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Messages - Victor Ivrii

Pages: 1 2 [3] 4 5 ... 169
31
Chapter 3 / Re: Section3.2
« on: February 14, 2020, 12:51:46 AM »
Sometimes, for some functions there are shortcuts, but generally not

32
Chapter 2 / Re: 2.2 home assignment question 18
« on: February 12, 2020, 03:12:59 PM »
Fixed. Thanks!

33
Chapter 2 / Re: 2.2 home assignment question 18
« on: February 11, 2020, 07:38:13 AM »
Please correct what you typed

34
Chapter 2 / Re: 2.3 question 3
« on: February 11, 2020, 07:37:38 AM »
Your answer is correct ($-\pi i$).

If you see discrepancies, report them

35
Chapter 2 / Re: 2.3 textbook
« on: February 09, 2020, 06:54:26 PM »
Yes, wrong sign  in the middle (should be $\drac{1}{2i}=-\dfrac{i}{2}$) but the final result is correct

36
Chapter 2 / Re: Section 2.6
« on: February 03, 2020, 08:18:07 AM »
Since $u$ is defined for $x<0$ and for $x>0$, you calculate limits from the left ($u|_{x=-0}$) and from the right ($u|_{x=+0}$). This is a standard notation

37
Quiz 2 / Re: Quiz2 TUT5101
« on: January 31, 2020, 10:49:26 PM »
In the tsecond/third lines should be $dt$, ..., not $\partial t$,...

38
Quiz 2 / Re: TUT5101 QUIZ2
« on: January 31, 2020, 09:05:11 AM »
You cannot use LH out of the box for complex variables. You need to consider $z=re^{i\theta}$ and consider $\bigl|\dfrac{\operatorname{Log}(z)}{z}\bigr|$ ...

39
Quiz 2 / Re: TUT5101 QUIZ2
« on: January 30, 2020, 06:39:52 AM »
Wrong reasoning

40
Quiz 2 / Re: TUT0701 QUIZ2
« on: January 30, 2020, 06:39:22 AM »
Wrong reasoning

41
Quiz 1 / Re: TUT5101
« on: January 26, 2020, 04:10:50 AM »
Escape cos, sin, log, .... : \cos, \sin, \log

42
Quiz 1 / Re: Quiz1 TUT5101
« on: January 24, 2020, 11:19:52 AM »
Correct. Please write partial derivatives as $\frac{\partial u}{\partial x}$ etc

43
Chapter 2 / Re: S2.1 online textbook problem #23
« on: January 22, 2020, 03:15:18 AM »
Since solving $x,y$ you get a circle of the constant radius $r$, you can parametrize it $x=r\cos(t)$, $y=r\sin(t)$; then integration will be easy. Don't forget in the end to get rid of $t,r$, leaving only $x,y$

44
Chapter 2 / Re: S2.4 online textbook
« on: January 21, 2020, 08:32:25 AM »
Indeed, it was a mistype. Corrected. Thanks.


Please change you screen name

45
Chapter 2 / Re: S2.2P Problem 2 (6)
« on: January 21, 2020, 08:27:19 AM »
Please, use MathJax for proper displaying equations. Also you need either repeat a problem here, or to provide a clickable link, like this

So, we have equation
\begin{equation}
u_t+3u_x-2u_y=x
\label{eqn-1}
\end{equation}
with the IVP
\begin{equation}
u |_{t=0}=0.
\label{eqn-2}
\end{equation}
Writing characteristics
\begin{equation}
\frac{dt}{1}=\frac{dx}{3}=\frac{dy}{-2}=\frac{du}{x}.
\label{eqn-3}
\end{equation}
Solving the first equality: $x-3t=c_1$, second $y+2t =c_2$ and the last one $u-\frac{x^2}{6}=C$, with $c_1, c_2, C$ constants along characteristics, which are marked by $c_1,c_2$. Then $C=\varphi(c_1,c_2)$ and finally
\begin{equation}
\boxed{u = \frac{x^2}{6} + \varphi (x-3t, y+2t)}
\label{eqn-4}
\end{equation}
is the general solution to (\ref{eqn-1}).

Plugging (\ref{eqn-4}) into (\ref{eqn-2}) we get $ \frac{x^2}{6} + \varphi (x, y) =0\implies \varphi(x,y)= -\frac{x^2}{6}$ and plugging into (\ref{eqn-4}) we get
\begin{equation}
\boxed{u = \frac{x^2}{6} - \frac{(x-3t)^2}{6} = xt - \frac{3}{2}t^2.}
\label{eqn-5}
\end{equation}

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