$a\frac{d^{3}x}{dt^{3}}+b\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=0$, on what condition of k that we have $\lim_{t\rightarrow\infty}x(t)=0$ $a\frac{d^{3}x}{dt^{3}}+b\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=0$, $a,b,c$ are constant, and  on what condition of k that we have $\lim_{t\rightarrow\infty}x(t)=0$
I completely have no idea how to solve this problem, can anyone could help me? Thanks very much!
What I did:
If we let $a=b=c=1$, then we have $r^{3}+r^{2}+r+k=0,x(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}+C_{3}e^{r_{3}t}$, since $\lim_{t\rightarrow\infty}x(t)=0$, we get $r_{1}<0,r_{2}<0,r_{3}<0$. Assume that $r^{3}+r^{2}+r+k=(r+r_{1})(r+r_{2})(r+r_{3})$, then expand it we get $r_{1}+r_{2}+r_{3}=1,r_{3}(r_{1}+r_{2})+r_{1}r_{2}=1,r_{1}r_{2}r_{3}=k$, then how can I get the condition on k? I can just get $k>0$

The characteristic polynomial is
$$ ar^3 + br^2 + cr + k = 0 $$
Vieta's formulas give us
$$ \left\{ \begin{matrix}
 r_1 + r_2 + r_3 = -\frac{b}{a} \\ 
r_1r_2 + r_2r_3 + r_1r_3 = \frac{c}{a} \\ 
r_1r_2r_3 = \frac{k}{a} 
\end{matrix} \right. $$
where $r_1, r_2, r_3$ are the 3 roots, which can be either all real or 1 real and 2 complex conjugates. In the case of 2 complex roots, their sum is a real number and their product is a positive real number.
In order for the limit to be $0$, the 3 roots must all have negative real parts, so $\frac{b}{a} > 0$. $\frac{c}{a} > 0$ and $\frac{k}{a} < 0$
All 3 conditions are needed, so you have to know more about $a, b, c$.
