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I came across this differential equation which I'm having trouble finding an analytic solution to:

$$\frac{dy}{dx}=\frac{A}{xy}+\frac{B}{(xy)^2}$$

I'm trying to solve for y. I have initial conditions as $x_0=0.02$ and $y_0=100000$, and A and B are known constants. I don't have a very heavy differential equations background so all I know is that I can't use separation of variables--what kind of method should I use to solve this equation?

Thank you!!

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    $\begingroup$ You can use a numerical method! $\endgroup$ – Dr. Sonnhard Graubner Aug 1 '19 at 14:15
  • $\begingroup$ Try something along the lines of using $z = \frac{1}{x y}$ as a new independent variable... $\endgroup$ – vonbrand Aug 1 '19 at 18:16
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Assume $A,B\neq0$ for the key case:

Hint:

Let $u=xy$ ,

Then $y=\dfrac{u}{x}$

$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$

$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\dfrac{A}{u}+\dfrac{B}{u^2}$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{u}{x^2}+\dfrac{A}{u}+\dfrac{B}{u^2}$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{(Au+B)x^2+u^3}{x^2u^2}$

$((Au+B)x^2+u^3)\dfrac{dx}{du}=xu^2$

Let $v=x^2$ ,

Then $\dfrac{dv}{du}=2x\dfrac{dx}{du}$

$\therefore\dfrac{(Au+B)x^2+u^3}{2x}\dfrac{dv}{du}=xu^2$

$((Au+B)x^2+u^3)\dfrac{dv}{du}=2u^2x^2$

$((Au+B)v+u^3)\dfrac{dv}{du}=2u^2v$

This belongs to an Abel equation of the second kind.

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