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The linear system:

$y”(t)+4y'(t)=4(\lambda -1)y(t)+z(t)$

$z'(t)=(\lambda -3)z(t)$

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Determine the stability of the system as a function of the parameter

$\lambda\in\mathbb{R}$.

How do i get started on this exercise? I am not asking for the solution, just anything that can help me on my way?

Should i rewrite the system as a system of first order equations? If so, how can i do this?

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Note that, the analysis of stability of a linear system is based on the steady state of the system. So, the question is : whether the steady state stable or unstable. First, you need to determine the steady state, then do the rest of the analysis. See here.

The system can be written as:

$f(x_{1},x_{2},x_{3})=x’_{1}=x_{2}$

$g(x_{1},x_{2},x_{3})=x’_{2}=-4\cdot (\lambda -1)\cdot x_{1}-4\cdot x_{2}+x_{3}$

$h(x_{1},x_{2},x_{3})=x’_{3}=(\lambda-3)\cdot x_{3}$

The critical point of this system is (0,0,0).

And the jacobian is $\left(\begin{matrix}\frac{\partial f}{\partial x_{1}} & \frac{\partial f}{\partial x_{2}} & \frac{\partial f}{\partial x_{3}}\\ \frac{\partial g}{\partial x_{1}} & \frac{\partial g}{\partial x_{2}} & \frac{\partial g}{\partial x_{3}}\\ \frac{\partial h}{\partial x_{1}} & \frac{\partial h}{\partial x_{2}} & \frac{\partial h}{\partial x_{3}}\\\end{matrix}\right)=\left(\begin{matrix}0 & 1 & 0\\ -4(\lambda -1) & -4 & 1\\ 0 & 0 & (\lambda -3)\end{matrix}\right)$

Then by checking the determinant and the trace of the jacobian i could determine the stability of the system as a function of the parameter $\lambda$.

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