Answer
$x(t)=\begin{bmatrix}
e^{5t}+2e^{t}\\
-e^{5t}+e^{t}
\end{bmatrix}$
Work Step by Step
Given: $\begin{bmatrix}
-1 & 4\\
2 & -3
\end{bmatrix}$
A straightforward calculation yields
$\det (A-\lambda I)=\begin{vmatrix}
-1-\lambda & 4\\
2 & -3-\lambda
\end{vmatrix}=(\lambda+5)(\lambda-1)$
so that A has eigenvalues $\lambda_1=-5\\
\lambda_2=1$
Eigenvalue $λ_1 =-5$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=-v_2\\
v_2=1\\
\rightarrow v_1=-1$
The solution is $v = r(-1,1)$. Therefore,
$v_1= e^{-5t}\begin{bmatrix}
-1\\
1
\end{bmatrix}$
Eigenvalue $\lambda_2 =1$: In this case, the system $(A − \lambda I)v = 0$ is $v_1=2v_2\\
v_2=1\\
\rightarrow v_1=2$
The solution is $v = r(2,1)$. Therefore,
$v_2=e^t\begin{bmatrix}
2\\
1\\
\end{bmatrix}$
Hence, the general solution to the given equation is
$x(t)=c_1v_1+c_2v_2=c_1e^{-5t}\begin{bmatrix}
-1\\
1
\end{bmatrix}+c_2e^t\begin{bmatrix}
2\\
1
\end{bmatrix}=\begin{bmatrix}
-c_1e^{-5t}+2c_2e^{t}\\
c_1e^{-5t}+c_2e^{t}
\end{bmatrix}$
Since \begin{bmatrix}
3\\
0
\end{bmatrix}
we obtain: $c_1+c_2=0\\
-c_1+2c_2=3$
then $c_1=-1\\
c_2=1$
Hence,
$x(t)=\begin{bmatrix}
e^{5t}+2e^{t}\\
-e^{5t}+e^{t}
\end{bmatrix}$
