Problems from Triangular systems of differential equations

Development:

Let's write the matrix of the system $$A(t)=\left(\begin{array}{cc}-1& 0\\ \sin (t)& -1\end{array}\right)$$ This is a matrix with non constant coefficients but triangular.

Therefore let's consider the first equation $$x_1^{\prime }=-x_1$$ This is a linear ODE; in fact, it is separable. Then: $$x_1^{\prime }=-x_1\Rightarrow {\displaystyle \frac{d{x}_1}{dt}}=-x_1\Rightarrow {\displaystyle \frac{d{x}_1}{x_1}}=-dt\Rightarrow \int {\displaystyle \frac{d{x}_1}{x_1}}=\int -dt\Rightarrow$$ $$\Rightarrow \ln |x_1(t)|=-t+C,C\in \mathbb{R}\Rightarrow |x_1(t)|=k_1\cdot e^{-t},k_1\ge 0\Rightarrow$$ $$\Rightarrow x_1(t)=k_1\cdot e^{-t},k_1\in \mathbb{R}$$ We insert this solution in the second equation: $$x_2^{\prime }=(\sin (t))\cdot x_1-x_2=-x_2+k_1\cdot \sin (t)\cdot e^{-t}$$ this is a non homogeneous linear ODE.

Let's solve the homogeneous part, (which is the same equation as the previous one): $$x_{2h}^{\prime }=-x_{2h}\Rightarrow x_{2h}(t)=k_2\cdot e^{-t},k_2\in \mathbb{R}$$ Let's look for a particular solution of the form $x_{2p}=u(t)\cdot e^{-t}$.

Let's designate as a solution: $$\begin{array}{l}{x}_{2p}^{\prime }=u^{\prime }\cdot e^{-t}-u\cdot e^{-t}\\ x_{2p}^{\prime }=k_1\sin (t)\cdot e^{-t}-u\cdot e^{-t}\end{array}\}\Rightarrow u^{\prime }(t)=k_1\sin (t)$$ 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:

$\left(\begin{array}{c}{x}_1(t)\\ x_2(t)\end{array}\right)=\left(\begin{array}{c}{k}_1\cdot e^{-t}\\ e^{-t}(k_2-k_1\cdot \cos (t))\end{array}\right),k_1,k_2\in \mathbb{R}$

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