Consider a 3rd order linear homogeneous DE of the form $$Lu=u'''+a_2(x)u''+a_1(x)u'+a_0(x)u=f(x) \ \ \ \ (1)$$ and for which $u_1=e^{-x}$ and $u_2=e^{-2x}$ are solutions to the homogeneous form of $(1)$.
Let $f(x)=10e^{-2x}$. Give an example of a form of $a_2, a_1$ and $a_0$ such that $(1)$ has a stable equilibrium point and an example such that $(1)$ has no stable equilibrium point.
My attempt:
When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced $(1)$ into a system of linear equations: \begin{align} \frac{du}{dt}&=y, \\ \frac{dy}{dt}&=z, \\ \frac{dz}{dt}&=-a_2z-a_1y-a_0u. \\ \end{align} This gives a corresponding matrix $$A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -a_0 & -a_1 & -a_2 \end{pmatrix}.$$ But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.
In particular, $u_1$ and $u_2$ are homogeneous solutions: \begin{aligned} -1 + a_2 - a_1 + a_0 &= 0 \\ -8 + 4 a_2 - 2 a_1 + a_0 &= 0 \end{aligned} i.e. $a_1 = 2 + \frac{3}{2} a_0$ and $a_2 = 3 + \frac{1}{2} a_0$. Now, it remains to examine the stability of the equilibrium points, which are the constants $u^*$ such that $a_0 u^* = f$. The eigenvalues of $A$ are $-2$, $-1$ and $-\frac{1}{2}a_0$. Therefore, if $a_0=\alpha u_2$ with $\alpha>0$, the system has a single stable equilibrium point $u^* = 10/\alpha$. However, if for instance $a_0 = \beta u_1$, the system has no equilibrium point.