How to show $ y′(t) + y(t) = h(t) $ has a bounded solution, if $h(t)$ is bounded I have a midterm tomorrow, and while studying for that I saw this question, however don't have any idea how to solve it. (I could not come up with a legitimate proof. All I could do was, by putting some functions, approving what the problem claims.) I will appreciate if you can help.
Suppose that $ h(t)$ is continuous and bounded on $(-\infty,\infty)$ and 
$$|h(t)| \leq M, \forall t\in (-\infty,\infty)$$
Show that equation
$$
y′(t) + y(t) = h(t)
$$
has one solution that is bounded on $(-\infty,\infty)$. Also, show that if $h(t)$ is a periodic function, then $y(t)$ is also periodic.
Regards,
Amadeus
 A: Multiplying by $e^{t}$, we have
$$ (e^{t}y(t))' = e^t h(t). $$
Integrating, we have
$$ y(t) = y(0)e^{-t} + e^{-t}\int_0^{t} e^{s}h(s) \, ds $$
Now we have to find a bounded solution. Since $h(s)\geqslant-M$, we can define $H(s)=h(s)+M$ positive, and hence we can also write the solution as
$$ y(t) = y(0)e^{-t} + e^{-t}\int_0^{t} e^{s}(H(s)-M) \, ds \\
= y(0)e^{-t} + e^{-t}\int_0^{t} e^{s}H(s) \, ds + M(e^t-1)e^{-t} \\
= M+(y(0)-M)e^{-t} + e^{-t}\int_0^{t} e^{s}H(s) \, ds $$
Now, the integrand is positive. Let us first check what happens as $t \to \infty$. The second term dies away. The integrand is between $0$ and $2M$, so for $t>0$,
$$ y(t) \leqslant  M + (y(0)-M) + 2M \int_0^t e^{s-t} \, ds = y(0)+ 2M(1-e^{-t}), $$
which is bounded. Now the problem comes, $-\infty$. As $t \to -\infty$, the integral tends to
$$ -A=\int_0^{-\infty} e^{s} H(s) \, ds = -\int_0^{\infty} e^{-s} H(-s) \, ds, $$
which converges because $H$ is bounded. Now, the obvious candidate for the bounded solution has
$$ y(0)-M-A=0, \tag{1} $$
because otherwise, there's no way that $y$ is bounded, since we'd have $y = O( e^{-t})$. Now, another way to write $y$ is
$$ y(t) = M + (y(0)-M-A)e^{-t} - e^{-t} \int_t^{-\infty} e^s H(s) \, ds $$
(I'm not convinced about my sign there, but it's irrelevant.) Now impose (1), so either way,
$$ \lvert y(-t)-M \rvert = e^{t} \int_{t}^{\infty} e^{-s} H(-s) \, ds, $$ 
where I've changed the sign of $t$ for convenience. All we have to do now is show that this is bounded. But $H(-s)<2M$, so the right-hand side is less than
$$ 2M \int_t^{\infty} e^{t-s} \, ds = 2M, $$
and so there is precisely one bounded solution.

Right, now the periodic bit. This is obviously false for general $y$, since I can choose what I like for $y(0)$ and get a nonperiodic $e^{-t}$, so we'll have to assume that $y(-\infty)$ is finite. Therefore, take
$$ y(t) = a + e^{-t} \int_{-\infty}^t e^{s}g(s) \, ds,  $$
where $g$ has integral zero over a period. (then
$$ y'(t) + y(t) = a+ e^{-t} \int_{-\infty}^t e^{s}g(s) \, ds- e^{-t} \int_{-\infty}^t e^{s}g(s) \, ds+g(t)=a+g(t), $$
so $a+g(s)=h(s)$, and this is a legitimate solution). Now suppose $g$ has period $T$. Then
$$ y(t+T)-y(t) = e^{-T} e^{-t} \int_{-\infty}^{t+T} e^{s}g(s) \, ds -e^{-t} \int_{-\infty}^t e^{s}g(s) \, ds \\
= e^{-t} \left( \int_{-\infty}^{t+T} e^{s-T} g(s) \, ds - \int_{-\infty}^t e^{s}g(s) \, ds \right) \tag{2} $$
Setting $u=s-T$ in the first integral, the upper limit becomes $t+T-T=t$, and we have
$$ \int_{-\infty}^{t} e^{u} g(u+T) \, du - \int_{-\infty}^t e^{s}g(s) \, ds $$
But $g(u+T)=g(u)$ by definition, so the bracket in (2) is zero, and $y$ is also periodic. Phew!
A: For $V=y^2$ we have
$$\frac{dV}{dt}=2y(t)\frac{dy}{dt}(t)=-2y^2(t)+2y(t)h(t)\leq -y^2(t)+h^2(t)=-V(t)+h^2(t)$$
Multiplying by $e^t$ we equivalently have
$$\frac{d}{dt}\left[V(t)e^t-\int_0^t{h^2(s)e^s ds}\right]\leq 0$$
If we now integrate over $[0,t]$ we obtain
$$y^2(t)\leq y^2(0)e^{-t}+e^{-t}\int_0^t{e^sM^2ds}=y^2(0)e^{-t}+M^2(1-e^{-t})$$
i.e. $y$ is bounded in $[0,\infty)$. This shows that the solution is bounded in $[0,\infty)$ for all initial conditions. A bounded solution on the whole real line  $(-\infty,\infty)$ is not possible for every initial condition at $t=0$. However as proved by Chappers in his answer there exists some bounded solution on the whole real line  $(-\infty,\infty)$ if the initial condition is chosen appropriately. 
