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### Messages - Meng Wu

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31
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 11:17:09 PM »
.

32
##### Final Exam / Re: FE-P1
« on: April 11, 2018, 10:54:49 PM »
Actually Professor I remember on the test the last term is -5y^4 instead of -5y^3. I'm not sure whether I'm correct.

It is $-5y^3$.

33
##### MAT244--Misc / Re: Final exam
« on: April 04, 2018, 12:52:08 PM »
Thanks for the clarification.

34
##### MAT244--Misc / Re: Final exam
« on: April 04, 2018, 11:35:59 AM »
2.7--2.8 covered, 2.9 not

Just want to make sure, because my Prof said there is nothing on modelling, numerical methods, exists and uniqueness theorem from chapter 2. (from Tues. B. Jacelon)

35
##### MAT244--Misc / Test Paper for Term Test#2
« on: April 03, 2018, 10:46:54 PM »
I saw the mark for test#2, now where can we pick up the test paper :O

36
##### Quiz-7 / Re: Q7-0901
« on: March 30, 2018, 02:25:18 PM »
Also, the critical points could be $(0,-2n\pi)$, where $0,1,2,\dots$ and $(2, -n\pi)$ where $n=1,3,5,\dots$.

37
##### Web Bonus Problems / Re: Phaseportrait
« on: March 28, 2018, 01:50:57 PM »
$\text{My attempt:}$
\begin{align}x'&=\sin(x)\cos(y)=F(x,y)\\y'&=-\cos(x)\sin(y)=G(x,y)\end{align}
Let \begin{align}\cases{F(x,y)=\sin(x)\cos(y)=0\\G(x,y)=-\cos(x)\sin(y)=0}\end{align}
$$\implies\cases{x=k\pi,y=k\pi\\x={\pi\over2}+k\pi,y={\pi\over2}+k\pi}$$
where $k$ can only be $-1,0,1$, since we have $x\in(-4,4)$ and $y\in(-4,4)$, $\pi\cong 3.14159265359$. $\\$
Thus we have the following critical points:$\\$
Case#1: $(0,0);(\pi,0);(-\pi,0);(0,\pi);(0,-\pi);(\pi,\pi);(-\pi,\pi);(-\pi,-\pi);(\pi,-\pi)$. $\\$
Case#2: $({\pi\over2},{\pi\over2});(-{\pi\over2},-{\pi\over2});({\pi\over2},-{\pi\over2});(-{\pi\over2},{\pi\over2})$.
$$J=\begin{pmatrix}F_x(x,y)&F_y(x,y)\\G_x(x,y)&G_x(x,y)\end{pmatrix}=\begin{pmatrix}\cos(x)\cos(y)&-\sin(x)\sin(y)\\\sin(x)\sin(y)&-\cos(x)\cos(y)\end{pmatrix}$$
When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ or $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, wtih eigenvalues $\lambda=\pm1$, eigenvectors $\xi=\begin{pmatrix}1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\end{pmatrix}$. $\\$
Thus, the critical points in Case#1 are all $\text{Saddle Points}$. $\\$
When $(x,y)=$ critical points in Case#2: we get matrices $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ with eigenvalues $\lambda=\pm i$, eigenvectors $\xi=\begin{pmatrix}1\\i\end{pmatrix}$ or $\begin{pmatrix}1\\-i\end{pmatrix}$. $\\$
Thus, the critical points in Case#2 are all $\text{Center}$. $\\$
Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that.

38
##### Term Test 2 / Re: TT2--P3M
« on: March 28, 2018, 09:11:15 AM »
Thanks for the correction. I made a mistake copying the result integral.
I got the same result.
I think there is a calculation error when computing the initial condition, hence the $c_1,c_2$ values. (if I am not mistaken )
$$c_1+c_2=-2-15\ln2$$ and $$-2c_2=3-15\ln2$$
Therefore, $$c_1=-{1\over2}-{45\over2}\ln2$$
$$c_2=-{3\over2}+{15\over2}\ln2$$

