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Solve : $ x \large\frac{dy}{dx}- \frac{y}{2}$$=x(\sin ^{-1}x) y^5$

$(a)\;y.x^2=\bigg(\sin ^{-1} x.\frac{x^3}{3} - \bigg(\frac{2}{3}(1-x^2)^{3/2}- 2\sqrt {1-x^2}\bigg)\bigg) \\ (b)\;y.x^2=-4 \bigg(\sin ^{-1} x.\frac{x^3}{3} - \frac{1}{6}\bigg(\frac{2}{3}(1-x^2)^{3/2}- 2\sqrt {1-x^2}\bigg)\bigg)\\ (c)\;y=-4 \bigg(\sin ^{-1} x .\frac{-1}{6} \bigg(\frac{2}{3}(1-x^2)^{3/2}- \sqrt {1-x^2}\bigg)\bigg)\\ (d)\;y=-4 \bigg (x^3. \sin ^{-1} x -\bigg((1-x^2)^{2}{3}-\sqrt {1-x^2}\bigg)\bigg) $

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$\large\frac{1}{y^5}\frac{dy}{dx}-\frac{1}{2xy^4}$$=(\sin ^{-1}x)$
$\large\frac{1}{y^4}$$=t$
$-4 y^{-5} \large\frac{dy}{dx}=\frac{dt}{dx}$
$-\large\frac{dt}{4 dx} -\frac{t}{2x}$$=\sin ^{-1}x$
$\large\frac{dt}{dx}+\frac{2t}{x}$$=-4 \sin ^{-1} x$
$I.f= e^{\int \large\frac{2}{x}}$$dx$
$\qquad= e^{2 \log x}$
$\qquad= x^2$
$y.x^2=-4 \int x^2 \sin ^{-1} x dx$
by parts $ \;y.x^2=-4 \bigg(\sin ^{-1} x.\frac{x^3}{3} - \frac{1}{6}\bigg(\frac{2}{3}(1-x^2)^{3/2}- 2\sqrt {1-x^2}\bigg)\bigg)$
Hence b is the correct answer.
answered Feb 7, 2014 by meena.p
 

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