The general solution of left(frac{dy}{dx}righ | vedclass

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Question

(A) Given the differential equation: $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$ Factor the right side: $\left(\frac{dy}{dx}\rig

Vedclass · Accepted Answer

$2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + C$

Vedclass · Answer

(A) Given the differential equation: $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$ Factor the right side: $\left(\frac{dy}{dx}\right)^{2} = (1 - x^{2}) - y^{2}(1 - x^{2}) = (1 - x^{2})(1 - y^{2})$ Taking the square root on both sides: $\frac{dy}{dx} = \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}$ Separate the variables: $\frac{dy}{\sqrt{1 - y^{2}}} = \sqrt{1 - x^{2}} dx$ Integrate both sides: $\int \frac{dy}{\sqrt{1 - y^{2}}} = \int \sqrt{1 - x^{2}} dx$ Using standard integrals $\int \frac{1}{\sqrt{1 - y^{2}}} dy = \sin^{-1} y$ and $\int \sqrt{1 - x^{2}} dx = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$: $\sin^{-1} y = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$ Multiply by $2$: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + 2C_1$ Let $C = 2C_1$,then: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + C$

The general solution of $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$ is

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