Is $xy' = y +xtg(y/x)$ a Bernoulli equation?

Question

I'm trying to find out whether or not $xy' = y +xtg(y/x)$ is a Bernoulli equation. I divided both sides by x but wasnt able to do much about $y/x$ existence of which undermines my confidence that this is indeed a solvable Bernoulli equation.

Answer 1

Not a Bernouilli's equation in y ... $$xy' = y +x\tan(y/x)$$ $$xy' - y =x\tan(y/x)$$ $$\left (\frac yx\right )'=\frac {\tan(y/x)}x$$ It's separable $$\int \frac {d\left (\frac yx\right )}{\tan(y/x)}=\int \frac {dx}x=\ln|x|+K$$ $$\ln|\sin(\frac yx)|=\ln|x|+K$$ $$\sin(\frac yx)=Kx$$ $$\boxed {y(x)=x\arcsin(Kx)}$$