39
##### Term Test 2 / Re: TT2--P3M
« on: March 28, 2018, 12:03:48 AM »
$\underline{\textbf{Solution:}}$ $\\$
$\textbf{(a)}$ $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-1-\lambda&1\\0&-3-\lambda\end{array}=0 \implies \lambda^2+4\lambda+3=0\implies \cases{\lambda_1=-1\\ \lambda_2=-3}$$
Find eigenvectors by $(A-\lambda I_2)\mathbf{x}=\boldsymbol 0$: $\\$
When $\lambda=-1$, eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}$. $\\$
When $\lambda=-3$, eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-2\end{pmatrix}$. $\\$
(Few steps omitted, since "Syed_Hasnain" got part(a) right already.)$\\$
Therefore, the general solution is $$\mathbf{x}(t)=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}$$
$\textbf{(b)}$ $\\$
$$W[\mathbf{x}^{(1)},\mathbf{x}^{(2)}](t)=\begin{array}{|c c|}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{array}=-2e^{-4t}\neq 0$$
Thus, $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}$$
For the non-homogeneous system, we have the general solution$$\mathbf{x}=\boldsymbol\Psi(t)\mathbf{u}(t)$$
Since we know $$\boldsymbol\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$$
$$\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{12\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
By row reduction: $$\begin{pmatrix}e^{-t}&0\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{18\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
Hence $$\cases{u_1'={18e^{t}\over e^t+1}\\u_2'={-6e^{t}\over e^t+1}} \implies \cases{u_1(t)=\int{{18e^{t}\over e^t+1}dt}=18\ln(e^t+1)+c_1\\u_2(t)=\int{{-6e^{t}\over e^t+1}dt}=-6\ln(e^t+1)-3e^{-2t}+6e^t+c_2}$$
Thus \begin{align}\mathbf{x}&=\boldsymbol\Psi(t)\mathbf{u}(t)\\&=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}\end{align}
Apply the given initial condition:$$\mathbf{x}(0)=\begin{pmatrix}12\ln2+c_1+c_2+3\\12\ln2-6-2c_2\end{pmatrix}=\begin{pmatrix}1-3\ln2\\-3-3\ln2\end{pmatrix}\implies \cases{c_1=-{1\over2}-{45\over2}\ln2\\c_2=-{3\over2}+{15\over2}\ln2}$$
Therefore, the general solution for the IVP:
$$\mathbf{x}=(-{1\over2}-{45\over2}\ln2)\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+(-{3\over2}+{15\over2}\ln2)\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}$$

40
##### Term Test 2 / Re: TT2--P3D
« on: March 27, 2018, 11:37:27 PM »
$\underline{\textbf{Solution:}}$ $\\$
$\textbf{(a)}$ $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-\lambda&1\\3&-2-\lambda\end{array}=0 \implies \lambda^2+2\lambda-3=0\implies \cases{\lambda_1=1\\ \lambda_2=-3}$$
Find eigenvectors by $(A-\lambda I_2)\mathbf{x}=\boldsymbol 0$: $\\$
When $\lambda=1$, eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1\end{pmatrix}$. $\\$
When $\lambda=-3$, eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-3\end{pmatrix}$. $\\$
(Few steps omitted, since "Syed_Hasnain" got part(a) right already.)$\\$
Therefore, the general solution is $$\mathbf{x}(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}$$
$\textbf{(b)}$ $\\$
$$W[\mathbf{x}^{(1)},\mathbf{x}^{(2)}](t)=\begin{array}{|c c|}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{array}=-4e^{-2t}\neq 0$$
Thus, $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{pmatrix}$$
For the non-homogeneous system, we have the general solution$$\mathbf{x}=\boldsymbol\Psi(t)\mathbf{u}(t)$$
Since we know $$\boldsymbol\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$$
$$\begin{pmatrix}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{e^{3t}\over e^t+1}\\{e^{3t}\over e^t+1}\end{pmatrix}$$
By row reduction: $$\begin{pmatrix}e^t&0\\0&-4e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{e^{3t}\over e^t+1}\\0\end{pmatrix}$$
Hence $$\cases{u_1'={e^{2t}\over e^t+1}\\u_2'=0} \implies \cases{u_1(t)=\int{{e^{2t}\over e^t+1}dt}=(e^t+1)-\ln(e^t+1)+c_1\\u_2(t)=\int{0dt}=c_2}$$
Thus \begin{align}\mathbf{x}&=\boldsymbol\Psi(t)\mathbf{u}(t)\\&=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}+\begin{pmatrix}1\\1\end{pmatrix}e^{2t}-\ln(e^t+1)\begin{pmatrix}1\\1\end{pmatrix}e^t\end{align}
Apply the given initial condition:$$\mathbf{x}(0)=\begin{pmatrix}c_1+c_2+1-\ln2\\c_1-3c_2+1-\ln2\end{pmatrix}=\begin{pmatrix}3-\ln2\\-1-\ln2\end{pmatrix}\implies \cases{c_1=1\\c_2=1}$$
Therefore, the general solution for the IVP:
$$\mathbf{x}=\begin{pmatrix}1\\1\end{pmatrix}e^t+\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}+\begin{pmatrix}1\\1\end{pmatrix}e^{2t}-\ln(e^t+1)\begin{pmatrix}1\\1\end{pmatrix}e^t$$

41
##### MAT244--Misc / Integrals
« on: March 26, 2018, 02:35:58 PM »
Will the Final Exam contain intergrals that are not listed in the posted anti-derivates sheet?
http://www.math.toronto.edu/courses/mat244h1/20181/LN/antiderivatives.pdf

42
##### MAT244--Misc / Re: Quiz 7 on Good Friday
« on: March 26, 2018, 02:27:45 PM »
You need to read an announcements. It was emailed to T0601, T0701 and posted

http://forum.math.toronto.edu/index.php?topic=1133.0

I think what she wants to know is the exact time. During April 5th(Thursday) tutorials or some other timeslots.

