The solution of the differential equation dy-dx - 3x-2-1-x-3-y - sin-2-x-1-x-3-is -y-1-x-3- - x - 1-2-sin-2x - c-2-tan-1 - x-y - - log-x-c - 0-log-y - -x-2 - y-2- - log-y-c - 0-sin-h-1 - x-y - - log-y- c -0-

The correct option is D sin h^ 1 ( x/y ) + log y+ c =0 dy/dx + 3x^2/1+x^3 y = sin^2 x/1+x^3 P = 3x^2/1+x^3⇒ I.F. = e^∫ p dx = e^log(1+x^3) = 1 + x^3 Thus the ...

The solution ...

Question

The solution of the differential equation $\frac{d y}{d x} + \frac{3 x^{2}}{1 + x^{3}} y = \frac{s i n^{2} x}{1 + x^{3}}$ is

$y (1 + x^{3}) = x + \frac{1}{2} s i n 2 x + c$

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$2 t a n^{- 1} (\frac{x}{y}) + l o g x + c = 0$

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$l o g (y + \sqrt{x^{2} + y^{2}}) + l o g y + c = 0$

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$s i n h^{- 1} (\frac{x}{y}) + l o g y + c = 0$

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Solution

The correct option is D
$s i n h^{- 1} (\frac{x}{y}) + l o g y + c = 0$

$\frac{d y}{d x} + \frac{3 x^{2}}{1 + x^{3}} y = \frac{s i n^{2} x}{1 + x^{3}} P = \frac{3 x^{2}}{1 + x^{3}} \Rightarrow I . F . = e^{\int p d x} = e^{l o g (1 + x^{3}) = 1 + x^{3}}$
Thus the solution is
$y . (1 + x^{3}) = \int \frac{s i n^{2} x}{1 + x^{3}} (1 + x^{3}) d x = \int \frac{1 - c o s 2 x}{2} d x \Rightarrow y (1 + x^{3}) = \frac{1}{2} x - \frac{s i n 2 x}{4} + c .$

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The solution of the differential equation $\frac{d y}{d x} + \frac{3 x^{2}}{1 + x^{3}} y = \frac{s i n^{2} x}{1 + x^{3}}$ is