The general solution of the differential equa | vedclass

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(C) Given differential equation is $\sqrt{1-x^{2} y^{2}} \cdot dx = y \cdot dx + x \cdot dy$.We know that $d(xy) = y \cdot dx + x \cdot dy$.Substituti

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$\sin (x + C) = xy$

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(C) Given differential equation is $\sqrt{1-x^{2} y^{2}} \cdot dx = y \cdot dx + x \cdot dy$.We know that $d(xy) = y \cdot dx + x \cdot dy$.Substituting this into the equation,we get $\sqrt{1-(xy)^{2}} \cdot dx = d(xy)$.Rearranging the terms,we have $dx = \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.Integrating both sides,we get $\int dx = \int \frac{d(xy)}{\sqrt{1-(xy)^{2}}}$.This results in $x = \sin^{-1}(xy) + C$.Taking the sine of both sides,we get $\sin(x - C) = xy$,which is equivalent to $\sin(x + C) = xy$ (where $C$ is an arbitrary constant).

The general solution of the differential equation $\sqrt{1-x^{2} y^{2}} \cdot dx = y \cdot dx + x \cdot dy$ is

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