The order and the degree (respectively) of the differential equation representing the family of curves $y^2=2c(x +\sqrt{c}), c > 0$ is a parameter are

Differentiating the given equation $y^2=2c(x+\sqrt{c})$

$2y\frac{dy}{dx}=2c\Rightarrow c=y\frac{dy}{dx}$                   (1)

Putting this value in (1) 

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

$\Rightarrow y-2x\frac{dy}{dx}=2{\left(y\frac{dy}{dx}\right)}^{3/2}$

$\Rightarrow {\left(y-2x\frac{dy}{dx}\right)}^2=4y^3{\left(\frac{dy}{dx}\right)}^3$

Thus the order of the differential equation is 1 and the degree is 3.