Question

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

$\mathit{x}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{12}& \mathbf{-}\mathbf{9}\\ \mathbf{4}& \mathbf{0}\end{array}\mathbf{\right)}\mathit{x}$

Short answer

Find the general solution of the given system.

$$\begin{gathered} x\text{'}=\left(\begin{array}{cc}12& -9\\ 4& 0\end{array}\right)x \\ A=\left(\begin{array}{cc}12& -9\\ 4& 0\end{array}\right) \\ \left|\begin{array}{cc}12-\lambda & -9\\ 4& 0-\lambda \end{array}\right|=0 \end{gathered}$$

This gives eigenvector and a corresponding solution vector:

$$\begin{gathered} \left(\begin{array}{cc}6& -9\\ 4& -6\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=0 \\ x=c_{1}k{e}^{\lambda 1t}+c_2\left(kt{e}^{\lambda 1t}+p{e}^{\lambda 1t}\right) \\ x=c_1\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+c_2\left[\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+\left(\begin{array}{c}\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right)e^{6t}\right] \end{gathered}$$

Step-by-step solution

definition of general solution

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

$$\begin{gathered} x\text{'}=\left(\begin{array}{cc}12& -9\\ 4& 0\end{array}\right)x \\ A=\left(\begin{array}{cc}12& -9\\ 4& 0\end{array}\right) \end{gathered}$$

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

$$\begin{gathered} \left|\begin{array}{cc}12-\lambda & -9\\ 4& 0-\lambda \end{array}\right|=0 \\ \left(12-\lambda \right)\left(0-\lambda \right)-\left(-36\right)=0 \\ {\lambda }^2-12\lambda +36=0 \\ \left(\lambda -6\right)\left(\lambda -6\right)=0 \end{gathered}$$

Our repeated eigenvalues are:

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

role="math" localid="1664037198622" $$\begin{gathered} \left(\begin{array}{cc}12-6& -9\\ 4& -6\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}6& -9\\ 4& -6\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation:

$$\begin{gathered} \frac{3}{2}{R}_2-R_1\to R_2 \\ \frac{1}{6}{R}_1\to R_1 \end{gathered}$$

$\left(\begin{array}{cc}6& -9\\ 0& 0\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

$K_1-\frac{3}{2}{K}_2=0\to X_1=\left(\frac{3}{2}\right)k_2$

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 gives eigenvector and a corresponding solution vector:

$k_1=2$Yields $k_1=3$

$$\begin{gathered} \left(A-\lambda I\right)p=K \\ \left(\begin{array}{cc}12-6& -9\\ 4& 0-6\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 2\end{array}\right) \\ \left(\begin{array}{cc}6& -9\\ 4& -6\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 2\end{array}\right) \end{gathered}$$

Apply row operation:

$$\begin{gathered} \frac{3}{2}{R}_2-R_1\to R_2 \\ \left(\begin{array}{cc}6& -9\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}3\\ 0\end{array}\right) \end{gathered}$$

Apply row operation:

$$\begin{gathered} \frac{1}{6}{R}_1\to R_1 \\ \left(\begin{array}{cc}1& -\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right) \end{gathered}$$

From the first Row:

$p_1-\frac{3}{2}{p}_2=\frac{1}{2}\to p_1=p_2+\frac{1}{2}$

Simplicity shall choose $p_2=0$so that$p_1=\frac{1}{2}$This gives eigenvector and a corresponding solution vector:

$p=\left(\begin{array}{c}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right)$

Corresponding solution vector:

$x_2=\left(\begin{array}{c}3\\ 2\end{array}\right)t{e}^{6t}+\left(\begin{array}{c}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right)e^{6t}$

General solution is given:

$$\begin{gathered} x=c_{1}k{e}^{\lambda 1t}+c_2\left(kt{e}^{\lambda 1t}+p{e}^{\lambda 1t}\right) \\ x=c_1\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+c_2[\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+\left(\begin{array}{c}\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right)e^{6t}] \end{gathered}$$

Hence,

$x=c_1\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+c_2[\left(\begin{array}{c}3\\ 2\end{array}\right)e^{6t}+\left(\begin{array}{c}\raisebox{1ex}{${\displaystyle 1}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0\end{array}\right)e^{6t}]$