Verify that the given function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$.
(N/A) Given equation: $y - \cos y = x$ ..........$(1)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)$
$y' - (-\sin y) y' = 1$
$y'(1 + \sin y) = 1$
$y' = \frac{1}{1 + \sin y}$ ..........$(2)$
Now,consider the $L.H.S.$ of the differential equation:
$L.H.S. = (y \sin y + \cos y + x) y'$
Substitute $x = y - \cos y$ from equation $(1)$:
$L.H.S. = (y \sin y + \cos y + y - \cos y) y'$
$L.H.S. = (y \sin y + y) y'$
$L.H.S. = y(1 + \sin y) y'$
Substitute $y'$ from equation $(2)$:
$L.H.S. = y(1 + \sin y) \cdot \frac{1}{1 + \sin y}$
$L.H.S. = y = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is indeed a solution of the differential equation.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)$
$y' - (-\sin y) y' = 1$
$y'(1 + \sin y) = 1$
$y' = \frac{1}{1 + \sin y}$ ..........$(2)$
Now,consider the $L.H.S.$ of the differential equation:
$L.H.S. = (y \sin y + \cos y + x) y'$
Substitute $x = y - \cos y$ from equation $(1)$:
$L.H.S. = (y \sin y + \cos y + y - \cos y) y'$
$L.H.S. = (y \sin y + y) y'$
$L.H.S. = y(1 + \sin y) y'$
Substitute $y'$ from equation $(2)$:
$L.H.S. = y(1 + \sin y) \cdot \frac{1}{1 + \sin y}$
$L.H.S. = y = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is indeed a solution of the differential equation.
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