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### Messages - Victor Ivrii

Pages: 1 ... 4 5  7 8 ... 152
76
##### 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

77
##### 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.

78
##### 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.

79
##### 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.

80
##### 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 $(0,0)$"

81
##### 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

82
##### 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,

83
##### 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
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }$$

84
##### 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

85
##### 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.

86
##### 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.

87
##### 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.

88
##### Term Test 2 / Problem 3 (noon)
« on: November 19, 2019, 04:22:43 AM »
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} 1 &2\\ 1 &0\end{pmatrix}\mathbf{x}+ \begin{pmatrix} 0 \\[1pt] \dfrac{6 e^{3t }}{e^{2t}+1}\end{pmatrix}.$$

89
##### Term Test 2 / Problem 3 (morning)
« on: November 19, 2019, 04:21:57 AM »
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} -2 &1\\ -1 &0\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} -2 &1\\ -1 &0\end{pmatrix}\mathbf{x}+ \begin{pmatrix} 0 \\ \dfrac{e^{-t}} {t^2+1} \end{pmatrix}.$$

90
##### Term Test 2 / Problem 3 (main sitting)
« on: November 19, 2019, 04:21:01 AM »
(a) Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 1 &1\\ -2 &4\end{pmatrix}\mathbf{x}$$
classify fixed point $(0,0)$ and sketch trajectories.

(b) Find the general solution
$$\mathbf{x}'=\begin{pmatrix} 1 &1\\ -2 &4\end{pmatrix}\mathbf{x}+ \begin{pmatrix} \dfrac{e^{4t }}{e^{2t}+1} \\ 0\end{pmatrix}.$$

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