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Show that if $\frac{N_x - M_y}{M} = Q$, where Q is a function of $y$ only, the the differential equation

$M + Ny' = 0$ (*)

has an integrating factor of the form

$u(y) = exp$$\int Q(y) \, dy$.

(The subscripts in the first equation are partial derivatives). I've done quite a few examples and see that it indeed works, but I'm not sure how to prove this general formula.

(So we need to show that multiplying through by $u(y)$ makes (*) exact.)

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For $M+Ny' =0$ to be exact we need,

$M_y - N_x = 0$

Multiplying the equation by $\mu(y)$, we have

$\mu(y) M + \mu(y) N y' = 0$. This is exact when:

$\mu'(y)M + M_y\mu(y) - \mu(y)N_x =0$ Or when

$\frac{\mu'(y)}{\mu(y)} = \frac{N_x - M_y}{M}$

Now if the RHS is a function of $y$ we can integrate both sides to obtain

$\mu(y) = e^{\int Q(y)}$ in your notation.

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  • $\begingroup$ Can you just briefly explain what happened from $μ(y)M+μ(y)Ny′=0$ to "This is exact when: $μ′(y)M+Myμ(y)−μ(y)Nx=0."$ Where did $y'$ go? Is this just chain rule or something? $\endgroup$
    – Wilson Brians
    Commented Oct 12, 2016 at 15:57
  • $\begingroup$ This is the definition of exact. You differentiate $\mu (y)M$ in $y$ and $\mu (y)N$ in $x$ noting that $\mu (y)$ is constant with respect to $x$ and using the product rule for the first. Then set their difference to zero (which is what exact means) and solve. $\endgroup$
    – 3-in-441
    Commented Oct 12, 2016 at 16:01

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