Question

In problem 21-30find the general solution of the given system.

$\mathit{x}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{5}& \mathbf{-}\mathbf{4}& \mathbf{0}\\ \mathbf{1}& \mathbf{0}& \mathbf{2}\\ \mathbf{0}& \mathbf{2}& \mathbf{5}\end{array}\mathbf{\right)}\mathit{x}$

Short answer

Find the general solution of the given system.

$x\text{'}=\left(\begin{array}{ccc}5& -4& 0\\ 1& 0& 2\\ 0& 2& 5\end{array}\right)x$

$$\begin{gathered} det\left(\left(\begin{array}{ccc}5& -4& 0\\ 1& 0& 2\\ 0& 2& 5\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right)=0 \\ \left|\begin{array}{ccc}5-\lambda & -4& 0\\ 1& 0-\lambda & 2\\ 0& 2& 5-\lambda \end{array}\right|=0 \\ \left(\begin{array}{ccc}5-5& -4& 0\\ 1& 0-5& 2\\ 0& 2& 5-5\end{array}\right)\left(\begin{array}{c}{P}_1\\ P_2\\ P_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left\{\begin{array}{l}-4P_2=2\\ P_1-5P_2+2P_3=0\to P_1+2P_3=\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 2P_2=-1\to P_2=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}\right. \end{gathered}$$

$$\begin{gathered} x=c_1\left(-4 \\ -5 \\ 2\right)+c_2\left(2 \\ 0 \\ -1\right)e^{5t}+c_3\left[\left(2 \\ 0 \\ -1\right)t{e}^{5t}+\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ -\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right. \\ -1\right)e^{5t}\right] \end{gathered}$$

Step-by-step solution

definition of general solution

We previously concluded that the given system's is general solution then its used the type of matrix form.

Given,

$x\text{'}=AX$

Equation for the eigen values,

localid="1664047359413" $$\begin{gathered} \det \left(A-\lambda I\right)=0 \\ det\left(\left(\begin{array}{ccc}5& -4& 0\\ 1& 0& 2\\ 0& 2& 5\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right)=0 \\ det\left(\left(\begin{array}{ccc}5-\lambda & -4& 0\\ 1& 0-\lambda & 2\\ 0& 2& 5-\lambda \end{array}\right)\right)=0 \\ \left|\begin{array}{ccc}5-\lambda & -4& 0\\ 1& 0-\lambda & 2\\ 0& 2& 5-\lambda \end{array}\right|=0 \end{gathered}$$

Multiply:

$$\begin{gathered} \left(5-\lambda \right)\left[\left(5-\lambda \right)\left(0-\lambda \right)-4\right]+5\left[\left(5-\lambda \right).1-0\right]+0\left[2.1-0\left(0-\lambda \right)\right]=0 \\ \left(0-\lambda \right){\left(5-\lambda \right)}^2-4\left(5-\lambda \right)+4\left[\left(5-\lambda \right)-0\right]+0=0 \\ \lambda {\left(5-\lambda \right)}^2=0 \\ \lambda =0,5,5 \end{gathered}$$

${\lambda }_1=0$we have,

$\left(A-{\lambda }_{1}I\right)K_1=0$

Multiply:

$\left(\begin{array}{ccc}5& -4& 0\\ 1& 0& 2\\ 0& 2& 5\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=0$

$\left\{\begin{array}{l}5K_1-4K_2=0\\ K_1+2K_3=0\\ 2K_2+5K_3=0\end{array}\right.$

Solving we get,

$$\begin{gathered} K_1=\left(-4 \\ -5 \\ 2\right) \end{gathered}$$

${\lambda }_2=5$have,

$\left(\mathrm{A}-{\mathrm{\lambda }}_2\mathrm{I}\right)=\mathrm{K}$

Subtraction:

$$\begin{gathered} \left(\begin{array}{ccc}5-5& -4& 0\\ 1& 0-5& 2\\ 0& 2& 5\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=0 \\ \left(\begin{array}{ccc}0& -4& 0\\ 1& -5& 2\\ 0& 2& 5\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=0 \\ \left\{\begin{array}{l}-4K_2=0\to K_2=0\\ K_1-5K_2+2K_3=0\to K_1+2K_3=0\\ 2K_2=0\end{array}\right. \end{gathered}$$

Corresponding solution vector

A vector is a mathematical object with a size, known as the magnitude, and a direction. This gives eigenvector and a corresponding solution vector:

Have solved for P For ${\lambda }_2=5$

$$\begin{gathered} \left(A-{\lambda }_{2}I\right)P=K \\ \left(\begin{array}{ccc}\left(5-5\right)& -4& 0\\ 1& \left(0-5\right)& 2\\ 0& 2& \left(5-5\right)\end{array}\right)\left(\begin{array}{c}{P}_1\\ P_2\\ P_2\end{array}\right)=\left(\begin{array}{c}2\\ 0\\ -1\end{array}\right) \\ \left(\begin{array}{ccc}0& -4& 0\\ 1& -5& 2\\ 0& 2& 0\end{array}\right)\left(\begin{array}{c}{P}_1\\ P_2\\ P_2\end{array}\right)=\left(\begin{array}{c}2\\ 0\\ -1\end{array}\right) \\ \left\{\begin{array}{l}-4P_2=2\\ P_1-5P_2+2P_3=0\to P_1+2P_3=\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 2P_2=-1\to P_2=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}\right. \end{gathered}$$

Solved,

$\left(\begin{array}{c}{P}_1\\ P_2\\ P_3\end{array}\right)=\left(\begin{array}{c}\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ -1\end{array}\right)$

Therefore the solution:

$$\begin{gathered} x=c_1\left(-4 \\ -5 \\ 2\right)+c_2\left(2 \\ 0 \\ -1\right)e^{5t}+c_3\left[\left(2 \\ 0 \\ -1\right)t{e}^{5t}+\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ -\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right. \\ -1\right)e^{5t}\right] \end{gathered}$$

Hence,

$$\begin{gathered} x=c_1\left(-4 \\ -5 \\ 2\right)+c_2\left(2 \\ 0 \\ -1\right)e^{5t}+c_3\left[\left(2 \\ 0 \\ -1\right)t{e}^{5t}+\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ -\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right. \\ -1\right)e^{5t}\right] \end{gathered}$$