Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y-\cos y=x:(y\sin y+\cos y+x)y^{\prime }=y$

Answer 1

Let $y-\cos y=x$
Differentiating both sides w.r.t. $x$ we get,
$\frac{d}{dx}[y-\cos y]=\frac{dx}{dx}$
$\frac{d\left(y\right)}{dx}-\frac{d\left[\cos y\right]}{dx}=1$
$\frac{dy}{dx}-(-\sin y)\frac{dy}{dx}=1$
$\frac{dy}{dx}+\sin y\frac{dy}{dx}=1$
$\frac{dy}{dx}[1+\sin y]=1$
$\frac{dy}{dx}=\frac{1}{1+\sin y}$
$\frac{dy}{dx}=y^{\prime }$
$y^{\prime }=\frac{1}{1+\sin y}$
Taking LHS
Putting $x=y-\cos y\left(y\sin y\right)+\cos y+x)y^{\prime }$
$=[y\sin y+\cos y+y-\cos y]y^{\prime }(y\sin y+\cos y+x)y^{\prime }=[y\sin y+y]y^{\prime }(y\sin y+\cos y+x)y^{\prime }=y(1+\sin y)y^{\prime }$
Putting
$y^{\prime \prime }=\frac{1}{1+\sin y}$
$(y\sin y+\cos ‘y+x)y^{\prime }$
$=y(1+\sin y)\left[\frac{1}{1+\sin y}\right]$
$(y\sin y+\cos y+x)y^{\prime }=y$
Thus $LHS=RHS$
Hence verified.

Final answer:
Hence, the fuction $y-\cos y=x$ is a solution of the differential equation $(y\sin y+\cos y+x)y^{\prime }=y$