Two trajectories $x_1(t)$ and $x_2(t)$ of the six dimensional system $\dot{x}=A_{6x6}x$ are given by $$x_1(t)=t^2e^{-2t}v_1+te^{-2t}v_2+e^{-2t}v_3+v_4 ,\> \> x_2(t)=v_5+e^{-t}v_6 \> \> \> (t\in \mathbb{R})$$ where $v_1,...,v_6$ are constant, nonzero vectors in $\mathbb{R^6}$ and $\{v_4,v_5\}$ is a linearly independent set. $(i)$ Prove that $v_i,i=1,...6$ are generalized eigenvectors of $A$ and identify the corresponding eigenvalues. $(ii)$ Prove that $\{v_1,...,v_6\}$ is a basis of $\mathbb{R^6}$. $(iii)$ What can you conclude about the limit as $t \rightarrow \infty$ of general trajectories of the system $\dot{x}=A_{6x6}x$?
Solution so far
$x_1(t)=t^2e^{-2t}v_1+te^{-2t}v_2+e^{-2t}v_3+v_4$; $\dot{x_1}=Ax_1(t)$. Calculate both sides to obtain $$t^2e^{-2t}(Av_1+2v_1)+te^{-2t}(Av_2+2v_2-2v_1+e^{-2t}(Av_3+2v_3-v_2)+Av_4=0 \> \> \forall t \in \mathbb{R}$$ Setting $t=0$,$Av_4=0 \implies v_4\in Null(A)$ is the eigenvector corresponding to eigenvalue 0. So now we get $t^2e^{-2t}(A+2)v_1+te^{-2t}((A+2)v_2-2v_1)+e^{-2t}((A+2)v_3-v_2)=0$. I need help showing that this implies $(A+2)v_1=0$,$((A+2)v_2-2v_1)=0$ $((A+2)v_3-v_2)=0$ I also need help showing that it follows that $v_1$ is an eigenvector corresponding to to eigenvalue -2 and $v_2$ $v_3$ are gegenralized eigenvector for -2.