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### Messages - Victor Ivrii

Pages: [1] 2 3 ... 166
1
##### Chapter 2 / Re: S2.4 online textbook
« on: Today at 08:32:25 AM »
Indeed, it was a mistype. Corrected. Thanks.

Please change you screen name

2
##### Chapter 2 / Re: S2.2P Problem 2 (6)
« on: Today at 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

u_t+3u_x-2u_y=x
\label{eqn-1}

with the IVP

u |_{t=0}=0.
\label{eqn-2}

Writing characteristics

\frac{dt}{1}=\frac{dx}{3}=\frac{dy}{-2}=\frac{du}{x}.
\label{eqn-3}

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

\boxed{u = \frac{x^2}{6} + \varphi (x-3t, y+2t)}
\label{eqn-4}

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

\boxed{u = \frac{x^2}{6} - \frac{(x-3t)^2}{6} = xt - \frac{3}{2}t^2.}
\label{eqn-5}

3
##### 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

4
##### 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

5
##### 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.

6
##### 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.

7
##### 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.

8
##### 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

9
##### 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

10
##### 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,

11
##### 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
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }$$

12
##### Term Test 2 / You may post solutions
« on: November 19, 2019, 04:25:47 AM »
After an (almost) perfect solution is posted, no need to post the same solution

13
##### Term Test 2 / Problem 4 (noon)
« on: November 19, 2019, 04:24:35 AM »
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 1 & 3\\ -2 &-3\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

14
##### Term Test 2 / Problem 4 (morning)
« on: November 19, 2019, 04:23:46 AM »
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 2 & -3\\ 4 &-2\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

15
##### Term Test 2 / Problem 4 (main sitting)
« on: November 19, 2019, 04:23:20 AM »
Find the general real solution to
$$\mathbf{x}'=\begin{pmatrix} 3 & 3\\ -2 &-1\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

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