Solve the following linear system:
$$\displaystyle \left \{ \begin{array}{rcl} x_1' & = & -x_1' \\ x_2' & = & x_1 \cdot \sin t-x_2\end{array} \right.$$
Development:
Let's write the matrix of the system $$$A(t)=\begin{pmatrix} -1 & 0 \\ \sin(t) & -1 \end{pmatrix}$$$ This is a matrix with non constant coefficients but triangular.
Therefore let's consider the first equation $$$x_1'=-x_1$$$ This is a linear ODE; in fact, it is separable. Then: $$$x_1'=-x_1 \Rightarrow \dfrac{dx_1}{dt}=-x_1 \Rightarrow \dfrac{dx_1}{x_1}=-dt \Rightarrow \int\dfrac{dx_1}{x_1}=\int-dt \Rightarrow$$$ $$$\Rightarrow \ln|x_1(t)|=-t+C, \ C\in\mathbb{R} \Rightarrow |x_1(t)|=k_1\cdot e^{-t}, \ k_1 \geq 0 \Rightarrow$$$ $$$\Rightarrow x_1(t)=k_1\cdot e^{-t}, \ k_1\in\mathbb{R}$$$ We insert this solution in the second equation: $$$x_2'=(\sin(t))\cdot x_1-x_2=-x_2+k_1\cdot\sin(t)\cdot e^{-t}$$$ this is a non homogeneous linear ODE.
Let's solve the homogeneous part, (which is the same equation as the previous one): $$$x_{2h}'=-x_{2h} \Rightarrow x_{2h}(t)=k_2\cdot e^{-t}, \ k_2\in\mathbb{R}$$$ Let's look for a particular solution of the form $$x_{2p}=u(t)\cdot e^{-t}$$.
Let's designate as a solution: $$$\left. \begin {array} {l} x_{2p}'=u'\cdot e^{-t}-u\cdot e^{-t} \\ x_{2p}'=k_1\sin(t)\cdot e^{-t}-u\cdot e^{-t} \end{array}\right\} \Rightarrow u'(t)=k_1\sin(t)$$$ Therefore, $$$u(t)=k_1\cdot\int\sin(t)\cdot dt=k_1(-\cos(t))=-k_1\cdot\cos(t) $$$ Finally $$$x_2(t)=x_{2h}(t)+x_{2p}(t)=k_2\cdot e^{-t}-k_1\cdot e^{-t}\cdot \cos(t)=e^{-t}(k_2-k_1\cdot\cos(t))$$$
Solution:
$$\begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}=\begin{pmatrix} k_1\cdot e^{-t} \\ e^{-t}(k_2-k_1\cdot\cos(t)) \end{pmatrix}, \ \ k_1,k_2\in\mathbb{R}$$