### 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 - Victor Ivrii

Pages: 1 ... 6 7  9 10 ... 120
106
##### Quiz 2 / Re: TUT0701 QUIZ2
« on: January 30, 2020, 06:39:22 AM »
Wrong reasoning

107
##### Quiz 1 / Re: TUT5101
« on: January 26, 2020, 04:10:50 AM »
Escape cos, sin, log, .... : \cos, \sin, \log

108
##### 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

109
##### 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 $x,y$

110
##### Chapter 2 / Re: S2.4 online textbook
« on: January 21, 2020, 08:32:25 AM »
Indeed, it was a mistype. Corrected. Thanks.

111
##### 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}

112
##### Final Exam / Ab solutely no posting before my command
« on: December 21, 2019, 06:31:02 AM »
All posts removed. Users who made them are not allowed to post on forum

113
##### Chapter 9 / Re: the stability characteristics of all periodic solutions
« on: December 16, 2019, 03:38:04 PM »
Limit cycles (not circles) are not covered by final exam. In contrast to spiral point these cycles have two sides: external and internal. See picture

114
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 24, 2019, 11:04:24 AM »
What everybody is missing

we see that characteristic roots $k_{1,2}=-1\pm \sqrt{2}i$ are complex, with negative real part. So, it is  stable focus  and with  clock-wise  orientation  since the bottom-left element is negative.

115
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 24, 2019, 11:00:36 AM »
What everybody is missing:

we see that characteristic roots $k_{1,2}= \pm \sqrt{8}i$ are purely imaginary. So, it is  center  and with  counter-clock-wise  orientation  since the bottom-left element is positive.

116
##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 24, 2019, 10:45:37 AM »
What everybody is missing

it is  unstable focus  and with  clock-wise  orientation  since the bottom-left element is negative.

117
##### Term Test 2 / Re: Problem 3 (noon)
« on: November 24, 2019, 10:00:42 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"

118
##### Term Test 2 / Re: Problem 3 (morning)
« on: November 24, 2019, 09:55:08 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"

stable improper node; since the bottom left element is negative, it is clockwise

119
##### Term Test 2 / Re: Problem 3 (main sitting)
« on: November 24, 2019, 09:41:32 AM »
What everybody is missing:
Part of the problem "classify fixed point $(0,0)$".
It is unstable node,

120
##### Term Test 2 / Re: Problem 1 (noon)
« on: November 24, 2019, 08:41:08 AM »
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+c_1 \Bigr)e^{t} + \Bigl( \arctan (e^t)+c_2\Bigr)e^{2t}. }$$
and
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }$$

Pages: 1 ... 6 7  9 10 ... 120