Let $a:[0,\infty) \to \mathbb R$ is continuous function.Consider the ODE $y'(x)=a(x)y(x),x>0, y(0)=y_0 \ne0.$ Then which of the following are true? 
Assume that $a:[0,\infty) \to \mathbb R$ is a continuous function. Consider the ordinary differential equation, $y'(x)=a(x)y(x),x>0, y(0)=y_0 \ne0.$ Then which of the following statement(s) are true ?
$1.$ If $\int_0^ \infty |a(x)|dx< \infty$, then $y$ is bounded.
$2.$ If $\int_0^ \infty |a(x)|dx< \infty$, then $\lim_{x \to \infty} y(x)$ exists.
$3.$ If $\lim_{x \to \infty}a(x)=1$, then $\lim_{x \to \infty} y(x)= \infty$
$4.$ If $\lim_{x \to \infty}a(x)=1$, then $y$ is monotonic.

I am not getting any clue how to approach this problem. Can anyone solve it.
 A: Given ordinary differential equation is $$y'(x)=a(x)y(x)\implies \dfrac{dy(x)}{y(x)}=a(x)dx$$Integrating between the limit $x_0$ to $x$, we have
$$\int_{x_0}^x\dfrac{dy(x)}{y(x)}=\int_{x_0}^xa(x)dx\implies y(x)=y(x_0)\exp\left(\int_{x_0}^xa(x)dx\right)$$Putting $~x_0=0~,$ we have $$y(x)=y_0\exp\left(\int_{0}^xa(x)dx\right)\tag1$$
For option $\bf{(1)}$ and  $\bf{(2)}$ :
If $~\int_{0}^x|a(x)|dx<\infty~,$ then $~y(x)<\infty\implies y~$ is bounded.$\implies \lim_{x\to\infty}y(x)~$ exists.Hence both options $(1)$ and $(2)$ are correct.
For option $\bf{(3)}$ : As $~\lim_{x\to\infty}a(x)=1\implies\exists~ k\in\mathbb R^+~$ such that $~a(x)>1/2~\forall~x>k~,$
$$\implies \int_k^\infty a(x)dx>\int_k^1 1/2 ~dx\implies \int_k^\infty a(x)dx\to\infty$$Now $$\int^\infty_0a(x)dx=\int_0^ka(x)dx+\int_k^\infty a(x)dx$$
As $~a(x)~$ is continuous on compact set $~[0,k]~,$ the  it is also bounded i.e.,
$~\int_0^k a(x) dx\le m\in\mathbb R~.$ So
$$\int^\infty_0 a(x)dx\to\infty.$$Therefore from $(1)$,  $$\lim_{x\to\infty} |y(x)|=y_0\exp\left(\int_{0}^xa(x)dx\right)\to\infty~.$$Hence option $(3)$ is also correct.
For option $\bf{(4)}$ : Take $$a(x)=1-\dfrac{8x}{1+4x^2}\implies \lim_{x\to\infty}a(x)=1$$
$$\implies y(x)=y_0\exp\left[\int_{0}^x\left(1-\dfrac{8x}{1+4x^2}\right)dx\right]$$
$$=y_0\exp\left[x-\ln(1+4x^2)\right]=y_0\dfrac{e^x}{1+4x^2}$$which is not monotonic as the function has both increasing and decreasing portions on $[0,\infty)$.
Hence option $(4)$ is also incorrect.
A: So right from the start, for $\lim_{x\to\infty}y(x)$ to be finite, $\lim_{x\to\infty}y'(x)$ should tend to $0$, i.e. we should have a horizontal asymptote as x tends to infinity. If $\lim_{x\to\infty}a(x)$ is $1$, then $\lim_{x\to\infty}y'(x)$ cannot be $0$. Therefore, option (c) is correct.
Nothing conclusive can be said about the monotonicity of the function $y(x)$ by the assertion that $\lim_{x\to\infty}a(x)$ is $1$, therefore, option (d) is incorrect.
Now, if we go ahead and solve the given differential equation, the solution obtained is:
$$y(x) = y_o \exp \left( \int_0^x a(t)dt\right)$$
If the integral is finite when $x \to \infty$, then $\lim_{x\to\infty}y(x)$ is a finite value that exists, therefore, (b) is correct
Since $a(x)$ is a continuous function, $y(x)$ is also a continuous function as the integral of a continuous function is continuous. If $\int_0^{\infty} a(t)dt$ tends to a finite value, then $y(x)$ is bounded. Therefore, option (a) is correct.
TL;DR: (a), (b) and (c) are correct answers
