First we start with the homogeneous system. The coefficient matrix is
\begin{equation}
A={
\left[\begin{array}{ccc}
1 & 1 \\
-1 & 1
\end{array}
\right ]},
\end{equation}
The eigenvalues are
\begin{equation}
\lambda_1=1+i
\end{equation}
\begin{equation}
\lambda_2=1-i
\end{equation}
The eigenvector corresponding to $\lambda_1$ is
\begin{equation}
\xi^(1)=(1,i)^{T}
\end{equation}
Then we can have the solution to the homogeneous system
\begin{equation}
(x,y)^T=c_1\cdot (e^t\cos(t), -e^t\sin(t))^T+c_2\cdot(e^t\sin(t), e^t\cos(t))^T
\end{equation}
We can get a fundamental matrix from here
\begin{equation}
\Psi={
\left[\begin{array}{ccc}
e^t\cos(t) & -e^t\sin(t) \\
e^t\sin(t) & e^t\cos(t)
\end{array}
\right ]},
\end{equation}
Note that we have
\begin{equation}
g(t)=(\frac{e^t}{\cos(t)},\frac{e^t}{\sin(t)})^T
\end{equation}
And suppose
\begin{equation}
\Psi \mu^\prime=g(t)
\end{equation}
We shoud have for the nonhomogeneous system
\begin{equation}
(x,y)^T=\Psi \mu
\end{equation}
This is the method of Variation of Parameters. By plugging in and integration, we have
\begin{equation}
\mu_1(t)=c_3
\end{equation}
\begin{equation}
\mu_2(t)=-\ln|cot(2t)+csc(2t)|+c_4
\end{equation}
Finally we get the required solution according to (6), (9), (10) and (11)
