show that all other solutions are bounded 
Suppose $G(x)$ is a solution of the differential equation $$x'(t)\ =\left(\begin{matrix}-5&2\\-4&1\end{matrix}\right) \ x(t)+ \ f(t)$$
  where $f(t)$ is a continous function and $Sup\{|G(x)|: t_0\le x <\infty\}<\infty$

show that all other solutions are bounded.
 A: This is false as it is written.
Let $f$ be the null function, let $G$ be the null function too (which ends up being a solution to the differential equation).
Now consider the map $t\mapsto e^{-3t}\begin{pmatrix} 1\\1\end{pmatrix}$. It certainly is unbounded near $-\infty$ and it is also a solution to the differential equation since, given $t\in \mathbb R$, it's derivative is $-3e^{-3t}\begin{pmatrix} 1\\1\end{pmatrix}$ and $$e^{-3t}\begin{pmatrix}-5&2\\-4&1\end{pmatrix}\begin{pmatrix} 1\\1\end{pmatrix}=e^{-3t}\begin{pmatrix} -3\\ -3\end{pmatrix}=-3e^{-3t}\begin{pmatrix} 1\\1\end{pmatrix}.$$

If instead you mean to prove that all solutions are bounded on all intervals of the form $[t_0,+\infty[$, then what's below is useful.
Hint: Every solution $y$ is of the form $y_h+G$, where $y_h$ is a solution of the associated homogeneous differential equation. You're given that $G$ is bounded, so it suffices to prove that all $y_h$'s are bounded on $[t_0, +\infty[$, for all $t_0$. To prove this, use Kim Jong Un's comment which states that the matrix $\begin{pmatrix}-5&2\\-4&1\end{pmatrix}$ has two negative eigenvalues.
