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Messages - kaye

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1
Quiz 3 / Quiz 3 line integral \int Re(z)dz
« on: October 09, 2020, 12:03:11 PM »
Problem. Compute the following line integral:
$$\int_{\gamma} Re(z)dz,$$

where $\gamma$ is the line segment from 1 to $i$.
\begin{align*}
\gamma(t) &= (1-t)\cdot 1 + t \cdot i\\
&= (1-t)+it \tag*{where $t \in [0,1]$}\\
\gamma'(t) &= i-1
\end{align*}

Then, we compute the line integral:
\begin{align*}
\int_{\gamma}Re(z)dz &= \int_0^1 Re[(1-t)+it]\cdot(i-1) \,dt\\
&= \int_0^1 (1-t)(i-1) \,dt\\
&= (i-1) \int_0^1 (1-t) \,dt\\
&= (i-1)\left[t-\frac{1}{2}t^2\right]_{t=0}^{t=1}\\
&= (i-1)\left(1-\frac{1}{2}\right)\\
&= (i-1)\frac{1}{2}
\end{align*}

The line integral $\int_{\gamma} Re(z)dz$ where $\gamma$ is the line segment from 1 to $i$ is $(i-1)\frac{1}{2}$.

2
Quiz 2 / Quiz 2 f(z)=(z-2)\log|z-2| at z_0=2
« on: October 02, 2020, 01:06:57 PM »
Problem(3pt). Find the limit of each function at the given point, or explain why it does not exist.
$$f(z)=(z-2)\log|z-2| \text{ at } z_0=2$$

Let $z'=z-2$, then
\begin{align*}
\lim_{z \to \infty} |f(z)| &=\lim_{z' \to 0} |f(z)|\\
&= \lim_{z' \to 0} |z'\log|z'||\\
&= \lim_{z' \to 0} \frac{\log|z'|}{\frac{1}{|z'|}}
\end{align*}
When $z' \to 0$, we get $\frac{\infty}{\infty}$, now use the L'Hospital's Rule we have:
\begin{align*}
\lim_{z \to \infty} |f(z)| &= \lim_{z' \to 0} \frac{\log|z'|}{\frac{1}{|z'|}}\\
&= \lim_{z' \to 0} \frac{\frac{1}{|z'|}}{\frac{-1}{|z'|^2}}\\
&= \lim_{z' \to 0} - |z'|\\
&= 0
\end{align*}
The limit of $f(z)$ at $z_0=2$ is 0.

3
Quiz 1 / Quiz 1 (Quiz-5101)
« on: September 25, 2020, 01:09:15 PM »
Problem(3pt). Find all solutions of the given equation:
$$z^5=-i .$$

Let $z=r(cos \theta + isin \theta)$, then $z^5 = r^5[cos(5\theta)+isin(5\theta)]$:
$$-i=1\left[cos\left(\frac{3\pi}{2}+2k\pi\right) +isin\left(\frac{3\pi}{2}+2k\pi\right)\right].$$

Therefore we get
$$r=1^{\frac{1}{5}}=1 \text{ and } \theta=\frac{1}{5}\left(\frac{3\pi}{2}+2k\pi\right) \text{ for } k=0,1,2,3,4$$

The five solutions of the given equation:
\begin{align*}
k&=0: \theta = \frac{3\pi}{10} \Rightarrow z=cos\left(\frac{3\pi}{10}\right) + isin\left(\frac{3\pi}{10}\right)\\
k&=1: \theta = \frac{7\pi}{10} \Rightarrow z=cos\left(\frac{7\pi}{10}\right) + isin\left(\frac{7\pi}{10}\right)\\
k&=2: \theta = \frac{11\pi}{10} \Rightarrow z=cos\left(\frac{11\pi}{10}\right) + isin\left(\frac{11\pi}{10}\right)\\
k&=3: \theta = \frac{3\pi}{2} \Rightarrow z=cos\left(\frac{3\pi}{2}\right) + isin\left(\frac{3\pi}{2}\right)=-i\\
k&=4: \theta = \frac{19\pi}{10} \Rightarrow z=cos\left(\frac{19\pi}{10}\right) + isin\left(\frac{19\pi}{10}\right)
\end{align*}

