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I am trying to solve differential equation of the following type: $$y y'=f(x)y+g(x)y'$$ This looks somewhat similar to Abel equations but not quite. Any ideas on how to approach it? Many thanks!

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    what do we know about $$f(x),g(x)$$? – 2017-02-13
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    Not much: they are continuously differentiable functions. Perhaps I can prove that $f(x)$ is decreasing and $g(x)$ is increasing, if it helps. – 2017-02-13
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    would help - since for nonlinear equations a solution a soluble up to integration. I mean if you by chance had $g(x) = \int_0^x f(s) ds$ then we could write your equation as $$\frac{d}{dx}y^2 = 2\dfrac{d}{dx} \left( y\int_0^x f(s)ds\right)$$ – 2017-02-13

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Hint:

$yy'=f(x)y+g(x)y'$

$(y-g(x))y'=f(x)y$

This belongs to an Abel equation of the second kind.

Let $u=y-g(x)$ ,

Then $y=u+g(x)$

$y'=u'+g'(x)$

$\therefore u(u'+g'(x))=f(x)(u+g(x))$

$uu'+g'(x)u=f(x)u+f(x)g(x)$

$uu'=(f(x)-g'(x))u+f(x)g(x)$