Question

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

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{-}\mathit{y}\mathbf{-}\mathit{z} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{y}\mathbf{-}\mathit{z} \\ \frac{\mathbf{dz}}{\mathbf{dt}}\mathbf{=}\mathit{x}\mathbf{-}\mathit{y}\mathbf{+}\mathit{z} \end{gathered}$$

Short answer

Find the general solution of the given system.

$$\begin{gathered} x\text{'}=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)x \\ A=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right) \end{gathered}$$

The matrix is

$$\begin{gathered} K_1-K_2-K_3=0 \\ {K}_1=K_2+K_3 \end{gathered}$$

$K_2=1$and$K_3=0$ and $K_2=0,K_3=1,K_1=0$, This give two eigenvector .

$K_2=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\text{and}{K}_3=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)$

$$\begin{gathered} x=c_1\left(1 \\ 1 \\ 0\right)e^t+c_2\left(1 \\ 1 \\ 0\right)e^{2t}+c_3\left(1 \\ 1 \\ 1\right)e^{2t} \end{gathered}$$

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,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=3x-y-z \\ \frac{\mathrm{dy}}{\mathrm{dt}}=x+y-z \\ \frac{\mathrm{dz}}{\mathrm{dt}}=x-y+z \end{gathered}$$

$$\begin{gathered} x\text{'}=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right)x \\ A=\left(\begin{array}{ccc}3& -1& -1\\ 1& 1& -1\\ 1& -1& 1\end{array}\right) \end{gathered}$$

$\det \left(A-\lambda I\right)=0$

$\left|\begin{array}{ccc}3-\lambda & -1& -1\\ 1& 1-\lambda & -1\\ 1& -1& 1-\lambda \end{array}\right|=0$

Multiply:

$$\begin{gathered} \left(3-\lambda \right)\left[\left(1-\lambda \right)\left(1-\lambda \right)-1\right]+1\left[\left(1-\lambda \right)+1\right]-1\left[-1-\left(1-\lambda \right)\right]=0 \\ \left(3-\lambda \right)\left[{\left(1-\lambda \right)}^2-1\right]+2\left(1-\lambda \right)+2=0 \\ \left(3-\lambda \right)\left(1-2\lambda +{\lambda }^2-1\right)+2\left(1-\lambda \right)+2=0 \\ \left(3-\lambda \right)\left({\lambda }^2-2\lambda \right)+2\left(1-\lambda \right)+2=0 \\ \left(3-\lambda \right)\left(\lambda -2\right)+2\left(1-\lambda \right)+2=0 \\ \left(3-\lambda \right)\lambda \left(\lambda -2\right)-\left(\lambda -2\right)-\left(\lambda -2\right)=0 \\ \left(\lambda -2\right)\left[\lambda \left(3-\lambda \right)-1-1\right]=0 \\ \left(\lambda -2\right)\left[\lambda \left(3-\lambda \right)-1-1\right]=0 \\ \left(\lambda -2\right)\left[-{\lambda }^2+3\lambda -2\right]=0 \\ \left(\lambda -2\right)\left(\lambda -2\right)\left(\lambda -1\right)=0 \end{gathered}$$

Our eigenvalues are:

${\lambda }_1=1,{\lambda }_1=2,{\lambda }_3=2$

For${\lambda }_1=1;$

$\left(A-\lambda I\right)K=0$

Subtraction:

$$\begin{gathered} \left(\begin{array}{ccc}3-1& -1& -1\\ 1& 1-1& -1\\ 1& -1& 1-1\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}2& -1& -1\\ 1& 0& -1\\ 1& -1& 0\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

Second and third row,

$$\begin{gathered} K_1-K_3=0\to K_1=K_3 \\ {K}_1-K_2=0\to K_1=K_2 \end{gathered}$$

$K_3=1$yields $K_2=1$and $K_1=1$

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:

This give eigenvector and a corresponding solution vector:

$$\begin{gathered} K_1=\left(1 \\ 1 \\ 1\right),x_1=\left(1 \\ 1 \\ 1\right)e^t \end{gathered}$$

For${\lambda }_2={\lambda }_3=2$

$$\begin{gathered} \left(A-2I\right)K=0 \\ \left(\begin{array}{ccc}3-2& -1& -1\\ 1& 1-2& -1\\ 1& -1& 1-2\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}1& -1& -1\\ 1& -1& -1\\ 1& -1& -1\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

From the matrix

$$\begin{gathered} K_1-K_2-K_3=0 \\ {K}_1=K_2+K_3 \end{gathered}$$

$K_2=1$and $K_3=0$and $K_2=0,K_3=1,K_1=0$This give two eigenvector .

$$\begin{gathered} K_2=\left(1 \\ 1 \\ 0\right)\text{and}{K}_3=\left(1 \\ 1 \\ 0\right) \end{gathered}$$

Two independent solution:

localid="1664041278221" $$\begin{gathered} x=c_1\left(1 \\ 1 \\ 0\right)e^t+c_2\left(1 \\ 1 \\ 0\right)e^{2t}+c_3\left(1 \\ 1 \\ 1\right)e^{2t} \end{gathered}$$

General solution is given :

$$\begin{gathered} x=c_1\left(1 \\ 1 \\ 0\right)e^t+c_2\left(1 \\ 1 \\ 0\right)e^{2t}+c_3\left(1 \\ 1 \\ 1\right)e^{2t} \end{gathered}$$

Hence,

$$\begin{gathered} x=c_1\left(1 \\ 1 \\ 0\right)e^t+c_2\left(1 \\ 1 \\ 0\right)e^{2t}+c_3\left(1 \\ 1 \\ 1\right)e^{2t} \end{gathered}$$