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(D) Given curve is ${y^2} = 2c(x + \sqrt{c}).$ Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2c$,which implies $c = y \frac{dy}{dx}.$

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Both $(a)$ and $(c)$

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(D) Given curve is ${y^2} = 2c(x + \sqrt{c}).$ Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2c$,which implies $c = y \frac{dy}{dx}.$ Substituting $c$ into the original equation: ${y^2} = 2 \left( y \frac{dy}{dx} \right) \left( x + \sqrt{y \frac{dy}{dx}} \right).$ Rearranging the terms: $\frac{y}{2(dy/dx)} - x = \sqrt{y \frac{dy}{dx}}.$ Squaring both sides: $\left( \frac{y}{2(dy/dx)} - x \right)^2 = y \frac{dy}{dx}.$ Multiplying by $4(dy/dx)^2$: $(y - 2x(dy/dx))^2 = 4y(dy/dx)^3.$ Expanding the square: ${y^2} - 4xy \frac{dy}{dx} + 4{x^2} \left( \frac{dy}{dx} \right)^2 = 4y \left( \frac{dy}{dx} \right)^3.$ Rearranging into the standard form: $4y \left( \frac{dy}{dx} \right)^3 - 4{x^2} \left( \frac{dy}{dx} \right)^2 + 4xy \frac{dy}{dx} - {y^2} = 0.$ The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $3$,so the degree is $3$. Thus,the differential equation is of order $1$ and degree $3$.

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

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