The differential equation of family of curves
${y^2} = 4a(x + a)$ is

Given that, ${y^2} = 4a(x + a)$
….(i)
On differentiating both sides w.r.t.
$x$,

we get

$2y\frac{{dy}}{{dx}} = 4a \Rightarrow 2y\frac{{dy}}{{dx}} = 4a$

$\Rightarrow$ $y\frac{{dy}}{{dx}} = 2a \Rightarrow a = \frac{1}{2}y\frac{{dy}}{{dx}}$

…(ii)
On putting the value of a from Eq. (ii) in Eq. (i),

we get ${y^2} = 2y\frac{{dy}}{{dx}}\left( {x + \frac{1}{2}y\frac{{dy}}{{dx}}} \right)$

$\Rightarrow$ ${y^2} = 2xy\frac{{dy}}{{dx}} + {y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2}$

$\Rightarrow$ $2x\frac{{dy}}{{dx}} + y{\left( {\frac{{dy}}{{dx}}} \right)^2} - y = 0$