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Question Number 129171 by benjo_mathlover last updated on 13/Jan/21

$$\sqrt{1-{\mathrm{y}}^2}\mathrm{dx}-\sqrt{1-{\mathrm{x}}^2}\mathrm{dy}=0$$$$\mathrm{y}\left(0\right)=\frac{\sqrt{3}}{2}$$

Answered by bramlexs22 last updated on 13/Jan/21

$$\Rightarrow \sqrt{1-{\mathrm{y}}^2}\mathrm{dx}=\sqrt{1-{\mathrm{x}}^2}\mathrm{dy}$$$$\Rightarrow \frac{\mathrm{dy}}{\sqrt{1-{\mathrm{y}}^2}}=\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}$$$$\Rightarrow \int \frac{\mathrm{dy}}{\sqrt{1-{\mathrm{y}}^2}}=\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}$$$$\Rightarrow {\sin }^{-1}\left(\mathrm{y}\right)={\sin }^{-1}\left(\mathrm{x}\right)+\mathrm{C}$$$$\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\sin ({\sin }^{-1}\left(\mathrm{x}\right)+\mathrm{C})$$$$\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\mathrm{x}\cos \mathrm{C}+\sqrt{1-{\mathrm{x}}^2}\sin \mathrm{C}$$$$\mathrm{y}\left(0\right)=\frac{\sqrt{3}}{2}\to \frac{\sqrt{3}}{2}=\sin \mathrm{C}\mathrm{then}\cos \mathrm{C}=\frac{1}{2}$$$$\mathrm{we}\mathrm{find}\mathrm{solution}:$$$$\therefore \mathrm{y}\left(\mathrm{x}\right)=\frac{x+\sqrt{3(1-x^2)}}{2}$$

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