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Find the particular solution which vanishes at $t=0\,$ and identify the Green's matrix $G_0(t,s)\,$ of the system $\begin{cases}x_1'=x_2+g_1(t) \\ x_2'=-x_1+g_2(t)\end{cases}\,$.

The homogeneous equations can be written $\vec x' = A(x)\vec x\,$ where

$A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\,$

Find the eigenvalues $\lambda = \pm i\,$.

The eigenvector for $\lambda_1=i\,$ is $\vec{u_1} = \begin{bmatrix}1\\i\end{bmatrix}\,$.

The real and imaginary parts are $\vec v = \begin{bmatrix}1\\0\end{bmatrix}, \vec w=\begin{bmatrix}0\\1\end{bmatrix}\,$.

So if $\lambda = \alpha+i\beta\,$ then

$e^{\lambda t} = e^{\alpha t} (\cos \beta t + i \sin \beta t)\,$

$\vec x = e^{\lambda t} \vec{u_1}\,$

$\vec x = e^0(\cos t + i\sin t)\begin{bmatrix}1\\i\end{bmatrix}\,$

$\vec x = \begin{bmatrix} \cos t + i \sin t \\ -\sin t + i\cos t \end{bmatrix}\,$

So two linearly independent solutions are the real and imaginary parts of $\vec x\,$.

$x_1 = \begin{bmatrix}\cos t\\-\sin t\end{bmatrix}, x_2 = \begin{bmatrix}\sin t\\\cos t\end{bmatrix}\,$

So the fundamental matrix is

$X(t) = \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}\,$

Now the particular solution is

$x_p = X(t)\int_0^t X^{-1}(s)g(s)ds\,$

Define Green's matrix

$G_0(t,s) = \begin{cases} 0 & t\le s\le b \\ X(t)X^{-1}(s) & a\le s < t \end{cases} \,$

$= \begin{bmatrix}\cos t\cos s + \sin t\sin s & -\cos t\sin s + \sin t \cos s \\ -\sin t\cos s + \cos t\sin s & \sin t\sin s + \cos t \cos s\end{bmatrix}\,$

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