Proving single solution of initial value problem is increasing Given the

Step 1
We use the simple fact that ${\displaystyle v\left(y\right)=y-\sin y}$ is increasing and positive on ${\displaystyle y>0}$
The fact that ${\displaystyle y_0\left(x\right)=0}$ is a solution implies that ${\displaystyle y\left(x\right)>0}$ for all x (by uniqueness). Then
${\displaystyle y^{\prime }\left(x\right)=v\left(y\left(x\right)\right)>0.}$
Thus ${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}y\left(x\right)}$ exists and is nonnegative. If it's not zero, then ${\displaystyle y\left(x\right)\ge y_1>0}$ for some fixed ${\displaystyle y_1}$ and so
${\displaystyle y^{\prime }\left(x\right)=v\left(y\left(x\right)\right)\ge v\left(y_1\right)>0.}$
That would imply for any ${\displaystyle x<0}$ (by the mean value theorem and that ${\displaystyle y\left(0\right)=1}$)
${\displaystyle y\left(x\right)=y\left(x\right)-y\left(0\right)+y\left(0\right)=y^{\prime }\left(x_0\right)x+1\le v\left(y_1\right)x+1}$
this is impossible as ${\displaystyle y\left(x\right)>0}$ for all x.

Step 1
Let us introduce the new variable z defined by ${\displaystyle z=-x}$. Then
${\displaystyle y^{\prime }\left(x\right)=\frac{dy}{dx}}$
${\displaystyle =\frac{dy}{dz}\frac{dz}{dx}}$
${\displaystyle =-\frac{dy}{dz}}$
Therefore, we have the differential equation with the reversed variable direction
${\displaystyle \frac{dy}{dz}=-y+\sin y}$
This is in fact the dynamics of a simple damped pendulum. Let ${\displaystyle V\left(y\right)=\frac{y^2}{2}}$
${\displaystyle \frac{dV}{dz}=y\frac{dy}{dz}}$
${\displaystyle =-y^2+y\sin y}$
and ${\displaystyle \frac{dV}{dz}<0}$ for all ${\displaystyle \frac{y\in \mathbb{R}}{\left\{0\right\}}}$ and ${\displaystyle \frac{dV}{dz}=0}$ at ${\displaystyle y=0}$. Hence, ${\displaystyle \mathbb{R}}$ is positively invariant with respect to the ODE, meaning that every solution starting at ${\displaystyle z=z_0}$ is defined for all ${\displaystyle zgiven{z}_0}$. Let v be the solution of the ODE with ${\displaystyle y\left(0\right)=1}$. Then, we have that ${\displaystyle \underset{z\Rightarrow \mathrm{\infty }}{lim}v\left(z\right)=0}$. Finally, we conclude that ${\displaystyle \underset{x\Rightarrow -\mathrm{\infty }}{lim}u\left(x\right)=0}$.
We can also show that ${\displaystyle v\left(z\right)}$ cannot change its sign and monotonically decreases whenever ${\displaystyle v\left(z\right)>0}$.