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

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Question

(A) Given the equation $y^2 = 4a(x+a)$.Differentiating both sides with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$y \frac{dy}{dx} = 2a$So,$a = \fra

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$y^2 = 2xy' + (y')^2$

Vedclass · Answer

(A) Given the equation $y^2 = 4a(x+a)$.Differentiating both sides with respect to $x$,we get:$2y \frac{dy}{dx} = 4a$$y \frac{dy}{dx} = 2a$So,$a = \frac{y}{2} \frac{dy}{dx}$.Substitute the value of $a$ back into the original equation:$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} ight) \left( x + \frac{y}{2} \frac{dy}{dx} ight)$$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} ight)$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} ight)^2$.

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

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