Matrix Differential Equations I am working on a practice problem with the following equation:
$$
\frac{d^3 x}{dt^3} + (k + 1)\frac{d^2x}{dt^2} + (k+1)\frac{dx}{dt} + kx = 0
$$
I understand the first part which is to convert to a matrix differential equation $u' = Au$ and then find the eigenvalues of $A$. I don't understand the second part which is to determine for which values of $k$ the system is stable. I would appreciate any help on that. Thanks.
 A: After setting
$$
u=\left(\begin{array}{c}x\\\frac{dx}{dt}\\\frac{d^2x}{dt^2}\end{array}\right),
$$
the DE should become:
$$
u'=Au,
$$
with
$$
A=\left(\begin{array}{ccc}0&1&0\\0&0&1\\-k&-k-1&-k-1\end{array}\right).
$$
The characteristic polynomial of $A$ is:
$$
p_A(\lambda)=\lambda^3+(k+1)\lambda^2+(k+1)\lambda+k
$$
You can check that
$$
(1-\lambda)p_A(\lambda)=(\lambda-1)[\lambda(\lambda^2+\lambda+1)+k(\lambda^2+\lambda+1)]=(\lambda+k)(\lambda^3-1),
$$
and therefore the eigenvalues of $A$ are:
$$
\lambda_1=-k,\,\lambda_2=e^{i\frac{2\pi}{3}},\,\lambda_3=\overline{\lambda_2}.
$$
Assume that the parameter $k\in \mathbb{R}$, the system $u'=Au$ is stable provided $\max\{-k,-\frac12\}\le 0$, i.e. $k\ge 0$.
A: Specifically,
$$A=\begin{pmatrix}0&1&0\\0&0&1\\-k&-k-1&-k-1\end{pmatrix}$$
whose eigenvalues are the roots of $\ p_k(\lambda)=\lambda^3+(k+1)\lambda^2+(k+1)\lambda+k$
The condition of stability is given by $\max(\text{Re}(\lambda_1),\text{Re}(\lambda_2),\text{Re}(\lambda_3))\le0$
Notice that 
$\lambda^3+(k+1)\lambda^2+(k+1)\lambda+k=(\lambda+k)(\lambda^2+\lambda+1)$
whose roots are...
