The solution of the differential equation x , | vedclass

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Question

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

Vedclass · Accepted Answer

$xy = \sin(x + c)$

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

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

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