Question

Question: In problem and use a CAS OR Linear algebra software as an aid in finding the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccccc}\mathbf{1}& \mathbf{0}& \mathbf{2}& \mathbf{-}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{5}\mathbf{.}\mathbf{1}& \mathbf{0}& \mathbf{-}\mathbf{1}& \mathbf{3}\\ \mathbf{2}& \mathbf{2}& \mathbf{-}\mathbf{3}& \mathbf{0}& \mathbf{0}\\ \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{8}& \mathbf{1}& \mathbf{-}\mathbf{3}\mathbf{.}\mathbf{1}& \mathbf{4}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{1}\mathbf{.}\mathbf{5}& \mathbf{1}\end{array}\mathbf{\right)}\mathit{X}$

Short answer

Finding the general solution of the given system.

$X\text{'}=\left(\begin{array}{ccccc}1& 0& 2& -1& 0\\ 0& 5.1& 0& -1& 3\\ 2& 2& -3& 0& 0\\ -1.8& 1& -3.1& 4& 0\\ 0& 0& 0& 1.5& 1\end{array}\right)X$

$A=\left(\begin{array}{ccccc}1& 0& 2& -1& 0\\ 0& 5.1& 0& -1& 3\\ 2& 2& -3& 0& 0\\ -1.8& 1& -3.1& 4& 0\\ 0& 0& 0& 1.5& 1\end{array}\right)$

$$\begin{gathered} X\left(t\right)=c_1\left(-0.2622 \\ -0.1627 \\ -0.8262 \\ -0.3464 \\ 0.3196\right)e^{-2.9236t}+c_2\left(-0.3013 \\ -0.4666 \\ 0.2221 \\ 0.0534 \\ -0.7996\right)e^{-0.1553t}+c_3\left(-0.0312 \\ -0.6418 \\ -0.3175 \\ -0.1738 \\ 0.5991\right)e^{2.0288t}+c_4\left(-0.0312 \\ -0.9491 \\ 0.2395 \\ 0.1958 \\ -0.0509\right)e^{5.0545t}+c_5\left(-0.2802 \\ -0.8366 \\ -0.2753 \\ 0.1760 \\ -0.3388\right)e^{4.0956t} \end{gathered}$$

Step-by-step solution

definition of Linear algebra

Linear algebra is the branch of mathematics that deals with linear equations and linear maps, as well as their representations in vector spaces and via matrices.

$X\text{'}=\left(\begin{array}{ccccc}1& 0& 2& -1& 0\\ 0& 5.1& 0& -1& 3\\ 2& 2& -3& 0& 0\\ -1.8& 1& -3.1& 4& 0\\ 0& 0& 0& 1.5& 1\end{array}\right)X$

$A=\left(\begin{array}{ccccc}1& 0& 2& -1& 0\\ 0& 5.1& 0& -1& 3\\ 2& 2& -3& 0& 0\\ -1.8& 1& -3.1& 4& 0\\ 0& 0& 0& 1.5& 1\end{array}\right)$

$>>A=\left[102-1.80;05.10-13;12-300;1-3.140;-2.8001.51\right]$

$A=\begin{array}{ccccc}1.0000& 0& 2.0000& -1.8000& 0\\ 0& 5.1000& 0& -1.0000& 3.0000\\ 1.0000& 2.0000& -3.0000& 0& 0\\ 0& 1.0000& -3.1000& 4.0000& 0\\ -2.8000& 0& 0& 1.5000& 1.0000\end{array}$

$>>\left[eigvec,eigval\right]=eig\left(A\right)$

$$\begin{gathered} eigvec=\begin{array}{ccccc}0.2622& -0.3013& 0.3132& -0.312& -0.2802\\ -0.167& 0.4666& -0.6418& -0.9491& -0.8366\\ -0.8262& 0.2221& -0.3175& -0.2395& -0.2753\\ -0.3464& 0.0534& -0.1738& -0.1958& 0.1760\\ 0.3196& -0.7996& 0.5991& -0.0509& 0.3388\end{array} \\ eigval=\begin{array}{ccccc}-2.9236& 0& 0& 0& 0\\ 0& -0.1553& 0& 0& 0\\ 0& 0& 2.0288& 0& 0\\ 0& 0& 0& 5.0545& 0\\ 0& 0& 0& 0& 4.0956\end{array} \end{gathered}$$

