The differential equation of the family of cu | vedclass

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Question

(B) Given the family of curves: ${y^2} = 4a(x + a) \quad (i)$Differentiating both sides with respect to $x$:$2y \frac{{dy}}{{dx}} = 4a$$a = \frac{y}{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the family of curves: ${y^2} = 4a(x + a) \quad (i)$Differentiating both sides with respect to $x$:$2y \frac{{dy}}{{dx}} = 4a$$a = \frac{y}{2} \frac{{dy}}{{dx}} \quad (ii)$Substitute the value of $a$ from equation $(ii)$ into equation $(i)$:${y^2} = 4 \left( \frac{y}{2} \frac{{dy}}{{dx}} ight) \left( x + \frac{y}{2} \frac{{dy}}{{dx}} ight)$${y^2} = 2y \frac{{dy}}{{dx}} \left( x + \frac{y}{2} \frac{{dy}}{{dx}} ight)$Divide by $y$ (assuming $y 
eq 0$):$y = 2 \frac{{dy}}{{dx}} \left( x + \frac{y}{2} \frac{{dy}}{{dx}} ight)$$y = 2x \frac{{dy}}{{dx}} + y {\left( \frac{{dy}}{{dx}} ight)^2}$Rearranging the terms:$y - y {\left( \frac{{dy}}{{dx}} ight)^2} = 2x \frac{{dy}}{{dx}}$$y \left[ {1 - {{\left( {\frac{{dy}}{{dx}}} ight)}^2}} ight] = 2x \frac{{dy}}{{dx}}$Thus,the correct option is $B$.

The differential equation of the family of curves ${y^2} = 4a(x + a)$,where $a$ is an arbitrary constant,is

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