Question

x*sqrt(1-y^2)dx+sqrt(1-x^2)dy=0

Answer 1

Ответ:

$x\cdot \sqrt{1-y^2}\, dx+\sqrt{1-x^2}\, dy=0\\\\\\x\cdot \sqrt{1-y^2}\, dx=-\sqrt{1-x^2}\, dy\ \ ,\ \ \ \ \displaystyle \int\frac{x\, dx}{\sqrt{1-x^2}}=-\int \frac{dy}{\sqrt{1-y^2}}\\\\\\\int \frac{dy}{\sqrt{1-y^2}}=\frac{1}{2}\int \frac{-2x\, dx}{\sqrt{1-x^2}}\\\\\\arcsin\, y=\frac{1}{2}\cdot 2\sqrt{1-x^2}+C\ \ ,\ \ \ \ arcsin\, y=\sqrt{1-x^2}+C\ \ ,\\\\\\y=sin(\sqrt{1-x^2}+C)$