

Connections

• Exact differential equations
• Total differentials and potential functions
• Linear first-order ODEs — the factor $e^{\int P\,dx}$ is a special integrating factor
• Separable equations
• Mixed partial derivatives (Clairaut's theorem)
• Conservative vector fields

Concept Map

Hinglish (regional understanding)