Let $$ \begin{cases} x'(t)=2x(t)-y(t), & \\ y'(t)=-x(t)+2y(t), & \\ \end{cases}$$
What are the solutions such that $\exists \lim_{t\to\infty} \textbf{x}(t)$?
I found the general solution $x(t)=c_1e^tv_1+c_2e^{3t}v_2$
Where $v_1=\pmatrix{1\\1} $ and $v_2=\pmatrix{1\\-1} $
The answer to my question should be: If $c_2 = 0$ the limit exists and $\lim_{t\to\infty} \textbf{x}(t) =\pmatrix{1\\1}$, otherwise it does not exist. I do not get why the limit exist iff $c_2 = 0$ and also why $c_2 = 0$ implies $\lim_{t\to\infty} \textbf{x}(t) =\pmatrix{1\\1}$. Could you help me with this, please?
Edit: If we were in the discrete case, instead, shouldn't the limit be $\lim_{t\to\infty} \textbf{x}(t) =\pmatrix{c_1\\c_1}$?
With the exponential functions given in $$x(t)=c_1e^tv_1+c_2e^{3t}v_2$$ there is no initial values which results in a bounded solution.
More than likely there is a typo in the question which makes both eigenvalues positive.
As you know we need non positive eigenvalues for stability of the system.