If x2y−x3dydx=y4cosx then x3y−3 equal to

x2y−x3dydx=y4cosxx2y3−x3y4dydx=cosx1xy3−1y4dydx=cosxx31y4dydx−1xy3=−1x3cosxLet t=−1y3 1y4dydx=13dtdx13dtdx+tx=−1x3cosxdtdx+3xt=−3x3cosxI.F=X3tx3=∫−3x3cosxx3dx+cy−3x3=3sinx+c

If x2y−x3dydx=y4cosx then x3y−3 equal to

x2y−x3dydx=y4cosxx2y3−x3y4dydx=cosx1xy3−1y4dydx=cosxx31y4dydx−1xy3=−1x3cosxLet t=−1y3 1y4dydx=13dtdx13dtdx+tx=−1x3cosxdtdx+3xt=−3x3cosxI.F=X3tx3=∫−3x3cosxx3dx+cy−3x3=3sinx+c

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If $x^{2} y - x^{3} \frac{d y}{d x} = y^{4} \cos x$ then $x^{3} y^{- 3}$ equal to

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3 cosx+c

2 sin x +c

sin x

3 sin x +c

answer is C.

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Detailed Solution

$x^{2} y - x^{3} \frac{d y}{d x} = y^{4} \cos x$
$\frac{x^{2}}{y^{3}} - \frac{x^{3}}{y^{4}} \frac{d y}{d x} = \cos x$
$\frac{1}{x y^{3}} - \frac{1}{y^{4}} \frac{d y}{d x} = \frac{\cos x}{x^{3}}$
$\frac{1}{y^{4}} \frac{d y}{d x} - \frac{1}{x y^{3}} = - \frac{1}{x^{3}} \cos x$
Let $t = \frac{- 1}{y^{3}}$
$\frac{1}{y^{4}} \frac{d y}{d x} = \frac{1}{3} \frac{d t}{d x}$
$\frac{1}{3} \frac{d t}{d x} + \frac{t}{x} = \frac{- 1}{x^{3}} \cos x$
$\frac{d t}{d x} + \frac{3}{x} t = \frac{- 3}{x^{3}} \cos x$
$I . F = X^{3}$
$t x^{3} = \int \frac{- 3}{x^{3}} \cos x x^{3} d x + c$
$y^{- 3} x^{3} = 3 \sin x + c$