The differential equation of the family of curve $y^2=4a(x+a)$ is

We have,

$\begin{array}{l}{y}^2=4a(x+a)\\ \Rightarrow 2y\frac{dy}{dx}=4a\Rightarrow a=\frac{1}{2}y\frac{dy}{dx}\end{array}$

Substituting the value of a in (i), we get

$\begin{array}{l}{y}^2=2y\frac{dy}{dx}\left(x+\frac{1}{2}y\frac{dy}{dx}\right)\\ \Rightarrow y^2=y\frac{dy}{dx}\left(2x+y\frac{dy}{dx}\right)\\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2+2xy\frac{dy}{dx}-y^2=0\end{array}$