Answer
$x_1(t)=-2C_1e^{-t}+C_2e^{5t}\\
x_2(t)=C_1e^{-t}+C_2e^{5t}$
Work Step by Step
The given system can be written as $x′ = Ax$,
where $A=\begin{bmatrix}
1 & 4\\
2 & 3
\end{bmatrix}$
The transformed system is $y′ = (S−1AS)y$ where $x = Sy$
To determine S, we need the eigenvalues and eigenvectors of A. The characteristic polynomial of A is
$p(λ) = \begin{bmatrix}
1-\lambda & 4\\
2 & 3 -\lambda
\end{bmatrix}=(1-\lambda)(3-\lambda)-8=(\lambda+1)(\lambda-5)$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=-1: v=r(-2,1)\\
\lambda_2=5: v=s(1,1)$
Set $S = \begin{bmatrix}
-2 & 1\\
1 & 1
\end{bmatrix}$
then from Theorem 7.3.4, $S−1AS = diag(-1, 5)$
so that the system becomes:
$\begin{bmatrix}
y_1'\\
y_2'
\end{bmatrix}=\begin{bmatrix}
-1 & 0\\
0 & 5
\end{bmatrix}\begin{bmatrix}
y_1\\
y_2
\end{bmatrix}$
Hence, $y_1'=-1y_1\\
y_2'=5y_2$
Both of these equations can be integrated to obtain $y_1(t)=C_1e^{-t}\\
y_2(t)C_2e^{5t}$
Return to the original variables, we have:
$x=Sy=\begin{bmatrix}
-2 & 1\\
1 & 1
\end{bmatrix}\begin{bmatrix}
C_1e^{-t}\\
C_2e^{5t}
\end{bmatrix} \\
\rightarrow x_1(t)=-2C_1e^{-t}+C_2e^{5t}\\
x_2(t)=C_1e^{-t}+C_2e^{5t}$
