# Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $y-\cos y=x\qquad:\;(y\sin y+\cos y+x)y'=y$

Toolbox:
• Differentiation of $\cos y$ is $- \sin y . y'$
Step 1:
Given: $y - \cos y = x$
Differentiating on both sides we get,
$y' - (-\sin y)y' = 1$
$y'(1 + \sin y) = 1$
$y' =\large\frac{ 1}{1+\sin y}$
Step 2:
Substituting y' in the given solution we get,
$\large\frac{(y\sin y + \cos y + x)}{1+\sin y} $$= y Now substituting for x we get \large\frac{[y\sin y + \cos y + -\cos y]}{1+\sin y} =\frac{ y(1+\sin y)}{1 + \sin y}$$= y$
Since the $LHS = RHS$, the given solution is verified
edited Aug 19, 2013