Find the unique solution to the system
\mathbf{x}^{\prime}=\left(\begin{array}{l}x_{1} \\x_{2}\end{array}\right)^{\prime}=\left(\begin{array}{ll}4 & 2 \\3 & 3\end{array}\right)\left(\begin{array}{l}x_{1} \\x_{2}\end{array}\right)+\left(\begin{array}{l}e^{t} \\e^{2 t}\end{array}\right)=A \mathbf{x}+\mathbf{f}(t), \quad \mathbf{x}(0)=\left(\begin{array}{l}1 \\2\end{array}\right).
A fundamental matrix solution for the homogeneous system is
\Phi(t)=\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right)(You should verify this.) Then
\Phi^{-1}(t)=\frac{1}{5}\left(\begin{array}{cc}e^{-t} & -e^{-t} \\3 e^{-6 t} & 2 e^{-6 t}\end{array}\right)and, by equation (12), we have the particular solution
\mathbf{x}^{\prime}=A(t) \mathbf{x}+\mathbf{f}(t), \quad \mathbf{x}\left(t_{0}\right)=\mathbf{x}_{0}\begin{aligned}\varphi_{p} &=\frac{1}{5}\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right) \int_{0}^{t}\left(\begin{array}{cc}e^{-s} & -e^{-s} \\3 e^{-6 s} & 2 e^{-6 s}\end{array}\right)\left(\begin{array}{l}e^{s} \\e^{2 s}\end{array}\right) d s \\&=\frac{1}{5}\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right) \int_{0}^{t}\left(\begin{array}{c}1-e^{s} \\3 e^{-5 s}+2 e^{-4 s}\end{array}\right) d s .\end{aligned}
Since the integral of a vector function is the vector of integrals,
\begin{aligned}\varphi_{p}(t) &=\frac{1}{5}\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right)\left(\begin{array}{c}t-e^{t}+1 \\\frac{11}{10}-\frac{3}{5} e^{-5 t}-\frac{e^{-4 t}}{2}\end{array}\right) \\&=\left(\begin{array}{c}\frac{2}{5} t e^{t}-\frac{1}{2} e^{2 t}+\frac{7}{25} e^{t}+\frac{11}{50} e^{6 t} \\\frac{-3}{5} t e^{t}+\frac{1}{2} e^{2 t}-\frac{18}{25} e^{t}+\frac{11}{50} e^{6 t}\end{array}\right) .\end{aligned}Note that \varphi_{p}(0)=0, which must be the case from the way in which we found \varphi_{p}. Next, from equation (9), we see that the unique solution is
\begin{aligned}\varphi(t) &=\Phi(t) \Phi^{-1}\left(t_{0}\right) \mathbf{x}_{0}+\Phi(t) \int_{t_{0}}^{t} \Phi^{-1}(s) \mathbf{f}(s) d s \\&=\varphi_{h}(t)+\varphi_{p}(t)\end{aligned}\begin{aligned}\varphi(t) &=\Phi(t) \Phi^{-1}(0) x_{0}+\varphi_{p}(t) \\&=\frac{1}{5}\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right)\left(\begin{array}{rr}1 & -1 \\3 & 2\end{array}\right)\left(\begin{array}{l}1 \\2\end{array}\right)+\varphi_{p} \\&=\frac{1}{5}\left(\begin{array}{rr}2 e^{t} & e^{6 t} \\-3 e^{t} & e^{6 t}\end{array}\right)\left(\begin{array}{r}-1 \\7\end{array}\right)+\varphi_{p}\end{aligned}
\begin{aligned}&=\left(\begin{array}{c}-\frac{2}{5} e^{t}+\frac{7}{5} e^{6 t} \\\frac{3}{5} e^{t}+\frac{7}{5} e^{6 t}\end{array}\right)+\left(\begin{array}{c}\frac{2}{5} t e^{t}-\frac{1}{2} e^{2 t}+\frac{7}{25} e^{t}+\frac{11}{50} e^{6 t} \\\frac{-3}{5} t e^{t}+\frac{1}{2} e^{2 t}-\frac{18}{25} e^{t}+\frac{11}{50} e^{6 t}\end{array}\right) \\&=\left(\begin{array}{c}\frac{2}{5} t e^{t}-\frac{1}{2} e^{2 t}-\frac{3}{25} e^{t}+\frac{81}{50} e^{6 t} \\\frac{-3}{5} t e^{t}+\frac{1}{2} e^{2 t}-\frac{3}{25} e^{t}+\frac{81}{50} e^{6 t}\end{array}\right)\end{aligned}
Note, as a check, that \varphi(0)=\left(\begin{array}{l}1 \\ 2\end{array}\right).