4
Quiz-4 / Quiz 4 TUT0402
« on: October 18, 2019, 02:00:00 PM »
Question:   Find the general solution of the given differential equation: $y'' - y' - 2y = cosh(2t)$    Hint: $cosh(t) = \frac{e^t + e^{-t}}{2}$

Write the differential equation into:
$$y'' - y' -2y = \frac{1}{2}e^{2t} + \frac{1}{2}e^{-2t}$$

homogeneous solution:
\begin{align}
r^2 - r - 2 &= 0 \notag\\
(r - 2)(r + 1) &= 0 \notag
\end{align}

$r_1 = 2$ and $r_2 = -1 \implies y_c = c_1e^{2t} + c_2e^{-t}$

$y'' - y' - 2y = \frac{1}{2}e^{2t}$:
\begin{align}
y_{p1} &= Ate^{2t}\notag\\
y' &= (A + 2At)e^{2t}\notag\\
y'' &= (4A + 4At)e^{2t}\notag
\end{align}
\begin{align}
(4A + 4At)e^{2t} - (A + 2At)e^{2t} - 2Ate^{2t} &= \frac{1}{2}e^{2t}\notag\\
4A + 4At - A - 2At - 2At &= \frac{1}{2}\notag\\
3A &= \frac{1}{2}\notag\\
A &= \frac{1}{6}\notag
\end{align}

$y'' - y' - 2y = \frac{1}{2}e^{-2t}$:
\begin{align}
y_{p2} &= Be^{-2t}\notag\\
y' &= (-2B)e^{2t}\notag\\
y'' &= (4B)e^{2t}\notag
\end{align}
\begin{align}
(4B)e^{2t} - (-2B)e^{2t} - 2Be^{-2t} &= \frac{1}{2}e^{-2t}\notag\\
4B + 2B - 2B &= \frac{1}{2}\notag\\
4B &= \frac{1}{2}\notag\\
B &= \frac{1}{8}\notag
\end{align}

from $y_c = c_1e^{2t} + c_2e^{-t}$, $y_{p1} = \frac{1}{6}te^{2t}$, and $y_{p2} = \frac{1}{8}e^{-2t}$ we get:
$$y = c_1e^{2t} + c_2e^{-t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t}$$

5
Quiz-3 / Quiz3 TUT0402
« on: October 11, 2019, 01:59:59 PM »
Question: Find the solution of the given initial value problem.
$$y''+5y'+3y=0, y(0)=1, y'(0)=0$$

Find roots of characteristic equation:

$$r = \frac{-5\pm\sqrt{5^2-4(1)(3)}}{2(1)} = \frac{-5\pm\sqrt{13}}{2}$$

So the general solution is:

\begin{align}
y &= c_1e^{\frac{-5+\sqrt{13}}{2}}+
c_2e^{\frac{-5-\sqrt{13}}{2}}\notag\\
\implies y' &= \frac{-5+\sqrt{13}}{2}c_1e^{\frac{-5+\sqrt{13}}{2}}
- \frac{5+\sqrt{13}}{2}c_2e^{\frac{-5-\sqrt{13}}{2}}\notag
\end{align}

Plug in the given initial value:
\begin{align}
y(0)=1 &\implies c_1+c_2=1\notag\\
y'(0)=0 &\implies \frac{-5+\sqrt{13}}{2}c_1- \frac{5+\sqrt{13}}{2}c_2=0\notag
\end{align}

Solving this system for $c_1$, $c_2$, we get:
\begin{align}
c_1 &= \frac{5+\sqrt{13}}{2\sqrt{13}}\notag\\
c_2 &= 1-\frac{5+\sqrt{13}}{2\sqrt{13}}\notag
\end{align}

