Let $f: [0,\infty) \rightarrow \mathbb{R} $ be a continuous function. Consider the ordinary differential equation
$$y'(x) = f(x) y(x) ,\quad x> 0 ,\quad y(0) = y_{0} \neq 0.$$
Which of the following statements are true?
$1$. If $\int_{0}^{\infty}|f(x)|dx < \infty$, then $y$ is bounded.
$2$. If $\int_{0}^{\infty}|f(x)|dx < \infty$, then $\lim_{x \to \infty} y(x) $ exists.
$3$. If $\lim_{x \to \infty} f(x) = 1 $, then $lim_{x \to \infty} |y(x)| = \infty$.
$4$. If $\lim_{x \rightarrow \infty} f(x) = 1 $ then $y $ is monotone.
My attempts: I know that options $1$, $2$, and $3$ are correct...but I'm doubting that option $4$ is true.
Please help me; any hints/solutions will be appreciated. Thank you.
Suppose that $f$ is continuous and $y_0>0$ and $y(x)=y_0\cdot e^{g(x)}$ where $g(x)=\int_0^xf(u)du.$ Suppose $f(1)=f(2)=0$ and that $f(x)>0$ for $x\in [0,1)\cup (2,\infty)$ and $f(x)<0$ for $x\in (1,2).$ And suppose that $\lim_{x\to \infty}f(x)=1.$
We have $y(2)-y(1)=$ $y_0\cdot e^{g(1)}\cdot (e^{g(2)-g(1)}-1).$ Now $g(2)-g(1)=\int_1^2f(u)du<0$ so $f(2)<f(1).$
We have $y(1)-y(0)=y_0\cdot e^{g(0)}\cdot (e^{g(1)-g(0)}-1).$ Now $g(1)-g(0)=\int_0^1 f(u)du>0$ so $f(1)>f(0).$
Remark. If $y>0$ then $yf=y'\implies f=y'/y=(\ln y)'$ which implies that $\ln y$ is an anti-derivative of $f$.