Question

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

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{-}\mathit{y}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{9}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}$

Short answer

Find the general solution of the given system.

$\left\{\frac{dx}{dt}\right.=3x-y\frac{dy}{dt}=9x-3y$

“written in the form”

role="math" localid="1664012055224" $$\begin{gathered} x\text{'}=\left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right)x \\ A=\left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right) \end{gathered}$$

Replace the value of$\lambda$ for$1$

Multiply:

role="math" localid="1664012183069" $$\begin{gathered} \left|\begin{array}{cc}3-\lambda & -1\\ 9& -3-\lambda \end{array}\right|=0 \\ \left(3-\lambda \right)\left(-3-\lambda \right)-\left(-9\right)=0 \\ -9-3\lambda +3\lambda +{\lambda }^2+9=0 \\ {\lambda }^2=0 \end{gathered}$$

role="math" localid="1664013440690" $X=c_1\left(\begin{array}{c}1\\ 3\end{array}\right)c_2\left[\left(\begin{array}{c}1\\ 3\end{array}\right)t+\left(\begin{array}{c}1/4\\ -1/4\end{array}\right)\right]$

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.

Give,

$\frac{\mathrm{dx}}{\mathrm{dt}}=3x-y\frac{\mathrm{dy}}{\mathrm{dt}}=9x-3y$

can be written in the form

$$\begin{gathered} x\text{'}=\left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right)x \\ A=\left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right) \end{gathered}$$

Replace the value of$\lambda$for$1$

Multiply:

role="math" localid="1664012226420" $$\begin{gathered} \left|\begin{array}{cc}3-\lambda & -1\\ 9& -3-\lambda \end{array}\right|=0 \\ (3-\lambda )(-3-\lambda )-(-9)=0 \\ -9-3\lambda +3\lambda +{\lambda }^2+9=0 \\ {\lambda }^2=0 \end{gathered}$$

So our repeated eigenvalues are${\lambda }_1=0$ and ${\lambda }_2=0$.

role="math" localid="1664012506974" $$\begin{gathered} \left(\begin{array}{cc}3-0& -1\\ 9& -3-0\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply the row:

$R_2-3R_1\to R_2$

role="math" localid="1664012966165" $\left(\begin{array}{cc}3& -1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{K}_1\\ K_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

First row:

$$\begin{gathered} 3K_1-K_2=0\to K_1=\frac{1}{3}{K}_2 \\ {K}_2=3,K_1=1 \end{gathered}$$

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:

$$\begin{gathered} K=\left(1 \\ 3\right) \end{gathered}$$and $$\begin{gathered} X_1=\left(1 \\ 3\right) \end{gathered}$$

$$\begin{gathered} \left(A-{\lambda }_{2}I\right)P=K\to \left(\begin{array}{cc}3-0& -1\\ 9& -3-0\end{array}\right)\left(\begin{array}{c}{P}_1\\ P_2\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right) \\ \left(\begin{array}{cc}3& -1\\ 9& -3\end{array}\right)\left(\begin{array}{c}{P}_1\\ P_2\end{array}\right)=\left(\begin{array}{c}1\\ 3\end{array}\right) \end{gathered}$$

Apply the row operation:

$R_2-3R_1\to R_2$

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

First row:

$3P_1-P_2=1$

$P_2=r_2$where $P_1=\frac{1-r_2}{3}$we can choose $r_2=-\frac{1}{4}$yields $P_2=-\frac{1}{4}$and $P_1=\frac{1-\left(\frac{1}{4}\right)}{3}=\frac{1}{4}$

This give eigenvector:

$P=\left(\begin{array}{c}1/4\\ -1/4\end{array}\right)$

Solution of vector:

$x_2=\left(\begin{array}{c}1\\ 3\end{array}\right)t+\left(\begin{array}{c}1/4\\ -1/4\end{array}\right)$

General solution is:

role="math" localid="1664013498508" $$\begin{gathered} x=c_{1}k{e}^{\lambda 1t}+c_2\left(kt{e}^{\lambda 1t}+p{e}^{\lambda 1t}\right) \\ X=c_1\left(\begin{array}{c}1\\ 3\end{array}\right)c_2[\left(\begin{array}{c}1\\ 3\end{array}\right)t+\left(\begin{array}{c}1/4\\ -1/4\end{array}\right)] \end{gathered}$$

Hence,

$X=c_1\left(\begin{array}{c}1\\ 3\end{array}\right)c_2[\left(\begin{array}{c}1\\ 3\end{array}\right)t+\left(\begin{array}{c}1/4\\ -1/4\end{array}\right)]$