Answer :
Answer:
The required position of the particle at time t is: [tex]x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}[/tex]
Step-by-step explanation:
Consider the provided matrix.
[tex]v_1=\begin{bmatrix}-3\\1 \end{bmatrix}[/tex]
[tex]v_2=\begin{bmatrix}-1\\1 \end{bmatrix}[/tex]
[tex]\lambda_1=4, \lambda_2=2[/tex]
The general solution of the equation [tex]x'=Ax[/tex]
[tex]x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}[/tex]
Substitute the respective values we get:
[tex]x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}[/tex]
[tex]x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}[/tex]
Substitute initial condition [tex]x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}[/tex]
[tex]\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}[/tex]
Reduce matrix to reduced row echelon form.
[tex]\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}[/tex]
Therefore, [tex]c_1=2.5,c_2=1.5[/tex]
Thus, the general solution of the equation [tex]x'=Ax[/tex]
[tex]x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}[/tex]
[tex]x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}[/tex]
The required position of the particle at time t is: [tex]x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}[/tex]