The differential equation for which y^2 = 4a( | vedclass

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(A) Given equation: $y^2 = 4a(x + a)$ ... $(i)$ On differentiating with respect to $x$, we get: $2y \frac{dy}{dx} = 4a(1 + 0) = 4a$ $2y y^{\prime} = 4

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$y - y y^{\prime 2} = 2x y^{\prime}$

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(A) Given equation: $y^2 = 4a(x + a)$ ... $(i)$ On differentiating with respect to $x$, we get: $2y \frac{dy}{dx} = 4a(1 + 0) = 4a$ $2y y^{\prime} = 4a$ $a = \frac{1}{2} y y^{\prime}$ Now, substitute the value of $a$ in equation $(i)$: $y^2 = 4 \left( \frac{1}{2} y y^{\prime} ight) \left( x + \frac{1}{2} y y^{\prime} ight)$ $y^2 = 2y y^{\prime} \left( \frac{2x + y y^{\prime}}{2} ight)$ $y^2 = y y^{\prime} (2x + y y^{\prime})$ Dividing both sides by $y$ (assuming $y 
eq 0$): $y = y^{\prime} (2x + y y^{\prime})$ $y = 2x y^{\prime} + y (y^{\prime})^2$ $y - y (y^{\prime})^2 = 2x y^{\prime}$

The differential equation for which $y^2 = 4a(x + a)$ (where $a$ is a parameter) is the general solution, is

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