For the plane autonomous system $$ x' = ax+by $$ $$ y' = cx+dy $$ If the solution to this system is, say, $ \binom{x}{y}= c_{1}\binom{1}{1}e^{-5t} + c_{2}\binom{1}{2}e^{-t} $, then it is understandable that as $t\rightarrow \infty$, the trajectory will approach the line $y=2x$ as the term $c_{2}\binom{1}{2}e^{-t}$ dominates the solution. As we can see from $$ \binom{x}{y}= c_{1}\binom{1}{1}e^{-5t} + c_{2}\binom{1}{2}e^{-t} = e^{-t}\left[ c_{1}\binom{1}{1}e^{-4t} + c_{2}\binom{1}{2} \right] $$
However, I was just wondering the case when both eigenvalues are positive. Specifically, if the solution to the same system is
$$ \binom{x}{y}= c_{1}\binom{1}{1}e^{5t} + c_{2}\binom{1}{2}e^{t} $$
So my question is :
Why doesn't the trajectory approach the line $y=x$ when $t\rightarrow \infty$, as the term $c_{1}\binom{1}{1}e^{5t}$ dominates the solution? As we may see this from $$ \binom{x}{y}= c_{1}\binom{1}{1}e^{5t} + c_{2}\binom{1}{2}e^{t} = e^{5t}\left[ c_{1}\binom{1}{1} + c_{2}\binom{1}{2}e^{-4t} \right] $$
Changed:
When you take your limit $t\to\infty$ you assume that $$ \lim_{t\to\infty} e^{5t}(c_1+c_2 e^{-{4t}})=c_1e^{5t}, $$ which is obviously not correct.