Completing the differential equation from exercise 10.23 in Tom Apostol's "Mathematical Analysis" I found this answer, outlining the exercise, to be interesting.  However, I have trouble solving the differential equation.
The question starts by attempting to solve the following integral without complex analysis:
$$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$
...So we let 
$$ F(y) = \int\limits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$$
Next, the portion I'm referring to (from which this question proceeds) starts with 
$$\displaystyle F''(y) - F(y) + \pi/2 = 0$$
I find part of a solution to the differential equation to be 
$$F(y)=\pi/2+e^y c_1 + e^{-y}c_2$$
I'm having trouble finding the constants.  Could someone please explain this step in great detail, as I'm somewhat of a novice.
 A: As we saw in the comments, we only have to show that $|F(y)|\leq M$ for some $M>0$ 
independent of $y$. Indeed, if its shown then $|c_1|e^y=|F(y)-\frac{\pi}2-e^{-y}c_2|\leq M+\frac{\pi}2+|c_2|$ for all $y$, therefore $c_1=0$ and $c_2=-\frac{\pi}2$.
We have $\displaystyle\int_0^{+\infty}\frac{\sin x}xdx=\int_0^{+\infty}\frac{\sin xy}xdx$ for 
all $y>0$ (make the substitution $t=xy$). We get 
\begin{align*}
F(y)-\int_0^{+\infty}\frac{\sin x}xdx&=\int_0^{+\infty}\frac{\sin(xy)}x
\left(\frac 1{1+x^2}-1\right)dx\\
&=-\int_0^{+\infty}\sin(xy)\frac{x}{1+x^2}dx,
\end{align*}
and integrating by parts 
\begin{align*}
F(y)-\int_0^{+\infty}\frac{\sin x}xdx&=-\left[\frac 1y\cos(xy)\frac x{1+x^2}\right]_{x=0}^{x\to+\infty}+\int_0^{+\infty}\frac 1y\cos(xy)\left(
\frac 1{1+x^2}-\frac{2x^2}{(1+x^2)^2}\right)dx\\
&=\frac 1y\int_0^{+\infty}\frac{\cos(xy)}{1+x^2}dx-\frac 1y\int_0^{+\infty}
\cos(xy)\frac{2x^2}{(1+x^2)^2}dx.
\end{align*}
Finally 
$$ \left|F(y)-\int_0^{+\infty}\frac{\sin x}xdx\right|\leq 
\frac 1y\frac{3\pi}2,$$
hence $F$ has a limit at $+\infty$ (and is continuous): it's bounded.
