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Suppose I have the separable equation: $$\frac{y}{f(x)} = \frac{dy}{dx}$$

Can I rearrange this to $\frac{dy}{dx} -\frac{y}{f(x)} = 0$ and use integrating factors even though there is a zero on the right hand side instead of a function of x?

I found this answer, Solve separable DE with integrating factor and homogeneous substitution, to work for $f(x)=x$ but can't make it work for other functions. I am trying to figure out if it is not possible for some functions or if my algebra skills are lacking.

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    You should be able to. Do you have a particular problem in mind? – 2017-02-05
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    Hi @TimThayer, I do have a problem in mind, but I think my professor would consider posting it as cheating. If yes is your answer, I'll run with it and keep trying! – 2017-02-05

2 Answers 2

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Consider the ODE:

$$y' + \frac{1}{f(x)}\cdot y = 0$$

We multiply through with an integrating factor $\mu(x) = \exp\left(\int\frac{dx}{f(x)}\right)$, then we have:

$$y'\exp\left(\int\frac{dx}{f(x)}\right) + \exp\left(\int\frac{dx}{f(x)}\right)\frac{y}{f(x)} = 0\Leftrightarrow \frac{d}{dx}\left[y\exp\left(\int\frac{dx}{f(x)}\right)\right] = 0$$

This implies that your solution is given by

$$y(x) = C\cdot\exp\left(-\int\frac{dx}{f(x)}\right)$$

The integral within the exponential function can't always be evaluated analytically, of course, but note that you'd have to compute the same integral when doing separation of variables, so that's not a potential problem unique to this approach.

Therefore, both approaches work in this case (though it should be noted that separation of variables is a lot more general, as it also allows you to tackle nonlinear first order equations).

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Yes, you can. Multiplying both sides by $ e^{-\int \frac{1}{f(x)} \mathrm dx} $:

$$ \left ( e^{-\int \frac{1}{f(x)} \mathrm dx} y \right)' = 0 \implies y = c_1 e^{\int \frac{1}{f(x)} \mathrm dx }, \ c_1 \in \mathbb{R}_+ $$

On the other hand, separating variables:

$$ \frac{1}{y} \frac{ \mathrm dy}{\mathrm dx} \mathrm dx = \frac{1}{f(x)} \mathrm dx \implies \ln (y) = \int \frac{1}{f(x)} \mathrm dx + c_1 \implies y = e^{c_2} e^{\int \frac{1}{f(x)} \mathrm dx}, \ c_2 \in \mathbb{R}$$

that is exactly the same solution.