Question:
$y-\cos y=x \quad:(y \sin y+\cos y+x) y^{\prime}=y$
Solution:
$y-\cos y=x$ ...(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. $=(y \sin y+\cos y+x) y^{\prime}$
$=(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y}$
$=y(1+\sin y) \cdot \frac{1}{1+\sin y}$
$=y$
$=$ R.H.S.
Hence, the given function is the solution of the corresponding differential equation.