Why solutions of 2nd order linear equations with constant coefficients goes to infinity, but solutions of 1st order only go asymptotic to 0? What I'm talking about is why does the solution of 2nd order $$y=k^2 y''$$ tends to blow up at both $x \rightarrow + \infty$ and  $x \rightarrow - \infty$; but the solution of first order $y=k^2 y'$ only blows up at $x \rightarrow + \infty$?
Also, why $y=-k^2 y''$ oscillates, but $y=-k^2 y'$ just go to $0$? I know the math checks out, but what could be the mathematical/analytical intuition here? What exactly makes the 2nd order differentiation so much stronger? I'm a Physics student and I understand them in a very physical way, which I believe is just reductive reasoning. 
Thanks a lot if someone could enlighten me on this! :D

What you say is not exactly true: $y=e^{-x}$ is a solution of $y=y''$ and goes to $0$ as $x\to+\infty$.
As for the second question, an equation of the form $y''= a\,y$ represents a mechanical system in which $y''$ is the acceleration, and $a\,y$ an force that is proportional to the displacement. If $a>0$ then the force points in the same direction as the acceleration, the velocity is always increasing and everything goes to $\infty$. However, if $a<0$, the direction of the force is the oposite to the direction of the displacement. The further away you go, the strongest is the force against you. This crates oscillation, as in a pendulum or a mass attached to a spring.