Therefore:
$$y= \frac{5+\sqrt{13}}{2\sqrt{13}}e^{\frac{-5+\sqrt{13}}{2}}+ \left(\frac{\sqrt{13}-5}{2\sqrt{13}} \right) e^{\frac{-5-\sqrt{13}}{2}}$$

6
Quiz-2 / QUIZ 2 TUT0402
« on: October 04, 2019, 02:00:02 PM »
Question: Solve the given equation $(2xy^2+2y)+(2x^2y+2x)y'=0$

\begin{align}
M(x,y)=2xy^2+2y &\implies M_y(x,y)=4xy+2\notag\\
N(x,y)=2x^2y+2x &\implies N_x(x,y)=4xy+2\notag
\end{align}

Since $M_y=N_x$, so we know that the given differential equation is exact, that is, there exist $\varphi(x,y)$ such that $\varphi_x=M$ and $\varphi_y=N$.

\begin{align}
\varphi_x(x,y) &= 2xy^2+2y \implies \varphi(x,y) = \int 2xy^2+2y dx = x^2y^2+2xy+f(y)\notag\\
\varphi_y(x,y) &= 2x^2y+2x+f'(y) \implies f'(y)=0 \implies f(y)=C\notag
\end{align}

So

$$\varphi(x,y)=x^2y^2+2xy+C$$
$$x^2y^2+2xy=C$$

7
Quiz-1 / QUIZ 1 TUT0402
« on: September 27, 2019, 02:00:09 PM »
Question: Find the general solution of the given function.
$$\frac{dy}{dx}=-\frac{(4x+3y)}{(2x+y)}$$

On RHS, divide both the numerator and the denominator by $x$:
$$\frac{dy}{dx} = -\frac{(4x+3y)}{(2x+y)} = -\frac{4\frac{x}{x}+3\frac{y}{x}}{2\frac{x}{x}+\frac{y}{x}}$$

Let $u = \frac{x}{y}$, so $y = ux$, then on LHS we have
$$\frac{dy}{dx} = \frac{d(ux)}{dx} = \frac{du}{dx}x + u$$

and on RHS we have
$$-\frac{4\frac{x}{x}+3\frac{y}{x}}{2\frac{x}{x}+\frac{y}{x}} = -\frac{(4+3u)}{(2+u)}$$

put them together:
\begin{align}
\frac{du}{dx}x+u & = -\frac{(4+3u)}{(2+u)}\notag\\
\frac{du}{dx}x & = -\frac{4+3u+2u+u^2}{2+u}\notag\\
& = -\frac{u^2+5u+4}{2+u}\notag\\
& = -\frac{(u+1)(u+4)}{2+u}\notag
\end{align}

rearrange this equation:
\begin{align}
-\frac{2+u}{(u+1)(u+4)}du &=\frac{1}{x}dx\notag\\
-\int\frac{2+u}{(u+1)(u+4)}du &=\int\frac{1}{x}dx\notag\\
-\int\frac{1}{3}\frac{1}{u+1}+\frac{2}{3}\frac{1}{u+4}du & =\int\frac{1}{x}dx\notag\\
\frac{1}{3}\ln{|u+1|}+\frac{2}{3}\ln{|u+4|}&=-\ln{|x|}+C\notag\\
\ln{|u+1|}+2\ln{|u+4|} &=-3\ln{|x|}+3C\notag\\
\ln{|\frac{y}{x}+1|}+2\ln{|\frac{y}{x}+4|} &=-3\ln{|x|}+3C\notag\\
\ln{|\frac{y+x}{x}|}+2\ln{|\frac{y+4x}{x}|} &=-3\ln{|x|}+3C\notag\\
\ln{|y+x|}-\ln{|x|}+2(\ln{|y+4x|}-\ln{|x|}) &=-3\ln{|x|}+3C\notag\\
\ln{|y+x|}+2\ln{|y+4x|} &=3C\notag\\
e^{\ln{|y+x|}}+e^{2\ln{|y+4x|}} &=e^{3C}\notag\\
|y+x||y+4x|^2 &=e^{3C}\notag
\end{align}

So, the general solution is $|y+x||y+4x|^2=C$.

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