Question

(√(1-y^2)dx)-(√(1-x^2)dy)=0, y(0)=√3/2)

Answer 1

$(\sqrt{1-y^2})dx-(\sqrt{1-x^2})dy=0\\ (\sqrt{1-y^2})dx=(\sqrt{1-x^2})dy\\ \frac{1}{\sqrt{1-y^2}}dy=\frac{1}{\sqrt{1-x^2}}dx\\ \text{Integrate both sides}\\ \sin^{-1}(y)=\sin^{-1}(x)+c\\ \text{But} y(0)=\frac{\sqrt{3}}{2}\\ \sin^{-1}(\frac{\sqrt{3}}{2})=\sin^{-1}(0)+c\\ 60^{\circ}=0+c\\ c=60^{\circ}\\ \text{Therefore}, \\ \sin^{-1}(y)=\sin^{-1}(x)+60°$