Author Topic: LEC0201-TT4-ALF-F-Q2  (Read 213 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}
$