The differential equation of the family of cu | vedclass

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Question

(A) Given the equation of the family of curves: $x^{2} = 4b(y+b)$.Differentiating both sides with respect to $x$,we get:$2x = 4b y^{\prime}$$b = \frac

Vedclass · Accepted Answer

$x(y^{\prime})^{2} = x + 2yy^{\prime}$

Vedclass · Answer

(A) Given the equation of the family of curves: $x^{2} = 4b(y+b)$.Differentiating both sides with respect to $x$,we get:$2x = 4b y^{\prime}$$b = \frac{2x}{4y^{\prime}} = \frac{x}{2y^{\prime}}$.Substitute the value of $b$ back into the original equation:$x^{2} = 4 \left( \frac{x}{2y^{\prime}} ight) \left( y + \frac{x}{2y^{\prime}} ight)$.Simplify the expression:$x^{2} = \frac{2x}{y^{\prime}} \left( \frac{2yy^{\prime} + x}{2y^{\prime}} ight)$.$x^{2} = \frac{2x(2yy^{\prime} + x)}{2(y^{\prime})^{2}}$.$x^{2} = \frac{x(2yy^{\prime} + x)}{(y^{\prime})^{2}}$.Dividing by $x$ (assuming $x 
eq 0$):$x(y^{\prime})^{2} = 2yy^{\prime} + x$.

The differential equation of the family of curves,$x^{2}=4 b(y+b), b \in R,$ is

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