The solutions of $y^{\prime \prime}+y=g$ are bounded Suppose that $g$ is a continuous differentiable, increasing and bounded real function.
How can one prove that the solutions of the differential equation $(E)$ 
$$y^{\prime \prime}+y=g$$ are bounded? And that $(E)$ has a unique solution having a finite limit at $\infty $?
 A: Since up to now no solution was posted, I will provide one even though I think that there might be a more direct approach to this problem. 
This is a linear dynamical system of the form
$$\dot{x}=Ax(t)+b(t)\\
y(t)=Cx(t)$$
with $x:=[y\quad\dot{y}]^T$ the state vector, $A=\left[\matrix{0 & 1 \\ -1 & 0}\right]$, $C=[1\quad 0]^T$ and $b(t)=[0\quad g(t)]^T$. For initial conditions $x(t_0)=x_0$, the solution is given by
$$x(t)=e^{A(t-t_0)}x_0+\int_{t_0}^t{e^{A(t-s)}b(s)ds}$$
For the specific form of $A$ we have
$$e^{At}=\left[\matrix{\cos t & \sin t\\ -\sin t & \cos t}\right]$$ and therefore
$$y(t)=Ce^{A(t-t_0)}x_0+\int_{t_0}^t{\sin(t-s)g(s)ds}$$
The first term in the r.h.s. of the above equation is a linear combination of $\sin t$ and $\cos t$ terms so the boundedness of $y(t)$ can be proved if $$\int_{t_0}^t{\sin(t-s)g(s)ds}$$ is proved bounded or equivalently if both terms
$$\int_{t_0}^t{\sin(s)g(s)ds}\\
\int_{t_0}^t{\cos(s)g(s)ds}$$
are proved bounded.
Also, since $g$ is increasing and bounded the limit $\lim_{t\rightarrow\infty}g(t)=g^*<\infty$ exists. We will now prove boundedness of the term $I(t):=\int_{t_0}^t{\cos(s)g(s)ds}$. We can proceed similarly for $\int_{t_0}^t{\sin(s)g(s)ds}$. We can write
$$I(t)=g^*(\sin t-\sin t_0)+\int_{t_0}^t{\cos(s)(g(s)-g^*)ds}$$
 Boundedness of $\int_{t_0}^t{\cos(s)(g(s)-g^*)ds}$ for all $t\geq t_0$ follows directly from the boundedness of the integral 
$$I'(k):=\int_{\pi/2}^{2k\pi+5\pi/2}{\cos(s)(g(s)-g^*)ds}$$
for all $k\in\mathbb{N}$.
In order to bound $I'(k)$ the idea is to decompose the integral over intervals where the sign of $\cos(s)(g(s)-g^*)$ changes and provide bounds for these terms. Specifically 
$$I'(k)=\sum_{m=0}^k{\int_{2m\pi+\pi/2}^{2m\pi+3\pi/2}{\cos(s)(g(s)-g^*)ds}}+\sum_{m=0}^k{\int_{2m\pi+3\pi/2}^{2m\pi+5\pi/2}{\cos(s)(g(s)-g^*)ds}}$$
$$I'(k)\leq \sum_{m=0}^k{(g(2m\pi+\pi/2)-g^*)\int_{2m\pi+\pi/2}^{2m\pi+3\pi/2}{\cos(s)ds}}+\sum_{m=0}^k{(g(2m\pi+5\pi/2)-g^*)\int_{2m\pi+3\pi/2}^{2m\pi+5\pi/2}{\cos(s)ds}}$$
Since
$$\int_{2m\pi+\pi/2}^{2m\pi+3\pi/2}{\cos(s)ds}=-2\\
\int_{2m\pi+3\pi/2}^{2m\pi+5\pi/2}{\cos(s)ds}=2$$
we have that
$$I'(k)\leq 2 \sum_{m=0}^k{(g(2m\pi+5\pi/2)-g^*)}-2 \sum_{m=0}^k{(g(2m\pi+\pi/2)-g^*)}=2[g(2k\pi+5\pi/2)-g(\pi/2)]\leq 2[g^*-g(\pi/2)]\quad \forall k\in\mathbb{N}$$
For the lower bound we have
$$I'(k)=\sum_{m=0}^k{\int_{2m\pi+\pi/2}^{2m\pi+3\pi/2}{\cos(s)(g(s)-g^*)ds}}+\sum_{m=0}^k{\int_{2m\pi+3\pi/2}^{2m\pi+5\pi/2}{\cos(s)(g(s)-g^*)ds}}\\ \geq  \sum_{m=0}^k{(g(2m\pi+3\pi/2)-g^*)\int_{2m\pi+\pi/2}^{2m\pi+3\pi/2}{\cos(s)ds}}+\sum_{m=0}^k{(g(2m\pi+3\pi/2)-g^*)\int_{2m\pi+3\pi/2}^{2m\pi+5\pi/2}{\cos(s)ds}}=0\quad \forall k\in\mathbb{N}$$
Thus $I'(k)$ is bounded $\forall k\in\mathbb{N}$ and therefore $I(t)$ is bounded for all $\geq t_0$. Similarly we can prove boundedness of $\int_{t_0}^t{\sin (s)g(s)ds}$ to result in the overall boundedness of $y$.
