The equation of a curve passing through the origin and satisfying the different equation $ \frac{dy}{dx} = (x-y)^2 ,$ is

$e^{2x}(1- x+y) = 1+ x- y$

The correct answer is option (1) : $e^{2x}(1- x+y) = 1+ x- y$

We have,

$\frac{dy}{dx} = (x- y)^2 $

Let $x-y = v .$ Then,

$1-\frac{dy}{dx} =\frac{dv}{dx}$

$⇒\frac{dy}{dx} = 1-\frac{dv}{dx}$

$∴\frac{dy}{dx} = (x-y)^2 $

$⇒1-\frac{dv}{dx} = v^2$

$⇒1-v^2 = \frac{dv}{dx}$

$⇒dx=\frac{1}{1-v^2 }dv$

$⇒2∫dx= 2∫\frac{1}{1-v^2 }dv$

$⇒2x=log \left(\frac{1+v}{1-v} \right) + kog C$

$⇒C \left(\frac{1+v}{1-v} \right) = e^{2x}$

$⇒C\left(\frac{x-y+1}{y-x+1}\right) =e^{2x}$

$⇒C(x-y+1) = e^{2x} (y-x+1)$