Explain why the IVP : $ x'(t)=\begin{bmatrix} -\cos t & 1 \\ 1 & -\sec t \end{bmatrix} x(t)+ \begin{bmatrix} 0 \\ -\sin t \end{bmatrix}; \ \ x(0)=\begin{bmatrix} 0 \\ 1 \end{bmatrix} $ has a unique solution of the form $ \ \begin{bmatrix} x \\ y \end{bmatrix}=X_h+\begin{bmatrix} 1 \\ \cos t \end{bmatrix} \ $ ,
where $ \ X_h \ $ is a solution of the homogeneous system $ \ x'(t) =\begin{bmatrix} -\cos t & 1 \\ 1 & -\sec t \end{bmatrix} x(t) \ $
Answer:
The given equation is inhomogeneous, thus the solution is of the form
$ X=X_h+X_{PI} \ $ , where $ \ X_{PI} \ $ is for the non-homogeneous part.
But I am unable to explain why the given solution is valid and unique.
Help me out