Q31 / 75 · JEE Main 2020 (08 Jan Shift 2) · Mathematics

The differential equation of the family of curves, $x^{2} = 4 b (y + b), b \in R$ , is.

Solution

x^{2} = 4 b (y + b)

Differentiate both sides,

$2 x = 4 b y'$

$\Rightarrow b = \frac{x}{2 y'}$

So, the differential equation is

$x^{2} = \frac{2 x}{y'} y + {(\frac{x}{y'})}^{2}$

$⟹ x {(\frac{d y}{d x})}^{2} = 2 y (\frac{d y}{d x}) + x$

$\Rightarrow x {(y^{'})}^{2} = x + 2 y y^{'}$

Hence, the correct answer is option: 'A'.