Question is to find the fundamental matrix $F(t)$ satisfying $F(0)=I$ for the given system of equation below.
$$
x'
=\left(\begin{array}{rr}2 & 3 \\ -1 & -2\end{array}\right)x
$$
My solution is:
det($\begin{bmatrix}
(2-r) & 3\\
-1 & -(2-r)
\end{bmatrix}) = r^2 -1$
$r_1=1$ and $r_2=-1$.
Eigenvector of $r_1$ is $\begin{bmatrix}
-3\\
1
\end{bmatrix}$ and eigenvector of $r_2$ is $\begin{bmatrix}
-1\\
1
\end{bmatrix}$. Therefore
$x(t) = c_1\begin{bmatrix}
-3\\
1
\end{bmatrix}e^t$ + $c_2\begin{bmatrix}
-1\\
1
\end{bmatrix}e^{-t}$
I got stuck there, what should I do?
Thanks.
Best Answer
First, recall that a fundamental matrix is one whose columns correspond to linearly independent solutions to the differential equation. Then, in our case, we have $$ \psi(t) = \pmatrix{-3e^t & -e^{-t} \\ e^t & e^{-t}} $$ To find a fundamental matrix $F(t)$ such that $F(0) = I$, we simply taking the product $$ F(t) = \psi(t)\psi^{-1}(0) = \pmatrix{-3e^t & -e^{-t} \\ e^t & e^{-t}}\pmatrix{-3 & -1 \\ 1 & 1}^{-1} = \frac{1}{2}\pmatrix{3e^t-e^{-t} & 3e^t-3e^{-t} \\ -e^t+e^{-t} & -e^t+3e^{-t}} $$ And we are ensured that $F(0) = \psi(0)\psi^{-1}(0) = I$.