Differential Equations · step 6 of 14

Exact equations & integrating factors

M(x,y)dx + N(x,y)dy = 0 is exact when it is the total differential of some potential F, and the test is ∂M/∂y = ∂N/∂x. The solution is then simply F(x,y) = C. When the test fails, an integrating factor may restore exactness — and if (∂M/∂y − ∂N/∂x)/N depends only on x, that factor is computable directly.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Write in the form M dx + N dy = 0Identify M and N clearly before differentiating anything.
Test exactnessCompute ∂M/∂y and ∂N/∂x. Equal means exact; unequal means you need an integrating factor.
Build F by integrating M with respect to xF = ∫M dx + g(y), where g is an unknown function of y alone.
Determine g by matching ∂F/∂y with NDifferentiate your F, set it equal to N, and integrate the leftover to find g(y).

Worked example

Solve (2xy + 3)dx + (x² − 1)dy = 0.

M = 2xy + 3 and N = x² − 1.
∂M/∂y = 2x and ∂N/∂x = 2x — equal, so the equation is exact.
F = ∫(2xy + 3)dx = x²y + 3x + g(y).
∂F/∂y = x² + g′(y), and this must equal N = x² − 1, so g′(y) = −1 and g(y) = −y.

Answer. x²y + 3x − y = C, which can be solved for y as y = (C − 3x)/(x² − 1).

Where marks get dropped

These are the specific errors that cost credit on exact equations & integrating factors questions — QED's rubric penalises each of them separately.

Differentiating M with respect to x instead of y. The test is ∂M/∂y versus ∂N/∂x — the cross partials.
Writing the integration "constant" as a number rather than g(y). It can be any function of the OTHER variable.
Declaring an equation non-exact and stopping. An integrating factor often fixes it.

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Exact equations & integrating factors — frequently asked questions

Why does the cross-partial test work?

Because if M dx + N dy is dF then M = ∂F/∂x and N = ∂F/∂y, and mixed partials commute for smooth F. Equality is therefore necessary, and on a simply connected domain also sufficient.

How do I find an integrating factor?

If (M_y − N_x)/N depends only on x, then μ = e^(∫that dx). If (N_x − M_y)/M depends only on y, integrate that in y instead.

Is every first-order ODE exact after some factor?

In principle yes, but finding the factor can be as hard as solving the equation. The two tests above cover the standard exam cases.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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