43
##### Term Test 2 / Re: TT2--P3
« on: March 21, 2018, 11:49:31 PM »
I honestly don't think I get part(b) right, correct me pelase

44
##### Term Test 2 / Re: TT2--P3
« on: March 21, 2018, 11:48:29 PM »
Part(b) $\\$
Now consider the non-homogeneous system: $\\$
First calculate the Wronskain $$W[\textbf{x}^{(1)},\textbf{x}^{(2)}](t)=\begin{array}{|c c|}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{array}=-3e^{-t}\neq 0$$
Thus, $\textbf{x}^{(1)}(t)$ and $\textbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}$$
Since the general solution for the non-homogeneous is:
$$\textbf{x}(t)=\boldsymbol{\Psi}(t)\boldsymbol{c}+\boldsymbol{\Psi}(t)\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds$$
Using Quick Formula from linear algebra:
Let $$\boldsymbol{\Psi}(t)=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
\begin{align}\boldsymbol{\Psi}^{-1}(t)&={1\over \det(\boldsymbol{\Psi}(t))}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\\&=-{1\over 3}e^t\begin{pmatrix}-2e^{-2t}&-e^{-2t}\\-e^{t}&e^{t}\end{pmatrix}\\&=\begin{pmatrix}{2\over3}e^{-t}&{1\over3}e^{-t}\\{1\over3}e^{2t}&-{1\over3}e^{2t}\end{pmatrix}\end{align}
Hence, \begin{align}\boldsymbol{\Psi}^{-1}(t)\boldsymbol{g}(t)=\begin{pmatrix}{2\over3}e^{-t}&{1\over3}e^{-t}\\{1\over3}e^{2t}&-{1\over3}e^{2t}\end{pmatrix}\begin{pmatrix}{e^{2t}\over e^t+1}\\{e^{2t}\over e^t+1}\end{pmatrix}=\begin{pmatrix}{e^{t}\over e^t+1}\\0\end{pmatrix}\end{align}
Thus, $$\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds=\int_{t_0}^{t}\begin{pmatrix}{e^{s}\over e^s+1}\\0\end{pmatrix}ds=\begin{pmatrix}\ln(e^t+1)\\k\end{pmatrix}$$
where $k$ is any arbitrary constant. $\\$Thus,
\begin{align}\boldsymbol{\Psi}(t)\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds&=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}{e^{s}\over e^s+1}\\0\end{pmatrix}\\&=\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+k\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}\end{align}
For conveniently chosen $t_0=t$, we have \begin{align}\textbf{c}&=\boldsymbol{\Psi}^{-1}(t_0)\textbf{x}^{0}\\&=\begin{pmatrix}{2\over3}&{1\over3}\\{1\over3}&-{1\over3}\end{pmatrix}\begin{pmatrix}3\\0\end{pmatrix}\\&=\begin{pmatrix}2\\1\end{pmatrix}\implies \cases{c_1=2\\c_2=1}\end{align}
Therefore, the general solution for IVP is $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+k\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$
where $k$ is any arbitrary constant.$\\$
If let $k=1$, we have $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$

45
##### Term Test 2 / Re: TT2--P3
« on: March 21, 2018, 11:48:11 PM »
$\underline{\text{Solution:}}$$\\ Part(a) \\ Find eigenvalues by \det(A-\lambda I_2)=0:$$\begin{array}{|c c|}-\lambda&1\\2&-1-\lambda\end{array}=0 \implies \lambda^2+\lambda-2=0=(\lambda-1)(\lambda+2)=0 \implies \cases{\lambda_1=1\\ \lambda_2=-2}$$Find eigenvectors by (A-\lambda I_2)\textbf{x}=\boldsymbol 0: \\ When \lambda=1, \\$$\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies\begin{pmatrix}-1&1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1\end{pmatrix}$$Thus, the eigenvector \boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1\end{pmatrix}, \textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1\end{pmatrix}e^t. \\ When \lambda=-2, \\$$\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies\begin{pmatrix}2&1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\-2\end{pmatrix}$$Thus, the eigenvector \boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-2\end{pmatrix}, \textbf{x}^{(2)}(t)=\begin{pmatrix}1\\-2\end{pmatrix}e^{-2t}.\\ Therefore, the general solution of given system is$$\textbf{x}(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-2t}$\$

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