General solution

K₁ and K₂… K_3,K₄, K₅and X₁and X₂, X₃, X₄, X₅.

$$\begin{gathered} {\mathrm{K}}_1=\left(-0.2622 \\ -0.1627 \\ -0.8262 \\ -0.3464 \\ 0.3196\right),X_1=\left(-0.2622 \\ -0.1627 \\ -0.8262 \\ -0.3464 \\ 0.3196\right)e^{-2.9236t} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_2=\left(-0.3013 \\ -0.4666 \\ 0.2221 \\ 0.0534 \\ -0.7996\right),X_2=\left(-0.3013 \\ -0.4666 \\ 0.2221 \\ 0.0534 \\ -0.7996\right)e^{-0.1553t} \end{gathered}$$$$\begin{gathered} {\mathrm{K}}_3=\left(-0.0312 \\ -0.6418 \\ -0.3175 \\ -0.1738 \\ 0.5991\right),X_3=\left(-0.0312 \\ -0.6418 \\ -0.3175 \\ -0.1738 \\ 0.5991\right)e^{2.0288t} \end{gathered}$$role="math" localid="1664004045229" $$\begin{gathered} {\mathrm{K}}_4=\left(-0.0312 \\ -0.9491 \\ 0.2395 \\ 0.1958 \\ -0.0509\right),X_4=\left(-0.0312 \\ -0.9491 \\ 0.2395 \\ 0.1958 \\ -0.0509\right)e^{5.0545t} \end{gathered}$$$$\begin{gathered} {\mathrm{K}}_5=\left(-0.2802 \\ -0.8366 \\ -0.2753 \\ 0.1760 \\ -0.3388\right),X_5=\left(-0.2802 \\ -0.8366 \\ -0.2753 \\ 0.1760 \\ -0.3388\right)e^{4.0956t} \end{gathered}$$

Therefore,

The general solution of the given system is

$$\begin{gathered} X\left(t\right)=c_1\left(-0.2622 \\ -0.1627 \\ -0.8262 \\ -0.3464 \\ 0.3196\right)e^{-2.9236t}+c_2\left(-0.3013 \\ -0.4666 \\ 0.2221 \\ 0.0534 \\ -0.7996\right)e^{-0.1553t}+c_3\left(-0.0312 \\ -0.6418 \\ -0.3175 \\ -0.1738 \\ 0.5991\right)e^{2.0288t}+c_4\left(-0.0312 \\ -0.9491 \\ 0.2395 \\ 0.1958 \\ -0.0509\right)e^{5.0545t}+c_5\left(-0.2802 \\ -0.8366 \\ -0.2753 \\ 0.1760 \\ -0.3388\right)e^{4.0956t} \end{gathered}$$

Hence,

$$\begin{gathered} X\left(t\right)=c_1\left(-0.2622 \\ -0.1627 \\ -0.8262 \\ -0.3464 \\ 0.3196\right)e^{-2.9236t}+c_2\left(-0.3013 \\ -0.4666 \\ 0.2221 \\ 0.0534 \\ -0.7996\right)e^{-0.1553t}+c_3\left(-0.0312 \\ -0.6418 \\ -0.3175 \\ -0.1738 \\ 0.5991\right)e^{2.0288t}+c_4\left(-0.0312 \\ -0.9491 \\ 0.2395 \\ 0.1958 \\ -0.0509\right)e^{5.0545t}+c_5\left(-0.2802 \\ -0.8366 \\ -0.2753 \\ 0.1760 \\ -0.3388\right)e^{4.0956t} \end{gathered}$$