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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


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\leq s\leq b\\X(t)X^{{-1}}(s)&a\leq 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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