Check for stability and asymptotic stability of the zero solution for a system

Question

I am trying to show that the zero solution for a system is unstable even though both eigenvalues are negative. This is the system: $$ \vec{y'} = \begin{pmatrix} -1 & e^{2t} \\ 0 & -1 \end{pmatrix} \vec{y}.$$ I know that I have to find the fundamental matrix, but I am stuck. This is what I have: $$|A-\lambda I| = \begin{vmatrix} -1- \lambda & e^{2t} \\ 0 & -1-\lambda \end{vmatrix} = (-1-\lambda)^2,$$ which means that both eigenvalues are equal to -1. The only eigenvector is: $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Is this the fundamental matrix: $$\begin{pmatrix} e^{-t} & e^{-t} \\ 0 &0 \end{pmatrix}$$ Do the zeros in the second imply that the zero solution is unstable, because there would be infinitely many solutions? I want to know if I am at least on the right track; I would really appreciate the help.

Answer 1

Your system is unusual in that the coefficients of the matrix $A$ in the equation $y' = Ay$ are not constant.

I think it is best to solve this "from scratch" as a system of equations. In terms of the separate components of $\vec y$, the system is $$ \frac{dy_1}{dt} = -y_1 + e^{2t}y_2\\ \frac{dy_2}{dt} = -y_2. $$ Solving the second equation yields $y_2 = C_1 e^{-t}$. Plugging this solution into the first system gives us $$ \frac{dy_1}{dt} = -y_1 + C_1 e^t \implies\\ \frac{dy_1}{dt} + y_1 = C_1 e^t \implies\\ e^t\frac{dy_1}{dt} + e^ty_1 = C_1 e^{2t} \implies\\ \frac d{dt}[e^t y_1] = C_1 e^{2t} \implies\\ e^t y_1 = \frac {C_1}{2} e^{2t} + C_2 \implies\\ y_1 = \frac{C_1}{2} e^t + C_2 e^{-t}. $$ In other words, every solution to this differential equation can be written in the form $$ \vec y = \pmatrix{e^{-t} & \frac 12 e^t \\0 & e^{-t}}\pmatrix{C_2\\C_1}. $$