The differential equation satisfied by the sy | vedclass

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(C) Given the equation of the parabola: $y^{2} = 4ax + 4a^{2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = 4a$$\Rightarrow a =

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

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(C) Given the equation of the parabola: $y^{2} = 4ax + 4a^{2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.Substitute the value of $a$ back into the original equation:$y^{2} = 4\left(\frac{y}{2} \frac{dy}{dx}ight)x + 4\left(\frac{y}{2} \frac{dy}{dx}ight)^{2}$.Simplify the equation:$y^{2} = 2xy \frac{dy}{dx} + 4 \cdot \frac{y^{2}}{4} \left(\frac{dy}{dx}ight)^{2}$.$y^{2} = 2xy \frac{dy}{dx} + y^{2} \left(\frac{dy}{dx}ight)^{2}$.Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}ight)^{2}$.Rearranging the terms,we get:$y \left(\frac{dy}{dx}ight)^{2} + 2x \frac{dy}{dx} - y = 0$.

The differential equation satisfied by the system of parabolas $y^{2} = 4a(x + a)$ is

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