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## Obtain the differential…

Kindly check my solution below. Do you have any answer key for each problem? It will be of great help to those who wish to answer if you post it here.

$Cx \sin y + x^2 y = Cy$

$x^2 y = C(y - x \sin y)$

$\dfrac{x^2 y}{y - x \sin y} = C$

$\dfrac{(y - x \sin y)(x^2 y' + 2xy) - x^2 y \left[ y' - (x \cos y y' + \sin y)\right]}{(y - x \sin y)^2} = 0$

$(y - x \sin y)(x^2 y' + 2xy) - x^2 y \left[ y' - (x \cos y y' + \sin y)\right] = 0$

$x^2 (y - x \sin y)y' + 2xy(y - x \sin y) - x^2 y y' + x^3 y \cos y y' + x^2 y \sin y = 0$

$\left[ x^2 (y - x \sin y) - x^2 y + x^3 y \cos y \right] y' + 2xy(y - x \sin y) + x^2 y \sin y = 0$

$(x^2 y - x^3 \sin y - x^2 y + x^3 y \cos y) y' + 2xy^2 - 2x^2y \sin y + x^2 y \sin y = 0$

$(-x^3 \sin y + x^3 y \cos y) y' + 2xy^2 - x^2 y \sin y = 0$

$x^3(y \cos y - \sin y) y' = x^2 y \sin y - 2xy^2$

$x^3(y \cos y - \sin y) y' = x(xy \sin y - 2y^2)$

$y' = \dfrac{x(xy \sin y - 2y^2)}{x^3(y \cos y - \sin y)}$

$y' = \dfrac{xy \sin y - 2y^2}{x^2(y \cos y - \sin y)}$