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If F(t) is continuous and Φ is a fundamental matrix of X ˙ =AX, then the inhomoge X ˙ =AX+F has solution X p ​ (t)=Φ(t)∫ t 0 ​ t ​ Φ −1 ( s)F(s)ds Use the result above to find a general solution of X ˙ =( 0 −1 ​ 1 0 ​ )X+( cost −sint ​ ) at t 0 ​ =0.

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The general solution of the system $\dot{X} = (0 -1 \ 1 0)X + (\cos t - \sin t)$ at $t_0 = 0$ is $X_p(t) = \begin{pmatrix} e^t \\ e^{-t} \end{pmatrix} \int_0^t e^{-s} (\cos s - \sin s) \, ds$.

The first step is to find the fundamental matrix $\Phi(t)$ of the system $\dot{X} = (0 -1 \ 1 0)X$. This can be done by finding the eigenvalues and eigenvectors of the matrix $A = (0 -1 \ 1 0)$.

The eigenvalues of $A$ are $0$ and $1$, and the corresponding eigenvectors are $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$. Therefore, the fundamental matrix is \Phi(t) = \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix}

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Now, we can use the result in the problem statement to find the solution of the inhomogeneous system $\dot{X} = (0 -1 \ 1 0)X + (\cos t - \sin t)$. The solution is X_p(t) = \Phi(t) \int_0^t \Phi^{-1}(s) (\cos s - \sin s) \, ds

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Plugging in the expression for $\Phi(t)$, we get

X_p(t) = \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} \int_0^t \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (\cos s - \sin s) \, ds

Evaluating the integral, we get

X_p(t) = \begin{pmatrix} e^t \\ e^{-t} \end{pmatrix} \int_0^t e^{-s} (\cos s - \sin s) \, ds

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This is the general solution of the system $\dot{X} = (0 -1 \ 1 0)X + (\cos t - \sin t)$ at $t_0 = 0$.

The fundamental matrix $\Phi(t)$ is the matrix that satisfies the following differential equation: \dot{\Phi}(t) = A \Phi(t)

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with the initial condition $\Phi(0) = I$. The solution of the inhomogeneous system $\dot{X} = AX + F$ can be written as X_p(t) = \Phi(t) \int_0^t \Phi^{-1}(s) F(s) \, ds, where $F(t)$ is the inhomogeneous term.

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