Consider the nonhomogeneous system for $\mathbf{x}(t)$: $$ \dot{\mathbf{x}} = \left[\begin{array}{cc} 1 & 1\\ 0 & -1 \end{array}\right] \mathbf{x} + \left[\begin{array}{c} t\\ \sin t \end{array}\right] . $$ The goal is to find the general solution for the non homogeneous linear system.
I've found general solution of the corresponding homogeneous system (although unsure) but it was $$ \mathbf{x}_h = c_1 e^{-t} \left[\begin{array}{c} 1\\ 2 \end{array}\right] + c_2 t e^{-t} \left[\begin{array}{c} 1\\ 2 \end{array}\right] $$ because my eigenvalue was $-1$ and the eigenvector I got was $[1,2]^\top$.
I'm onto finding particular solution which I guessed (could be wrong) to be in the form: $$ \mathbf{x}_p = a \cos(t) + b \sin(t) + c $$ and \begin{aligned} &LHS= \mathbf{x}_p' = -a \sin(t) + b \cos (t) \\ &RHS = A \mathbf{x}_p+ g = A a \cos(t) + A b \sin(t) + Ac + \left[\begin{array}{c} t\\ \sin t \end{array}\right] \end{aligned}
Now I'm finding trouble matching the coefficients as I can't separate the $t $ out of my $g(t)$ which was $[t, \sin(t)]^\top$.
I'm stuck there. Thanks.
I'm not too sure about the nonhomogeneous part, but that matrix should have two distinct, real eigenvalues, not just one, namely $1$ and $-1$. Further, the eigenvectors should be $ \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] $ and $ \left[ \begin{matrix} 1 \\ -2 \end{matrix} \right] $ , respectively, of course allowing for multiplication by an arbitrary constant. So perhaps double check your homogeneous solution.