### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Huyi Xiong

Pages: [1]
1
##### Quiz 5 / QUIZ5 TUT0101
« on: March 04, 2020, 08:08:09 PM »
Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.
\begin{align*}
\frac{e^z-1}{z^2}, z_0=0
\end{align*}

\begin{align}
\frac{e^z-1}{z^2} &= \frac{\sum_{n=0}^{\infty}{\frac{z^n}{n!}}-1}{z^2} \\
&= \frac{\sum_{n=1}^{\infty}{\frac{z^n}{n!}}}{z^2} \\
&= \sum_{n=1}^{\infty}{\frac{z^{n-2}}{n!}} \\
&= \frac{1}{z}+\frac{1}{2}+\frac{1}{6}z +...\\
residue = 1
\end{align}

2
##### Quiz 2 / Re: TUT0701 QUIZ2
« on: March 03, 2020, 08:17:50 PM »

3
##### Quiz 2 / TUT0701 QUIZ2
« on: January 29, 2020, 07:20:32 PM »
Find the limit at $\infty$ of the given function, or explain why it does not exist.
\begin{align*}
h(z)=\frac{z}{|z|^2}, z \neq 0
\end{align*}

\begin{align}
\lim_{z\to\infty} h(z) &=\lim_{z\to\infty} \frac{z}{|z|^2} \\
&=\lim_{z\to0}\frac{|z|^2}{z} && {\text{since z $\neq 0$}}\\
&=\lim_{z\to0} \frac{z\overline{z}}{z} \\
&=\lim_{z\to0} \overline{z}\\
&=\lim_{(x,y)\to(0,0)}x-iy \\
&=0
\end{align}

4
##### Quiz 1 / QUIZ1 TUT0101
« on: January 28, 2020, 10:11:03 AM »
Describe the locus of points z satisfying the given equation.
\begin{align}
Im(2iz) = 7  \\
z = x + iy,  x,y \in \mathbb{R} \\
Im[2i(x+iy)]&=7 \\
Im(2xi-2y) &= 7 \\
2x &= 7 \\
x &= \frac{7}{2}
\end{align}
The locus of point z is a vertical line cross $(\frac{7}{2}, 0)$.

5
##### Quiz-4 / TUT0801 QUIZ4
« on: October 18, 2019, 02:02:26 PM »

Solve the initial value problem.
\begin{align*}
y'' -6y' +9y = 0\\
y(0) = 0, y'(0) = 2
\end{align*}

\begin{align}
r^2 - 6r +9 = 0 \\
(r-3)^2 = 0\\
r_1 = r_2 = 3\\\
y(t) = c_1e^{3t}+c_2te^{3t}\\
y(0)=c_1e^0+0=0\\
c_1=0
\end{align}
\begin{align}
y'(0)&=3c_1e^{3t}+c_2(e^{3t}+3te^{3t})\\
&=3c_1e^{3t}+c_2e^{3t}+3c_2 te^{3t} \\
&=3c_1e^0+c_2e^0+0\\
&=3c_1+c_2=2 \\
c_2&=2-3c_1\\
&=2-0=2\\
y(t)&=2te^{3t}
\end{align}

6
##### Quiz-4 / QUIZ 4 TUT0801
« on: October 18, 2019, 01:59:38 PM »
Solve the initial value problem.
\begin{align*}
y'' -6y' +9y = 0\\
y(0) = 0, y'(0) = 2
\end{align*}

\begin{align}
r^2 - 6r +9 = 0 \\
(r-3)^2 = 0\\
r_1 = r_2 = 3\\\
y(t) = c_1e^{3t}+c_2te^{3t}\\
y(0)=c_1e^0+0=0\\
c_1=0
\end{align}
\begin{align}
y'(0)&=3c_1e^{3t}+c_2(e^{3t}+3te^{3t})\\
&=3c_1e^{3t}+c_2e^{3t}+3c_2 te^{3t} \\
&=3c_1e^0+c_2e^0+0\\
&=3c_1+c_2=2 \\
c_2&=2-3c_1\\
&=2-0=2\\
y(t)&=2te^{3t}
\end{align}

7
##### Quiz-3 / TUT0801 QUIZ3
« on: October 11, 2019, 02:07:53 PM »
Find the Wronskian of the given pair of functions.
\begin{align*}
cost, sint
\end{align*}

\begin{align}
W =
\begin{vmatrix}
cost & sint \\
-sint & cost  \\
\end{vmatrix}
&= cos^2 t - (-sin^2 t)\\
&=1
\end{align}

8
##### Quiz-2 / TUT 0801 QUIZ2
« on: October 04, 2019, 02:01:48 PM »
\begin{align}
1 + (\frac{x}{y}-sin(y))y' = 0
\end{align}

$M = 1$    $N = \frac{x}{y} - siny$    $M_y = 0$     $N_x = \frac{1}{y}$

$\implies M_y \neq N_x$

\begin{align}
\textbf{Find integrating factor} \\
R_1 &= \frac{M_y-N_x}{M} = \frac{0-\frac{1}{y}}{1} = - \frac{1}{y} \\
\mu &= e^{-\int R_1 dy} = e^{-\int -\frac{1}{y} dy} = e^{lny} = y \\
\textbf{Multiply both sides by $\mu$} \\
y + (\frac{x}{y} - siny)yy' &= 0 \\
y + (x-ysiny)y' &= 0
\end{align}

$P = y$    $Q = x-ysiny$

\begin{align}
f &= \int {P dx} = \int {y dx} = xy + g(y) \\
f_y &= Q = x - ysiny = x + g'(y) \\
g'(y) &= -ysiny \\
g(y) &= -\int {ysiny dy} \\
\textbf{Integrating by parts}
\end{align}

$u = y$    $du = dy$    $v = -cosy$    $dv = sinydy$

\begin{align}
g(y) &= -[-ycosy-\int {-cosy dy]} +c \\
&= ycosy - siny + c \\
\implies f = xy+ycosy-siny=c
\end{align}

9
##### Quiz-1 / TUT0801 QUIZ1
« on: September 27, 2019, 02:50:13 PM »
\begin{align}
y'+\frac{1}{t}y &= 3cos(2t)\\
p(t) &= \frac{1}{t} \\
m &= e^{-\int {p(t)} dt} \\
&= e^{-\int {\frac{1}{t}} dt} \\
&= e^{-lnt} \\
&= \frac{1}{t} \\
n &= \int {\frac{3cos(2t)}{\frac{1}{t}}} dt \\
&=  \int {3tcos(2t)} dt \\
\textbf{Integrating by parts:} \\
u = t, du = dt, v = \frac{1}{2}sin(2t), dv = cos(2t)dt \\
n &= 3[\frac{1}{2}tsin(2t)- \int {\frac{1}{2}sin(2t)} dt] + c \\
&= 3[\frac{1}{2}tsin(2t)+ \frac{1}{4}cos(2t)] + c \\
y &= mn \\
&= \frac{1}{t}[3[\frac{1}{2}tsin(2t)+ \frac{1}{4}cos(2t)] + c] \\
&= \frac{3}{2}sin(2t) + \frac{3cos(2t)}{4t} + \frac{c}{t}
\end{align}

Pages: [1]