I have this question, and the working out below is as far as I can get: $$ x \frac{dy}{dx} - y = y^2 \\ p(x) = -\frac{1}{x} \\ q(x) = \frac{y^2}{x} \\ u(x)= e^{\int -\frac{1}{x}dx} \rightarrow \frac{1}{x} \\ y = \frac{\int \left( \frac{1}{x} \right) \left( \frac{y^2}{x} \right)dx }{x^{-1}} $$
How do I find $$\int \left( \frac{1}{x} \right) \left( \frac{y^2}{x^2} \right)dx $$ Wolfram aplha says the indefinite integral is equal to $$-\frac{y^2}{x} + C$$ But I have no clue on how they got that answer, unless you treat $y^2$ as a constant? Can someone please show me how to integrate it, or tell me which method to use? If there was no $y$ terms I would have just used integration by parts, but I got confused. Thank you in advance.
As far as I can see you are using the formula for a first order linear DE. But this DE is not linear (because of the $y^2$) and so you cannot use this method.
Hint. Write the DE as $$x\frac{dy}{dx}=y+y^2\ .$$ Then you should be able to recognise this as a special type of DE, probably the first type you learnt, and apply the standard solution procedure for this type. Good luck!
Another comment: going from $$y = \frac{\int \left( \frac{1}{x} \right) \left( \frac{y^2}{x} \right)dx }{x^{-1}} $$ to $$\int \left( \frac{1}{x} \right) \left( \frac{y^2}{x^2} \right)dx\ ,$$ you have taken the factor $x^{-1}$ inside the integral sign (sort of). But you can't do this because $x^{-1}$ is not a constant.