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Solve : $\large\frac{dy}{dx}$$=\sqrt {16 x^2 y -4x^2 y^2}\qquad y(2)=1$

$(a)\;y=2 \sin ^2 (x+c) \\ (b)\;y= 4 \sin ^2 (x+c) \\ (c)\;y= 2 (1- \sin (x^2+c)) \\ (d)\;y= 2 (1- \sin x)+c $

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$\large\frac{dy}{dx}$$=2 x \sqrt {4y-y^2}$
$\large\frac{dy}{\sqrt {4y-y^2}}$$=2x dx$
$\large\frac{dy}{\sqrt {4 -(y-2)^2}}$$=2xdx$
$\sin ^{-1} \bigg(\large\frac{y-2}{2}\bigg)=\large\frac{2x^2}{2}$$+c$
$y-2= 2 \sin (x^2+c)$
$y= 2 ( 1- \sin (x^2+c))$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
 

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