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

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

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

Answered by bramlexs22 last updated on 13/Jan/21

⇒ (√(1−y^2 )) dx = (√(1−x^2 )) dy ⇒ (dy/( (√(1−y^2 )))) = (dx/( (√(1−x^2 )))) ⇒∫ (dy/( (√(1−y^2 )))) = ∫ (dx/( (√(1−x^2 )))) ⇒ sin^(−1) (y)=sin^(−1) (x)+C ⇒ y(x)= sin (sin^(−1) (x)+C) ⇒y(x) = x cos C + (√(1−x^2 )) sin C y(0)=((√3)/2) → ((√3)/2) = sin C then cos C = (1/2) we find solution : ∴ y(x)= ((x+(√(3(1−x^2 ))))/2)

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

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