### Author Topic: LEC0201-TT4-ALF-F-Q2  (Read 2462 times)

#### RunboZhang

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##### LEC0201-TT4-ALF-F-Q2
« on: December 03, 2020, 02:20:22 PM »
$\textbf{Problem1:} \\$
$\textbf{(a)} \text{ Find the general solution of }$ $$x'=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}x$$
$\text{classify fixed point (0, 0) and sketch trajectories.} \\$
$\textbf{(b)} \text{ Find the general solution}$ $$x'=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}x + \begin{bmatrix} 0\\ \frac{150e^{25t}}{e^{30t}+1} \end{bmatrix}x$$

$\textbf{Solution:} \\$
$\textbf{(a)}$

$\text{Let }$ $$A=\begin{bmatrix} 13 & -9\\ 6 & -8 \end{bmatrix}$$

$\text{Then }$ $$det(A-\lambda I) = (13-\lambda)(-8-\lambda)-6\cdot(-9)=0$$

$\text{Solve for } \lambda:$ $$\lambda_1 = 10 , \lambda_2 = -5$$

$\text{When }\lambda = 10,$ $$A-\lambda I = \begin{bmatrix} 3 & -9\\ 6 & -18 \end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix} 1 & -3\\ 0 & 0 \end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_1}=\begin{bmatrix} 3\\ 1 \end{bmatrix}$$

$\text{When }\lambda = -5,$ $$A-\lambda I = \begin{bmatrix} 18 & -9\\ 6 & -3 \end{bmatrix} \xrightarrow{\text{ref}}\begin{bmatrix} 2 & -1\\ 0 & 0 \end{bmatrix}$$
$\text{Hence the eigenvector is }$ $$\vec{v_2}=\begin{bmatrix} 1\\ 2 \end{bmatrix}$$

$\text{Therefore we have }$ $$x=c_1e^{10t}\begin{bmatrix} 3\\ 1 \end{bmatrix} + c_2e^{-5t}\begin{bmatrix} 1\\ 2 \end{bmatrix}$$

$\text{(graph is attached below)}$

$\textbf{(b)}$

$\text{Let }$ $$\varphi (t)=\begin{bmatrix} 3e^{10t} & e^{-5t}\\ e^{10t} & 2e^{-5t} \end{bmatrix}, u'(t) = \begin{bmatrix} u_1'(t)\\ u_2'(t) \end{bmatrix}$$

$\text{Thus we have }$ $$\left[ \begin{array}{cc|c} 3e^{10t} & e^{-5t} & 0 \\ e^{10t} & 2e^{-5t} & \frac{150e^{25t}}{e^{30t}+1} \end{array}\right] \xrightarrow{R_2=3R_2-R_1} \left[ \begin{array}{cc|c} 3e^{10t} & e^{-5t} & 0 \\ 0 & 5e^{-5t} & \frac{450e^{25t}}{e^{30t}+1} \end{array}\right]$$

$\text{Observe the second row, we can get }$
\begin{gather} \begin{aligned} 5e^{-5t}\cdot u_2'(t) &= \frac{450e^{25t}}{e^{30t}+1} \\ \\ u_2'(t) &= \frac{90e^{30t}}{e^{30t}+1}\\ \\ u_2(t) &=\int{\frac{90e^{30t}}{e^{30t}+1}} \,dt \\ \\ &= 90\int{\frac{e^{30t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\ &= 3ln(e^{30t}+1)+c_2 \end{aligned} \end{gather}

$\text{Then compute the first row}$
\begin{gather} \begin{aligned} 3e^{10t}u_1'(t)+e^{-5t}u_2'(t) &= 0 \\ \\ 3e^{10t}u_1'(t)+e^{-5t}\frac{90e^{30t}}{e^{30t}+1} &= 0 \\ \\ u_1 &= -30\int{\frac{e^{15t}}{e^{30t}+1}} \,dt \ \ \ \text{(integral by substitution)}\\ \\ &=-30\int{\frac{\frac{1}{15}}{u^2+1}} \,du \\ \\ &= -2arctan(e^{15t})+c_1 \end{aligned} \end{gather}

$\text{Therefore }$
\begin{gather} \begin{aligned} x&=\varphi{(t)}\cdot u(t) \\ \\ &= \begin{bmatrix} 3e^{10t} & e^{-5t}\\ e^{10t} & 2e^{-5t} \end{bmatrix} \cdot \begin{bmatrix} -2arctan(e^{15t})+c_1\\ 3ln(e^{30t}+1)+c_2 \end{bmatrix} \\ \\ &= c_1\begin{bmatrix} 3e^{10t}\\ e^{10t} \end{bmatrix} +c_2\begin{bmatrix} e^{-5t}\\ 2e^{-5t} \end{bmatrix} + \begin{bmatrix} -6e^{10t}arctan(e^{15t})+3e^{-5t}ln(e^{30t}+1)\\ -2e^{10t}arctan(e^{15t})+6e^{-5t}ln(e^{30t}+1) \end{bmatrix} \end{aligned} \end{gather}