Problems from ODE's homogeneous systems with constant coefficients

Development:

We observe that we are dealing with a linear systems with constant coefficients, where $$A=\left(\begin{array}{ccc}-1& -1& 1\\ 0& -1& -1\\ 0& 1& -3\end{array}\right)$$ The eigenvalues of this matrix are:

$$\begin{gathered} {\lambda }_1=-1 \\ {\lambda }_2={\lambda }_3=-2 \end{gathered}$$

and a basis of eigenvectors is: $$v_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),v_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),v_3=\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)$$ Notice that this matrix does not diagonalize. Therefore we already calculated: $$J=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -2& 0\\ 0& 1& -2\end{array}\right);S=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)$$ We know that the fundamental matrix of the system is: $$\varphi (t)=S\cdot e^{tJ}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{ccc}{e}^{-t}& 0& 0\\ 0& e^{-2t}& 0\\ 0& t\cdot e^{-2t}& e^{-2t}\end{array}\right)=$$ $$=\left(\begin{array}{ccc}{e}^{-t}& e^{-2t}& 0\\ 0& e^{-2t}+t\cdot e^{-2t}& e^{-2t}\\ 0& t\cdot e^{-2t}& e^{-2t}\end{array}\right)=$$ $$=\left(\begin{array}{ccc}{e}^{-t}& e^{-2t}& 0\\ 0& (t+1)e^{-2t}& e^{-2t}\\ 0& t\cdot e^{-2t}& e^{-2t}\end{array}\right)$$

Multiplying by a vector of constants, we obtain our solution.

Solution:

$$\left(\begin{array}{c}x(t)\\ y(t)\\ z(t)\end{array}\right)=\left(\begin{array}{ccc}{e}^{-t}& e^{-2t}& 0\\ 0& (t+1)e^{-2t}& e^{-2t}\\ 0& t\cdot e^{-2t}& e^{-2t}\end{array}\right)\cdot \left(\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right)$$

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