Recipe

Find and apply an integrating factor for a non-exact equation

Use when: an equation M\,dx+N\,dy=0 fails the exactness test (M_y\ne N_x), and at least one of the two special-case ratios below simplifies to a function of a single variable

Goal: a factor \mu that restores exactness, so the exact-equation method solves the repaired equation

Taught in Integrating factors for non-exact equations

1. 1

   Compute M_y and N_x; if they don't match, exactness fails and an integrating factor is needed.

   Example

   For 2xy\,dx+(y^2-x^2)\,dy=0: M=2xy, N=y^2-x^2. M_y=2x, N_x=-2x — these don't match, so the equation isn't exact as written.

   Why: there's no point hunting for \mu on an equation that's already exact.

2. 2

   Compute this ratio and simplify; if every y cancels, \mu(x)=e^{\int (M_y-N_x)/N\,dx} works and Step 4 is next.

   Example

   \dfrac{M_y-N_x}{N}=\dfrac{2x-(-2x)}{y^2-x^2}=\dfrac{4x}{y^2-x^2} — this still has both x and y in it after simplifying, so it's not a function of x alone.

   • the ratio simplifies to a function of x alone — \mu(x) is found; skip to Step 4.
   • the ratio still involves y — move on to Step 3 and test the other ratio.

   Why: this ratio is exactly what demanding exactness after multiplying by \mu(x) forces \mu to satisfy — when it collapses to one variable, that variable's own equation for \mu is solvable by separation.

3. 3

   Compute this second ratio; if every x cancels, \mu(y)=e^{\int (N_x-M_y)/M\,dy} works instead.

   Example

   \dfrac{N_x-M_y}{M}=\dfrac{-2x-2x}{2xy}=\dfrac{-4x}{2xy}=\dfrac{-2}{y} — every x cancelled, leaving a function of y alone.

   Why: this is the mirror image of Step 2's test — a non-exact equation can be repaired from EITHER side, and typically only one of the two ratios simplifies.

4. 4

   Integrate whichever single-variable expression simplified, exponentiate to get \mu, multiply the ENTIRE original equation by it, then recheck M_y=N_x on the new M and N.

   Example

   \mu(y)=e^{\int(-2/y)\,dy}=e^{-2\ln y}=y^{-2}=\dfrac1{y^2}. Multiplying through: new M=\dfrac{2x}{y}, new N=1-\dfrac{x^2}{y^2}. Checking: M_y=-\dfrac{2x}{y^2}, N_x=-\dfrac{2x}{y^2} — they match, so \boxed{\mu(y)=\dfrac1{y^2}\ \text{repairs the equation}}

   Why: the whole point of finding \mu was to reach an exact equation the previous two sections already know how to solve — this check confirms the repair actually worked before moving on.