The differential equation of all circles whic | vedclass

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Question

(A) The general equation of a circle passing through the origin with its center on the $y$-axis is given by $x^{2} + (y-a)^{2} = a^{2}$,where $a$ is a

Vedclass · Accepted Answer

$(x^{2}-y^{2}) \frac{dy}{dx}-2xy=0$

Vedclass · Answer

(A) The general equation of a circle passing through the origin with its center on the $y$-axis is given by $x^{2} + (y-a)^{2} = a^{2}$,where $a$ is an arbitrary constant.Expanding this,we get $x^{2} + y^{2} - 2ay + a^{2} = a^{2}$,which simplifies to $x^{2} + y^{2} - 2ay = 0$.To eliminate the arbitrary constant $a$,we differentiate with respect to $x$:$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$.Dividing by $2$,we get $x + y \frac{dy}{dx} - a \frac{dy}{dx} = 0$,which implies $a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} = x \frac{dx}{dy} + y$.Substituting $a$ back into the original equation $x^{2} + y^{2} = 2ay$:$x^{2} + y^{2} = 2(x \frac{dx}{dy} + y)y = 2xy \frac{dx}{dy} + 2y^{2}$.Rearranging,$x^{2} - y^{2} = 2xy \frac{dx}{dy}$.Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$,we have $x^{2} - y^{2} = \frac{2xy}{\frac{dy}{dx}}$.Thus,$(x^{2} - y^{2}) \frac{dy}{dx} = 2xy$,or $(x^{2} - y^{2}) \frac{dy}{dx} - 2xy = 0$.

The differential equation of all circles which pass through the origin and whose centres lie on the $y$-axis is:

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