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 respectively:

Given the equation: $y^2=2c(x+\sqrt{c})$ ... (i)Differentiating with respect to $x$:$2y{y}^{\prime }=2c(1)⟹c=y{y}^{\prime }$ ... (ii)Substitute the value of $c$ from (ii) into (i):$y^2=2(y{y}^{\prime })(x+\sqrt{y{y}^{\prime }})$$y=2y^{\prime }x+2y^{\prime }\sqrt{y{y}^{\prime }}$$y-2x{y}^{\prime }=2y^{\prime }\sqrt{y{y}^{\prime }}$ Squaring both sides to remove the radical:$(y-2x{y}^{\prime }{)}^2=(2y^{\prime }{)}^2(y{y}^{\prime })=4y{y}^{\prime 3}$ The resulting differential equation is:$4y{\left(\frac{dy}{dx}\right)}^3-{\left(y-2x\frac{dy}{dx}\right)}^2=0$ The highest order derivative is $y^{\prime }$ (order 1) and its highest power is 3 (degree 3).