Solve this differential equation $\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}$ 
Solve this differential equation:
  $$\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}$$

How to solve this?
 A: $\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}\implies(p-1)(y-xp)=p$, where $p\equiv \frac{dy}{dx}$
$\implies y-xp=\frac{p}{p-1}=\implies y=xp+\frac{p}{p-1}$ . . . . $(1)$
which is known as Clairaut’s equation.
So the general solution of the differential equation $(1)$ is
$y=cx+\frac{c}{c-1}$, where $c$ is the integrating constant.

If you once show that a differential equation is of Clairaut’s form i.e., of the form $$y=px+f(p)$$
Then for general solution of this kind of ODE, just replace $p\equiv \frac{dy}{dx}$ by the integrating constant $c$. i.e., its general solution is $$y=cx+f(c)$$
For more details about the solution of Clairaut’s equation, you may follow the reference:
"http://www.library.gscgandhinagar.in/assets/admin/images/MAT-102(UNIT1,2).pdf"
A: We have $y(x)=px+\frac{p}{-1+p}$. Where $p=\frac{dy}{dx}$ This is the Clairaut's equation . The solution will he of the form $y=cx+f(c)$ .Differentiating we have $dy=pdx+xdp+\frac{dp}{(1-p)^2}$ we can write $dy=pdx$ the equation becomes $-xdp=\frac{dp}{(1-p)^2}$ assuming $dp\neq 0$ we have $x=\frac{1}{(1-p)^2}$ put this value in the first equation and get an expression for $y(p)$. The $x(p),y(p) $ are called singular solutions.
A: Since nmasanta already posted the answer I don't edit my calculus. 
I only add the singular solution $y=x+1\pm 2\sqrt{x}$ and a graph of the complete solutions of the ODE :

A: Hint: Write $(y-x)'\left(\dfrac{x}{y}\right)'=\dfrac{y'}{y^2}$ and with substitutions $u=y-x$ and $\dfrac{x}{y}=v$ then
$$\dfrac{dv}{1-v}=\dfrac{du}{u^2-u}$$
