(1−y2)dx−(1−x2)dy=0(1−y2)dx=(1−x2)dy11−y2dy=11−x2dxIntegrate both sidessin−1(y)=sin−1(x)+cButy(0)=32sin−1(32)=sin−1(0)+c60∘=0+cc=60∘Therefore,sin−1(y)=sin−1(x)+60°(\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°(1−y2)dx−(1−x2)dy=0(1−y2)dx=(1−x2)dy1−y21dy=1−x21dxIntegrate both sidessin−1(y)=sin−1(x)+cButy(0)=23sin−1(23)=sin−1(0)+c60∘=0+cc=60∘Therefore,sin−1(y)=sin−1(x)+60°
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