The solution of the differential equation x d | vedclass

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Question

(B) Given the differential equation: $x dy + y dx - \sqrt{1 - x^2 y^2} dx = 0$. Rearranging the terms,we get: $x dy + y dx = \sqrt{1 - x^2 y^2} dx$. W

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$xy = \sin(x + c)$

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(B) Given the differential equation: $x dy + y dx - \sqrt{1 - x^2 y^2} dx = 0$. Rearranging the terms,we get: $x dy + y dx = \sqrt{1 - x^2 y^2} dx$. We know that $d(xy) = x dy + y dx$. Substituting this into the equation: $d(xy) = \sqrt{1 - (xy)^2} dx$. Dividing both sides by $\sqrt{1 - (xy)^2}$,we get: $\frac{d(xy)}{\sqrt{1 - (xy)^2}} = dx$. Integrating both sides: $\int \frac{d(xy)}{\sqrt{1 - (xy)^2}} = \int dx$. This results in: $\sin^{-1}(xy) = x + c$. Taking the sine of both sides: $xy = \sin(x + c)$.

The solution of the differential equation $x dy + y dx - \sqrt{1 - x^2 y^2} dx = 0$ is

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