Find the general integral of the ODE $xy'^2+2xy'-y=0$ I have to find the general integral of the following ODE
$$
xy'^2+2xy'-y=0
$$
The book says that, integrating w.r.t. $y'$, we get two homogeneous equations
$$
y'=-1+\sqrt{1+\frac{y}{x}},\qquad y'=-1-\sqrt{1+\frac{y}{x}}
$$
defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are
$$
\left(\sqrt{1+\frac{y}{x}}-1\right)^2=\frac{C}{x},\qquad \left(\sqrt{1+\frac{y}{x}}+1\right)^2=\frac{C}{x}.
$$
I do not understand how the book deduces such general integrals. Can someone help me?
Thank You
 A: Hint. Let $u(x):=\sqrt{1+\frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$
Hence
$$y'(x)=-1\pm\sqrt{1+\frac{y(x)}{x}}$$
becomes
$$u^2(x)+2xu'(x)u(x)-1=-1\pm u(x)$$
that is
$$u^2(x)+2xu'(x)u(x)=\pm u(x)$$
or
$$u(x)+2xu'(x)=\pm 1$$
which is a linear ODE of first order. Can you take it from here?
A: Substitute $$y(x)=xv(x)$$ then we get $$x\left(x\frac{dv(x)}{dx}+v(x)\right)^2+2x\left(x\frac{dv(x)}{dx}+v(x)\right)-xv(x)=0$$ and we get
$$\frac{dv(x)}{dx}=\frac{-x-\sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
$$\frac{dv(x)}{dx}=\frac{-x+\sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$
Can you finish?
A: Set $y=u-x$, then $y'=u'-1$ and 
$$
u=y+x=x(y'^2+2y'+1)=xu'^2\tag1
$$
which has a derivative
$$
u'=u'^2+2xu'u''\implies u'(2xu''+u'-1)=0
$$
Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.
For $x>0$ the second equation solves as
$$
(\sqrt xu')'=\frac{2xu''+u'}{2\sqrt x}=\frac1{2\sqrt x}\implies \sqrt xu'=\sqrt{x}+C
$$
and the original equation in form (1) gives
$$
u(x)=xu'^2=x+2C\sqrt x+C^2
$$
