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dy-dx-2cos-2-x-sin-2-x-y-2-2cos-x-y-0-1-amp-y-1-sin-x-




Question Number 146911 by EDWIN88 last updated on 16/Jul/21
   (dy/dx) = ((2cos^2 x−sin^2 x+y^2 )/(2cos x))     y(0)=−1 & y(1)=sin x
$$\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{y}^{\mathrm{2}} }{\mathrm{2cos}\:\mathrm{x}} \\ $$$$\:\:\:\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1}\:\&\:\mathrm{y}\left(\mathrm{1}\right)=\mathrm{sin}\:\mathrm{x}\: \\ $$

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