Solution of differential equation $x^{2} y - x^{3} \frac{d y}{d x} = y^{4} \cos x$ is

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$x^{2} y^{3} = 3 \sin x + c x^{2} y$

$x^{2} y^{- 3} = 2 \sin x + c$

$x^{2} y^{- 3} = 3 \cos x + c$

$x^{3} y^{- 3} = 3 \sin x + c$

answer is C.

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Detailed Solution

$\frac{1}{y^{4}} \frac{d y}{d x} + (- \frac{1}{x}) \frac{1}{y^{3}} = \frac{- \cos x}{x^{3}} ..... (1) P u t \frac{1}{y^{3}} = ϑ \Rightarrow \frac{1}{y^{4}} \frac{d y}{d x} = \frac{- 1}{3} \frac{d ϑ}{d x} ∴ (1) \Rightarrow \frac{d ϑ}{d x} + \frac{3}{x} ϑ = 3 \cos x$

Which is linear is $ϑ$
IF $= e^{3 \int \frac{1}{x} d x} = e^{3 \log x} = x^{3}$
Solution is $ϑ x^{3} = 3 \int \cos x d x + c$
$x^{3} y^{- 3} = 3 \sin x + c$