Find bounded solutions of this ODE The function $y(x)$ satisfies the following ode:
$L[y]=\frac{d}{dx}(x^2\frac{dy}{dx})-x^2y= f(x)$
By substituting $y(x)=\frac{z(x)}{x}$ we can solve the homogeneous problem $(L[y]=0)$ and obtain the answer 
$y(x)=\frac{Ae^x}{x}+\frac{Be^{-x}}{x}$.
How would you show that the solution given by $sinh(x)/x$ is bounded as $x \rightarrow 0$?
Furthermore how would you find a solution that remains bounded as $x \rightarrow \infty$?

Hint on the first question:
$$f(x) = \frac{\sinh(x)}{x} = \frac{1}{2x}(e^x - e^{-x})$$
We can show that this function has limit at $x=0$ as follows:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$
$e^{-x} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}$
\begin{align*}\frac{\sinh(x)}{x} &= \frac{1}{2x} \left(\sum_{n=0}^\infty \frac{x^n}{n!} - \sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}\right)\\
&= \frac{1}{2x} \sum_{n=0}^\infty (1-(-1)^n)\frac{x^n}{n!}\\
&= \frac{1}{2x} \sum_{n=0}^\infty \frac{2 x^{2n+1}}{(2n+1)!}\\
&=\sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!}
\end{align*}
So we conclude that
$$\lim_{x\to 0^+} \frac{\sinh(x)}{x} = \lim_{x\to 0^+} \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!} = 1$$
This tells us that $\lim_{x\to 0^+} f(x) = 1$, hence the function remains bounded as $x \to 0$.
Hint on the second question:
$$\lim_{x\to\infty} \frac{e^x}{x} = +\infty$$
What this means is that any nonzero constant in front of the term $\frac{e^x}{x}$ will result in an unbounded solution as $x\to\infty$.
