Question

Determine the stability of the system

$\frac{\mathbf{d}\overrightarrow{x}}{\mathrm{dt}}\mathbf{=}\left[\begin{array}{cc}-1& 2\\ 3& -4\end{array}\right]\overrightarrow{x}$

Short answer

Thus, the solution is unstable equilibrium.

Step-by-step solution

Given in the question. 

The given system is:

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}-1& 2\\ 3& -4\end{array}\right]\overrightarrow{x}$

Step 2:Solve for Eigen values. 

The given determinant is solved as follows to obtain Eigen values.

$\begin{array}{l}\det (\left[\begin{array}{cc}-1& 2\\ 3& -4\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right])=0\\ \det \left(\left[\begin{array}{cc}-1-\lambda & 2\\ 3& -4-\lambda \end{array}\right]\right)=0\\ {\lambda }^2+5\lambda -2=0\\ \lambda =\frac{-5\pm \sqrt{25-4\left(1\right)(-2)}}{2}\\ \lambda =\frac{-5\pm \sqrt{33}}{2}\end{array}$

Thus, there are two real roots that is one positive and one negative. But both are not negative so, the solution is unstable equilibrium.

Hence, the solution is unstable equilibrium.