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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.

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1 Answer

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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 $F(0) = \psi(0)\psi^{-1}(0) = I$.

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