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91
Test 2 / Re: Possible typo in 2018 Version B Q3(b)
« Last post by Xuefeng Fan on December 07, 2020, 02:34:56 PM »
After find what is Vx and Vy, Vy=Ux Vx=-Uy
92
Test 1 / Re: 2020 Night Sitting #1
« Last post by Xuefeng Fan on December 07, 2020, 02:28:14 PM »
After finding out that e^z=-\frac{1}{2}\:i\frac{\sqrt{3}}{2}
We can take ln and get the answer that z =i(+- (2/3)pi +2kpi)
93
Test 4 / test 4 main D
« Last post by yantian4 on December 07, 2020, 03:50:58 AM »
1 b)   x(t)=[-arctan(e^3t)+c1][2 1]e^-3t+[In(e^6t+1)+c2]+[1 2]e^-6t
94
Test 4 / MAT244 TT4 AlT-E Q2
« Last post by Jin Qin on December 03, 2020, 07:20:06 PM »
Hi, here is my solution for TT4 ALT-E Q2.
95
Test 4 / MAT244 TT4 AlT-E Q1b
« Last post by Jin Qin on December 03, 2020, 07:19:26 PM »
Hi, here is my solution for TT4 ALT-E Q1b.
96
Test 4 / MAT244 TT4 AlT-E Q1a
« Last post by Jin Qin on December 03, 2020, 07:18:42 PM »
hi, here is my solution for TT$ ALT-E Q1a.
97
Quiz 6 / Mat244 Quiz6 5A
« Last post by Jin Qin on December 03, 2020, 07:15:20 PM »
This is my answer for Quiz6 5a. I hope that this can clarify you.
98
Test 4 / LEC0201-TT4-ALF-F-Q2
« Last post by RunboZhang 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}
$
99
Test 4 / LEC0201-TT4-ALF-F-Q2
« Last post by RunboZhang on December 03, 2020, 01:24:53 PM »
$\textbf{Problem2:}$

$\text{Find the general solution of }$ $$x' = \begin{bmatrix}
-3 & 25\\
-9 & 3
\end{bmatrix} x$$
$\text{classify fixed point (0,0) (type, is stable or not, orientation if applicable) and sketch trajectories.}$

$\textbf{Solution:}$

$ \text{Let }$ $$A=  \begin{bmatrix}
-3 & 25\\
-9 & 3
\end{bmatrix}$$

$\text{Then we have}$ $$det(A-\lambda I) = (-3-\lambda)(3-\lambda)-25\cdot(-9)=0$$
$\text{Solve for } \lambda ,$  $$ \lambda_1 = 6 \sqrt{6}i ,\ \lambda_2=-6\sqrt{6}i $$

$\text{Take } \lambda = 6\sqrt{6}i \text{, then }$
$$A-\lambda I = A-6\sqrt{6}iI=\begin{bmatrix}
-3-6\sqrt{6}i & 25\\
-9 & 3-6\sqrt{6}i
\end{bmatrix} \xrightarrow{\text{ref}} \begin{bmatrix}
-3-6\sqrt{6}i & 25\\
0 & 0
\end{bmatrix}$$

$\text{Hence we have }$ $$null(A-\lambda I) = \begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix}$$

$\text{and}$ $$x=e^{6\sqrt{6}it}\begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix}$$

$\text{Further expand it and we get }$

$
\begin{gather}
\begin{aligned}

x &= [cos(6\sqrt{6}t)+isin(6\sqrt{6}t)]\begin{bmatrix}
25 \\
3+6\sqrt{6}i
\end{bmatrix} \\\\
&=\begin{bmatrix}
25cos(6\sqrt{6}t)+i25sin(6\sqrt{6}t) \\
3cos(6\sqrt{6}t)+i3sin(6\sqrt{6}t)+i6\sqrt{6}cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t)
\end{bmatrix} \\\\
&=c_1 \begin{bmatrix}
25cos(6\sqrt{6}t)\\
3cos(6\sqrt{6}t)-6\sqrt{6}sin(6\sqrt{6}t)
\end{bmatrix} + c_2 \begin{bmatrix}
25sin(6\sqrt{6}t) \\
3sin(6\sqrt{6}t)+6\sqrt{6}cos(6\sqrt{6}t)
\end{bmatrix}

\end{aligned}
\end{gather}
$

$\text{(graph is attached in the pic below)}$
100
Test 4 / TT4 Alt G Q2
« Last post by Xiao Lu on December 03, 2020, 12:41:49 PM »
Questions and solutions attached below.
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