106
Quiz 2 / Re: TUT0701 QUIZ2
« on: January 30, 2020, 06:39:22 AM »
Wrong reasoning
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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.
Please change you screen name
Please change you screen name
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}
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.
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.
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.
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 $(0,0)$"
saddle
In problem got lost "classify point $(0,0)$"
saddle
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
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,
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
$$
\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }
$$
\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}. }
$$