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

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(C) 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 $1$ and variable centres along the $x$-axis

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(C) 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 $du = -2y dy$,which implies $y dy = -\frac{1}{2} du$.The integral becomes: $-\frac{1}{2} \int u^{-1/2} du = x + C$.$-\frac{1}{2} \cdot 2u^{1/2} = x + C \Rightarrow -\sqrt{1-y^2} = x + C$.Squaring both sides: $1-y^2 = (x+C)^2$.Rearranging the terms: $(x+C)^2 + y^2 = 1$.This represents a family of circles with a fixed radius of $1$ and centres at $(-C, 0)$,which lie along the $x$-axis.

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

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