The differential equation for the family of c | vedclass

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Question

(C) Given the family of curves is ${x^2} + {y^2} - 2ay = 0$ ..... $(i)$ Differentiating both sides with respect to $x$,we get: $2x + 2yy' - 2ay' = 0$

Vedclass · Accepted Answer

$({x^2} - {y^2})y' = 2xy$

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(C) Given the family of curves is ${x^2} + {y^2} - 2ay = 0$ ..... $(i)$ Differentiating both sides with respect to $x$,we get: $2x + 2yy' - 2ay' = 0$ $2ay' = 2x + 2yy'$ $2a = \frac{2x + 2yy'}{y'} = \frac{2x}{y'} + 2y$ ..... $(ii)$ Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$: ${x^2} + {y^2} - (\frac{2x}{y'} + 2y)y = 0$ ${x^2} + {y^2} - \frac{2xy}{y'} - 2{y^2} = 0$ ${x^2} - {y^2} - \frac{2xy}{y'} = 0$ Multiplying by $y'$,we get: $({x^2} - {y^2})y' - 2xy = 0$ $({x^2} - {y^2})y' = 2xy$.

The differential equation for the family of curves ${x^2} + {y^2} - 2ay = 0,$ where $a$ is an arbitrary constant,is

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