Consider the ODE $y'(x)=a(x)y(x),\; x>0, \;y(0)=y_{0}\neq 0$. Which of the following statements are true? 
Assume that $a:[0,\infty) \to \mathbb{R}$ is a continuous function. Consider the ODE $$y'(x)=a(x)y(x),\quad x>0, \;y(0)=y_0 \neq 0$$ Which of the following statements are true?

*

*If $\int_{0}^{\infty}\vert a(x)\vert dx<\infty$, then $y$ is bounded.

*If $\int_{0}^{\infty}\vert a(x)\vert dx<\infty$, then $\lim_{x \to \infty}y(x)$ exists.

*If $\lim_{ x \to \infty}a(x)=1$, then $\lim_{x \to \infty}\vert y(x) \vert=\infty$

*If $\lim_{ x \to \infty}a(x)=1$, then $y$ is monotone.


My attempt:
$f(x,y)=a(x)y(x)$ then $f(x,y)$ is continuous and $\frac{\partial f}{\partial y}=a(x)$ which is continuous at $(0,y_{0})$. So, it follows that there exists unique solution using Picard uniqueness theorem. I didn't get how to explicitly solve for option 1 and 2.
Also for option 3, i solved through an example. So, if $\lim_{ x \to \infty}a(x)=1$, then choose $a(x)=1$, clearly satisfies the hypothesis. Then the solution is $y=y_{0}e^x$ and it satisfies options $3$ and $4$. But can someone please show me how to work on these questions explicitly, without using some examples etc? Thanks in advance.
(Also can someone please edit the tags, there does not exists tag for uniqueness-existence of solution and the famous picard's theorem for existsence and uniqueness).
 A: This ODE can be easily solved explicitly, by the same method used as for $a(x)$ constant
\begin{align*}
y'(x) &= a(x) y(x) \\
\implies \frac{y'(x)}{y(x)} &= a(x) \\
\implies \int_{y_0}^y \frac{dy}{y} &= \int_0^x a(x) dx \\
\implies y(x) &= y_0 e^{\int_0^x a(x) dx}
\end{align*}
From this, options $1$ and $2$ are trivial - $\int_{0}^{\infty}\vert a(x)\vert dx<\infty \implies \lim_{x\to\infty}y(x) $ exists, and thus $y(x)$ is bounded.
Also, if $\lim_{x\to\infty}a(x)=1$, then by the definition of a limit, $\forall\epsilon > 0\ \exists\  x_0$ such that for $x>x_0$, $1-\epsilon<a(x)<1+\epsilon$.
Thus, taking $C =y_0 e^{\int_0^{x_0} a(x) dx}$ (which is a constant), we get
\begin{align*}
y(x) &= y_0 e^{\int_0^x a(x) dx} \\
&= y_0 e^{\int_0^{x_0} a(x) dx}e^{\int_{x_0}^x a(x) dx} \\
\implies C e^{\int_{x_0}^x (1+\epsilon) dx} > y(x)&> C e^{\int_{x_0}^x (1-\epsilon) dx} \\
\implies C e^{(1+\epsilon)(x-x_0)dx} > y(x)&> C e^{(1-\epsilon)(x-x_0)dx} \\
\end{align*}
Hence, as $x\to \infty$, so does $|y(x)|\to \infty$
I still can't figure out Option 4 though, unfortunately.
