$a\frac{d^{3}x}{dt^{3}}+b\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=0$, 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 $k>0$

$\bf hint:$ the differential equation has a solution of the form $x = e^{\lambda t}$ if 
$$ a \lambda^3 + b \lambda ^2 + c \lambda + k = 0$$ for all solutions to have $\lim_{ t \to\infty} x = 0,$ the real part of $\lambda$ must be negative.
you can't too far without knowing more about the coefficients $a, b, c$ and $k.$
