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$y-\cos y=x \quad:(y \sin y+\cos y+x) y^{\prime}=y$


$y-\cos y=x$                                      ...(1)

Differentiating both sides of the equation with respect to x, we get:

$\frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x)$

$\Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1$

$\Rightarrow y^{\prime}(1+\sin y)=1$

$\Rightarrow y^{\prime}=\frac{1}{1+\sin y}$

Substituting the value of $y^{\prime}$ in equation (1), we get:

L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$

$=(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y}$

$=y(1+\sin y) \cdot \frac{1}{1+\sin y}$


$=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.





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