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(D) Given equation: $y^2=2 c(x+\sqrt{c}) \dots (i)$Differentiating with respect to $x$:$2 y \frac{dy}{dx} = 2c \implies c = y \frac{dy}{dx} \dots (ii)

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order $1$,degree $3$

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(D) Given equation: $y^2=2 c(x+\sqrt{c}) \dots (i)$Differentiating with respect to $x$:$2 y \frac{dy}{dx} = 2c \implies c = y \frac{dy}{dx} \dots (ii)$Substituting $(ii)$ into $(i)$:$y^2 = 2 \left(y \frac{dy}{dx}ight) \left(x + \sqrt{y \frac{dy}{dx}}ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2 \frac{dy}{dx} \left(x + \sqrt{y \frac{dy}{dx}}ight)$$y - 2x \frac{dy}{dx} = 2 \frac{dy}{dx} \sqrt{y \frac{dy}{dx}}$Squaring both sides:$(y - 2x \frac{dy}{dx})^2 = 4 \left(\frac{dy}{dx}ight)^2 \left(y \frac{dy}{dx}ight)$$(y - 2x \frac{dy}{dx})^2 = 4y \left(\frac{dy}{dx}ight)^3$The highest order derivative is $\frac{dy}{dx}$,so the order is $1$. The power of the highest order derivative is $3$,so the degree is $3$.

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

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