The differential equation for the family of curves $x^2-y^2-2 a y=0$, where $a$ is an arbitrary constant, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\left(x^2+y^2\right) \frac{d y}{d x}=2 x y$
Given: $x^2 - y^2 - 2ay = 0$
Differentiate w.r.t. $x$:
$2x - 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$
$2x = 2(y + a) \frac{dy}{dx}$
From original equation:
$x^2 - y^2 = 2ay \Rightarrow a = \frac{x^2 - y^2}{2y}$
Substitute:
$2x = 2 \left( y + \frac{x^2 - y^2}{2y} \right) \frac{dy}{dx}$
$2x = \frac{x^2 + y^2}{y} \frac{dy}{dx}$
$2xy = (x^2 + y^2) \frac{dy}{dx}$
Hence, Option (1) is correct.