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(D) Given the family of curves: $x^2 = c(y+c)^2$. Taking the square root on both sides,we get $x = \sqrt{c}(y+c)$ (assuming $x > 0$ for simplicity,as

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$x\left(\frac{dy}{dx}\right)^3-y\left(\frac{dy}{dx}\right)^2-1=0$

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(D) Given the family of curves: $x^2 = c(y+c)^2$. Taking the square root on both sides,we get $x = \sqrt{c}(y+c)$ (assuming $x > 0$ for simplicity,as the parameter $c$ absorbs the sign). Differentiating with respect to $x$: $1 = \sqrt{c} \frac{dy}{dx}$,which implies $\sqrt{c} = \frac{dx}{dy}$. Substituting $\sqrt{c} = \frac{dx}{dy}$ into the equation $x = \sqrt{c}(y+c)$: $x = \frac{dx}{dy} \left( y + \left( \frac{dx}{dy} \right)^2 \right)$. Multiplying by $\left( \frac{dy}{dx} \right)^3$: $x \left( \frac{dy}{dx} \right)^3 = \left( \frac{dx}{dy} \cdot \frac{dy}{dx} \right) \left( y \left( \frac{dy}{dx} \right)^2 + 1 \right)$. Since $\frac{dx}{dy} \cdot \frac{dy}{dx} = 1$,we have: $x \left( \frac{dy}{dx} \right)^3 = y \left( \frac{dy}{dx} \right)^2 + 1$. Rearranging the terms,we get $x \left( \frac{dy}{dx} \right)^3 - y \left( \frac{dy}{dx} \right)^2 - 1 = 0$. Thus,option $D$ is correct.

If $c$ is a parameter,then the differential equation of the family of curves $x^2=c(y+c)^2$ is

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