Using theorem 4.2 and problem 1 determine the stability of the following linear system
$z'_1 =-7z_1+10t^2(1+t^2)^{-1}z_2$
$z'_2=-(4+t^{-1})z_1+5z_2$
Theorem 4.2: Suppose a constant matrix $A_0$ has eigenvalues $\lambda_j$ such that Re$\lambda_j$<0 for all eigenvalues. Let $A(t)=A_0+B(t)$ where B(t) is continuous for all t and $|B(t)| \rightarrow 0$ as $t \rightarrow \infty$. Then the zero solution is uniformly and asymptotically stable.
So in the previous problem we used the matrix $\begin{bmatrix}-7 & 10 \\ -4 & 5 \end{bmatrix}$ which has complex eigenvalues whose reals are negative (and I showed that it was asymptotically and uniformly stable). I am able to separate the matrix into
$A_0+B(t)=\begin{bmatrix} - 7& ? \\ -4 &5 \end{bmatrix}+\begin{bmatrix}0 & ? \\ -\frac{1}{t} & 0 \end{bmatrix} $
But I cannot split that 10 in the proper way. I assume it should split with $A_0$ as the previous problem, since the solution says it is still both kinds of stable, so is that ? split a misprint or am I missing something? Thanks.
$$\frac{10t^2}{1+t^2}z_{2}=\frac{10t^2}{1+t^2}z_{2} + \frac{10}{1+t^2}z_{2} - \frac{10}{1+t^2}z_{2} = 10z_{2} - \frac{10}{1+t^2}z_{2}$$
So you get:
$${A_0} + B\left( t \right) = \left[ {\begin{array}{*{20}{c}}{ - 7}&{10}\\{ - 4}&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0&{ - \frac{{10}}{{1 + {t^2}}}}\\{ - \frac{1}{t}}&0\end{array}} \right].$$