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I want to solve this differential equation :

$$xy'+y=y^2$$

So with conditions $y \neq 0,y\neq1$ and $x \neq0$ I can rewrite it as :

$$\frac{y'}{y(y-1)}=\frac{1}{x}$$

So after observing that also $y=0$ and $y=1$ are solutions I can use a theorem on separable differential eq. to write :

$$\int\frac{dy}{y(y-1)}=\int\frac{dx}{x} +K$$

My question is about how to treat the condition $x \neq 0 $ formally :

Should I integrate over the two open intervals $(-\infty,0)$ and $(0,+\infty)$,and then observing that the solution is :

$$y = \frac{1}{1-Cx} $$

I can say that it can include $0$ because $y(0) = \frac{1}{1-C\cdot0} = 1$ is a solution of the equation?

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    $\begingroup$ You could also use $$\frac{(xy)'}{(xy)^2}=\frac1{x^2}.$$ $\endgroup$ Aug 10 '20 at 10:39
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It doesn't matter how you got the solution $y = \frac{1}{1-Cx}$. You can make all the assumptions you want.

But, if you can show that it satisfies the differential equation for all values of $y(0)$, you've got a solution that works for all values of $y(0)$ - even those you neglected while deriving $y(x)$. And in the end, that's all that matters.

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Yes you may say that. Normally in differential equations we are looking for solutions that are functions defined in domains (topos). So your solution will be valid in the interval $(-\infty,1/C)$ or $(1/C,\infty)$.

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When you are insisting $y\ne 0$ and $y\ne 1$, then $x=0$ cannot be in the domain of the solution.

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