The differential equation of the family of cu | vedclass

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(B) The length of the normal to a curve is given by the formula $L = |y|\sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is a constan

Vedclass · Accepted Answer

${\left( {y\frac{{dy}}{{dx}}} \right)^2} = {k^2} - {y^2}$

Vedclass · Answer

(B) The length of the normal to a curve is given by the formula $L = |y|\sqrt{1 + (\frac{dy}{dx})^2}$.Given that the length of the normal is a constant $k$,we have $|y|\sqrt{1 + (\frac{dy}{dx})^2} = k$.Squaring both sides,we get $y^2(1 + (\frac{dy}{dx})^2) = k^2$.Expanding the equation,we get $y^2 + y^2(\frac{dy}{dx})^2 = k^2$.Rearranging the terms,we obtain $y^2(\frac{dy}{dx})^2 = k^2 - y^2$,which can be written as $(y\frac{dy}{dx})^2 = k^2 - y^2$.

The differential equation of the family of curves for which the length of the normal is equal to a constant $k$,is given by

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