Question

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

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{+}\mathbf{4}\mathit{z} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{2}\mathit{z} \\ \frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{+}\mathbf{3}\mathit{z} \end{gathered}$$

Short answer

Find the general solution of the given system.

$$\begin{gathered} x\text{'}=\left(\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right)x \\ A=\left(\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right) \end{gathered}$$

$$\begin{gathered} \det \left(A-\lambda I\right)=0 \\ \left|\begin{array}{ccc}3-\lambda & 2& 4\\ 2& 0-\lambda & 2\\ 4& 2& 3-\lambda \end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} x_2=\left(1 \\ -2 \\ 0\right)e^{-t}\text{and}{x}_3=\left(0 \\ -2 \\ 1\right)e^{-t} \end{gathered}$$

$$\begin{gathered} x=c_1\left(2 \\ 1 \\ 2\right)e^{8t}+c_2\left(1 \\ -2 \\ 0\right)e^{-t}+c_3\left(0 \\ -2 \\ 1\right)e^{-t} \end{gathered}$$

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,

$$\begin{gathered} \frac{dx}{dt}=3x+2y+4z \\ \frac{dy}{dt}=2x+2z \\ \frac{dz}{dt}=4x+2y+3z \end{gathered}$$

Given, matrix form

$$\begin{gathered} x\text{'}=\left(\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right)x \\ A=\left(\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right) \end{gathered}$$

$$\begin{gathered} \det (A-\lambda I)=0 \\ \left|\begin{array}{ccc}3-\lambda & 2& 4\\ 2& 0-\lambda & 2\\ 4& 2& 3-\lambda \end{array}\right|=0 \end{gathered}$$

Multiply:

$$\begin{gathered} \left(3-\lambda \right)\left[\left(0-\lambda \right)\left(3-\lambda \right)-4\right]-2\left[2\left(3-\lambda \right)-8\right]+4\left[4-4\left(0-\lambda \right)\right]=0 \\ \left(3-\lambda \right)\left[-3\lambda +{\lambda }^2-4\right]-2\left[6-2\lambda -8\right]+4\left[4+4\lambda \right]=0 \\ \left(3-\lambda \right)\left[{\lambda }^2-3\lambda -4\right]-2\left[-2-2\lambda \right]+16\left[1+\lambda \right]=0 \\ \left(3-\lambda \right)\left[\left(\lambda -4\right)\left(\lambda +1\right)\right]+4\left[1+\lambda \right]+16\left[1+\lambda \right]=0 \\ \left(\lambda +1\right)\left[\left(3-\lambda \right)\left(\lambda -4\right)+4+16\right]=0 \\ \left(\lambda +1\right)\left[3\lambda -12-{\lambda }^2+4\lambda +20\right]=0 \\ \left(\lambda +1\right)\left[-{\lambda }^2+7\lambda +8\right]=0 \\ \left(\lambda +1\right)\left(\lambda +1\right)\left(\lambda -8\right)=0 \end{gathered}$$

${\lambda }_1=8,{\lambda }_1={\lambda }_3=-1$

For${\lambda }_1=1;$

$\left(A-8I\right)K=0$

Subtraction:

localid="1664043342967" $$\begin{gathered} \left(\begin{array}{ccc}3-8& 2& 4\\ 2& 0-8& 2\\ 4& 2& 3-8\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}-5& 2& 4\\ 2& -8& 2\\ 4& 2& -5\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation:

localid="1664043445213" $$\begin{gathered} R_3-2R\to R_3 \\ \left(\begin{array}{ccc}-5& 2& 4\\ 2& -8& 2\\ 0& 2& -9\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}-45& 0& 45\\ 2& -8& 2\\ 4& 18& -9\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

from first and third row,

$$\begin{gathered} -45K_1+45K_3=0\to K_1=K_3 \\ 18K_2-9K_3=0\to K_3=2K_2 \end{gathered}$$

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:

$K_2=1,K_1=2,K_2=2$

$$\begin{gathered} K_1=\left(2 \\ 1 \\ 2\right),x_1=\left(2 \\ 1 \\ 2\right)e^{8t} \end{gathered}$$

For ${\lambda }_2={\lambda }_3=-1$

$$\begin{gathered} \left(\begin{array}{ccc}3-\left(-1\right)& 2& 4\\ 2& 0-\left(-1\right)& 2\\ 4& 2& 3-\left(-1\right)\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}4& 2& 4\\ 2& 1& 2\\ 4& 2& 4\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

$$\begin{gathered} 4K_1+2K_2+4K_3=0 \\ 2K_1+K_2+2K_3=0 \\ {K}_2=-2K_1+2K_3 \end{gathered}$$

$$\begin{gathered} K=\left(\begin{array}{c}{K}_1\\ K_2\\ K_3\end{array}\right) \\ =\left(\begin{array}{c}{K}_1\\ -2K_1-2K_3\\ K_3\end{array}\right) \\ =\left(\begin{array}{c}{K}_1\\ -2K_1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ -2K_3\\ K_3\end{array}\right) \\ =K_1\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)+K_3\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right) \end{gathered}$$

Two linear independent vector here,

Eigenvector$\lambda =-1$Two corresponding solution vector

localid="1664044555297">x2=1-20e-tandx3=0-21e-t

General solution is given :

$$\begin{gathered} x=c_1\left(2 \\ 1 \\ 2\right)e^{8t}+c_2\left(1 \\ -2 \\ 0\right)e^{-t}+c_3\left(0 \\ -2 \\ 1\right)e^{-t} \end{gathered}$$

Hence,

$$\begin{gathered} x=c_1\left(2 \\ 1 \\ 2\right)e^{8t}+c_2\left(1 \\ -2 \\ 0\right)e^{-t}+c_3\left(0 \\ -2 \\ 1\right)e^{-t} \end{gathered}$$