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

The correct option is B

order $1$, degree $3$

Explanation for the correct option:

Finding the order and degree:

Given, equation is,
$$\begin{gathered} y^2=2c(x+\sqrt{c}) \\ \Rightarrow y^2=2cx+2c^{\frac{3}{2}}\dots \dots \text{(i)} \end{gathered}$$
On differentiating with respect to $x$, we get
$\begin{array}{rcl}2y\frac{dy}{dx}& =& 2c\\ c& =& y\left(\frac{dy}{dx}\right)\dots \dots \left(ii\right)\end{array}$
On substituting equation $\left(ii\right)$ in $\left(i\right)$, we get
$\begin{array}{rcl}{y}^2& =& 2xy\left(\frac{dy}{dx}\right)+2{\left(y\frac{dy}{dx}\right)}^{\frac{3}{2}}\\ \Rightarrow {\left(y^2-2xy\left(\frac{dy}{dx}\right)\right)}^2& =& {\left(2\right)}^2{y}^3{\left(\frac{dy}{dx}\right)}^3\\ \Rightarrow {\left(y^2-2xy\left(\frac{dy}{dx}\right)\right)}^2& =& 4y^3{\left(\frac{dy}{dx}\right)}^3\\ & & \end{array}$
Order = Highest differential equation,
Degree = Highest degree of order.
$\therefore \text{orde}r=1\text{and degree}=3$
Hence, option (B) is the correct answer.