Messages

16

Thanksgiving bonus / Re: Thanksgiving bonus 1

« on: October 06, 2018, 10:52:20 AM »

Let$$u(x,y)=\frac{x}{x^2+y^2}, v(x,y)=\frac{y}{x^2+y^2}$$
$$\frac{\partial{u}}{\partial{x}}=\frac{-(x^2-y^2)}{(x^2+y^2)^2}, \frac{\partial{u}}{\partial{y}}=\frac{2xy}{-(x^2+y^2)^2}$$
$$\frac{\partial{v}}{\partial{x}}=\frac{-2xy}{(x^2+y^2)^2}, \frac{\partial{v}}{\partial{y}}=\frac{x^2-y^2}{(x^2+y^2)^2}$$
$$\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{y}}=0, \frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}=0, x^2+y^2>0$$
Thus,  given function represents a locally sourceless and irrotational flow.

17

Quiz-2 / Re: Q2 TUT 0301

« on: October 05, 2018, 06:39:52 PM »

By Ratio Test:
$$\begin{align}\sum_{n=1}^\infty \Biggl(\frac{1+2i}{\sqrt{6}}\Biggl)^n &= \lim_{n\to\infty}\Biggl|\frac{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n+1}}{\Big(\frac{1+2i}{\sqrt{6}}\Big)^{n}}\Biggr|\\&=\lim_{n\to\infty}\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\Biggl|\frac{1+2i}{\sqrt{6}}\Biggr|\\&=\sqrt{\Big(\frac{1}{\sqrt{6}}\Big)^2+\Big(\frac{2}{\sqrt{6}}\Big)^2}\\&=\sqrt{\frac{5}{6}}<1\end{align}$$
Therefore,  the given infinite series converges.

18

Quiz-1 / Re: Q1: TUT 0301

« on: September 28, 2018, 05:51:56 PM »

Just notice, I posted the answer before the 18:00 you mentioned. Is this gonna be an issue? Just wondering.

19

Quiz-1 / Re: Q1: TUT 0301

« on: September 28, 2018, 04:34:22 PM »

Let $$z=x+iy,\text{where} \space x,y \in \mathbb{R}.$$Given equation becomes:
$$|x+iy-1|^2=|x+iy+1|^2+6$$
Since,$$|x+iy-1|=\sqrt{(x-1)^2+y^2} \\ |x+iy+1|=\sqrt{(x+1)^2+y^2}$$
Hence, we have$$\require{cancel}\begin{align}\bigg(\sqrt{(x-1)^2+y^2}\bigg)^2&=\bigg(\sqrt{(x+1)^2+y^2}\bigg)^2+6 \\(x-1)^2+y^2&=(x+1)^2+y^2+6 \\ \cancel{x^2}-2x+\cancel{y^2}&=\cancel{x^2}+2x+\cancel{y^2}+6\\-4x&=6\\x&=-\frac{3}{2}\end{align}$$
$\\$
$\\$
Therefore, the locus of points $z$ are the straight line(vertical line) $x=-\frac{3}{2}.$

20

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:56:10 PM »

Also for $(20)$, $\ln(e^x)$ can be simplified as $\ln(e^x)=x$.

21

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:54:19 PM »

Small Error: $W_2(x)$ should be $2e^{4x}$.

For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

Oh, you're right. My mistake.

22

Final Exam / Re: FE-P3

« on: April 11, 2018, 11:47:31 PM »

Small Error: $W_2(x)$ should be $2e^{4x}$.

23

Final Exam / Re: FE-P2

« on: April 11, 2018, 11:29:00 PM »

Most of the answer is correct, there is one small computational error.$\\$
For $Y_2(t)$, it should be $Y_2(t)=-4\cos(t)+2\sin(t)$.$\\$

24

Final Exam / Re: FE-P1

« on: April 11, 2018, 11:17:09 PM »

.

25

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 $-5y^3$.

26

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 $n=1,3,5,\dots$.

27

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, $$c_1=-{1\over2}-{45\over2}\ln2$$
$$c_2=-{3\over2}+{15\over2}\ln2$$

28

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:
$$\mathbf{x}=(-{1\over2}-{45\over2}\ln2)\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+(-{3\over2}+{15\over2}\ln2)\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}$$

29

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:
$$\mathbf{x}=\begin{pmatrix}1\\1\end{pmatrix}e^t+\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}+\begin{pmatrix}1\\1\end{pmatrix}e^{2t}-\ln(e^t+1)\begin{pmatrix}1\\1\end{pmatrix}e^t$$

30

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