I am coursing differential equations and recently encountered with the concept of integrating factors. I have seen them to solve two types of ODEs: inexact and linear. A linear equation is an ODE in the form: $$\frac{dy}{dx}+p(x)y = q(y)$$ the integrating factor ends up being: $$u(x) = e^{\int p(x)dx}$$ so that the equation comes to: $$\frac{d}{dx}(uy) = q(x)u(x)$$ the equation can now be solved if $\int q(x)u(x)dx$ can be computed.

An inexact equation is an equation in the form $$A(x,y)dx+B(x,y)dy=0$$, where $$A_y \ne B_x$$ (i.e. $Adx+Bdy$ is not an exact differential)

The integrating factor for these equations (I will call it $\mu$ for inexact equations) is a function such that $$(\mu A)_y = (\mu B)_x$$ Expanding, $$\mu A_y + \mu_y A = \mu B_x + \mu_x A$$ I have read the Wikipedia article, which says that to solve this equation where $\mu = \mu (x,y)$ requires partial differential equations, but if $\mu = \mu(x)$ or $\mu = \mu(y)$, then there is a straightforward formula for both, in terms of $A$ and $B$ (and their partial derivatives, respectively). But here is the important part: it says

"[...] in which case we only need to find $\mu$ with a first-order linear differential equation or a separable differential equation [...]"

Does this mean that this method can only be used for linear ODEs? In that case, I think the first method is way faster.

Article mentioned


1 Answer 1


No, it is referring to how to find $\mu$, not the form of the original D.E.

Consider the equation you quote:-$$\mu A_y + \mu_y A = \mu B_x + \mu_x A$$

If $\mu = \mu(x)$ then this equation becomes $$\mu A_y = \mu B_x + \mu_x A$$ $$\mu (A_y - B_x) = \mu_x A$$ This is the 'straightforward equation' referred to in the article.

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    $\begingroup$ Alright. I was really confused and thought for a moment that the phrase meant that the method would be actually "useless" and that the fact that the factor was a single-variable function implied that the differential equation was linear or separable. This was really useful. Thanks! $\endgroup$
    – user685013
    Sep 20, 2019 at 22:07

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