Solving $\frac{dy}{dx} (x + y) x = y (x - y)$? I need to find the general solution to the second order differential equation $$ \frac{dy}{dx} (x + y) x = y (x - y)$$ I tried to rewrite this in the form of the Bernoulli's equation, which has the form $$ \frac{dy}{dx} + P(x) y(x) = Q(x) y^n (x) $$ but then I see this can't be done. Should I look for a substitution of some kind?
 A: Notice, $$\frac{dy}{dx}(x+y)x=y(x-y)$$
$$\frac{dy}{dx}=\frac{y(x-y)}{x(x+y)}=\frac{y}{x}\frac{\left(1-\frac{y}{x}\right)}{\left(1+\frac{y}{x}\right)}$$
Now, let $y=vx\implies \frac{dy}{dx}=v+x\frac{dv}{dx}$ $$v+x\frac{dv}{dx}=v\frac{1-v}{1+v}$$
$$x\frac{dv}{dx}=\frac{-2v^2}{1+v}$$
$$\frac{(1+v)dv}{v^2}=-2\frac{dx}{x}$$
$$\left(\frac{1}{v^2}+\frac{1}{v}\right)dv=-2\frac{dx}{x}$$
I hope you can take it from here.
A: Due to homogeneity of degree consider substitution  $ y = v x $ in
$$ \frac{dy}{dx}= \frac {x (x + y)  } { y (x - y)} $$ 
A: You need to use a substitution in this case and change the variables in order to transform the equation into a variable separable one. Firstly, transform the equation by dividing both numerator and denominator of RHS by $x$ as follows:
$$ \frac{dy}{dx} = \frac{y(x-y)}{x(x+y)}= \frac{y}{x} \frac{\frac{x}{x}-\frac{y}{x}}{\frac{x}{x}-\frac{y}{x}}= \frac{y}{x} \frac{1-\frac{y}{x}}{1-\frac{y}{x}}$$
Now let $t = \frac{y}{x}$
it follows that $\frac{dt}{dx} = -\frac{y}{x^2}+ \frac{1}{x}\frac{dy}{dx}$, $\implies \ x\frac{dt}{dx} + t =\frac{dy}{dx} $
The differential equation now becomes:
$$x\frac{dt}{dx} + t = \frac{t(1-t)}{1+t}$$
You can now solve this using separation of variables:
$$ \left( \frac{1}{t^2} + \frac{1}{t}\right)dt = -2\frac{1}{x} dx$$
You can solve this by integrating both sides and then re-substituting $t$ with $\frac{y}{x}$
