The differential equation representing the family of the curves $y^2=2c\left(x+\sqrt{c}\right)$ where $c$is a positive parameter, is of

Given the family of curve, $y^2=2c(x+\sqrt{c}) ...........\left(1\right)$
Differentiate both sides of given equation, we have,
$$\begin{gathered} 2y\frac{dy}{dx}=2c(1+0) \\ c=y\frac{dy}{dx}. \end{gathered}$$
from equation $\left(1\right)$, we have.
$$\begin{gathered} y^2=2y\frac{dy}{dx}\left[x+{\left(y\frac{dy}{dx}\right)}^{1/2}\right] \\ \Rightarrow \left(y^2-2xy\frac{dy}{dx}\right)=2{\left(y\frac{dy}{dx}\right)}^{3/2} \end{gathered}$$
Squaring both sides cure obtain
$$\begin{gathered} {\left(y^2-2xy\frac{dy}{dx}\right)}^2={\left(2{\left(y\frac{dy}{dx}\right)}^{3/2}\right)}^2 \\ \Rightarrow {\left(y^2-2xy\frac{dy}{dx}\right)}^2=4{\left(y\frac{dy}{dx}\right)}^3. \end{gathered}$$
that means, order $=1$ {order of a differential equation is the order of the highest derivative present in the Equation }.

And Degree $=3$  {the degree  of differential  equation is represent by the power of the highest order derivative in the given differential equation }