Question

Solve the system with initial value condition:

Answer 1

where at you stuck?

Answer 2

characteristic equation give you 2 roots \(\lambda = 0 \) and \(\lambda =-6\)

Answer 3

\(\lambda =0\) gives out the eigenvector \(\left[\begin{matrix}1\\1\end{matrix}\right]\)
so the solution \(X_1 (t) = \left[\begin{matrix}1\\1\end{matrix}\right]\)
\(\lambda =-6\) gives out the eigenvector \(\left[\begin{matrix}4\\1\end{matrix}\right]\)
so the solution \(X_2 (t) = e^{-6t} \left[\begin{matrix}4\\1\end{matrix}\right]\)
therefore, general solution is
\(X(t) = C_1\left[\begin{matrix}1\\1\end{matrix}\right] + C_2 e^{-6t} \left[\begin{matrix}4\\1\end{matrix}\right]\) = \(\left[\begin{matrix}C_1 + 4 C_2 e^{-6t}\\C_1 + C_2e^{-6t}\end{matrix}\right]\)
\(X(0) = \left[\begin{matrix}C_1 + 4 C_2 e^{-6*0}\\C_1 + C_2e^{-6*0}\end{matrix}\right]\) = \(\left[\begin{matrix}19\\4\end{matrix}\right]\)

Answer 4

that gives you \(C_1 = -1\) and \(C_2 = 5\)
you know how to replace, right?