The differential equation frac{dy}{dx} = frac | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ Separating the variables,we get: $\int \frac{y}{\sqrt{1-y^2}} dy = \int

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fixed radius of $1$ unit and variable centres along the $X$-axis

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ Separating the variables,we get: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$ Let $u = 1-y^2$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$. Substituting this into the integral: $-\frac{1}{2} \int u^{-1/2} du = x + C$ $-\frac{1}{2} \cdot 2u^{1/2} = x + C$ $-\sqrt{1-y^2} = x + C$ Squaring both sides: $1-y^2 = (x+C)^2$ $(x+C)^2 + y^2 = 1$ This is the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$,where the centre is $(-C, 0)$ and the radius $r = 1$. Since $C$ is an arbitrary constant,the centre $(-C, 0)$ varies along the $X$-axis,while the radius remains fixed at $1$ unit.

The differential equation $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

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