Author Topic: Problem 3 (noon)  (Read 1238 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2550
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
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
$$
\mathbf{x}'=\begin{pmatrix} 1 &2\\
1 &0\end{pmatrix}\mathbf{x}+
\begin{pmatrix} 0 \\[1pt]
\dfrac{6 e^{3t }}{e^{2t}+1}\end{pmatrix}.$$

Changhao Jiang

  • Jr. Member
  • **
  • Posts: 8
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #1 on: November 19, 2019, 05:15:16 AM »
(a)
To find eigenvalues, $(1-\lambda x)(-\lambda)-2=0$,we can get $\lambda = 2$ or $\lambda = -1$
To find eigenvectors, when $\lambda=2$,
$\begin{pmatrix}
-1 & 2 \\
1 & -2
\end{pmatrix}
~
\begin{pmatrix}
1 & -2 \\
0 & 0
\end{pmatrix}$
let $x_2=t, x_1=2t$, so the eigenvector is \begin{bmatrix}1 \\ 2\end{bmatrix}
when $\lambda = -1$
$\begin{pmatrix}
2 & 2 \\
1 & 1
\end{pmatrix}$
~
$\begin{pmatrix}
1 & 1 \\
0 & 0
\end{pmatrix}$
let $x_2=-t, x_1=t$, so the eigenvector is \begin{bmatrix}1 \\ -1\end{bmatrix}
Therefore, the general solution is y=$c_1 \begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t}$ + $c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$

(b)
from (a), we can know $\phi(t)=
\begin{pmatrix}
e^{2t} & e^{-t} \\
2e^{2t} & -e^{-t}
\end{pmatrix}$
then
$\begin{pmatrix}
e^{2t} & e^{-t} \\
2e^{2t} & -e^{-t}
\end{pmatrix}
\begin{bmatrix}u_1'\\ u_2'\end{bmatrix}
=
\begin{bmatrix} 0 \\ \frac{6e^{3t}}{e^{2t}+1}\end{bmatrix}$
By ref form, we can know $u_1'= \frac{6e^t}{e^2t+1}, u_2' = 0$, then by integrating, we can get $u_1=6arctan(e^t)+c_1, u_2=c_2$
then the  general solution is $x(t)=(6arctan(e^t)+c_1)\begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t} + c_2\begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$
« Last Edit: November 19, 2019, 06:00:07 AM by Changhao Jiang »

xuanzhong

  • Jr. Member
  • **
  • Posts: 12
  • Karma: 1
    • View Profile
Re: Problem 3 (noon)
« Reply #2 on: November 19, 2019, 06:00:45 AM »
Here's the solution for sketching.

Jingjing Cui

  • Full Member
  • ***
  • Posts: 16
  • Karma: 5
    • View Profile
Re: Problem 3 (noon)
« Reply #3 on: November 19, 2019, 06:59:22 AM »
$$
a)det(A-\lambda I)=0\\
det\begin{vmatrix}
1-\lambda&2\\
1&-\lambda\\
\end{vmatrix}=0\\
(1-\lambda)(-\lambda)-2=0\\
\lambda_1=-1 \;\; \lambda_2=2\\
(A-\lambda I)x=0\\
\\
when\; \lambda=-1\\
\begin{pmatrix}
2&2\\
1&1\\
\end{pmatrix}->
\begin{pmatrix}
2&2\\
0&0\\
\end{pmatrix}\\
so\; 2x_1+2x_2=0\\
let\; x_2=t\;\;\;then\;x_1=-t\\
so\;the\;corresponding\;eigenvector\;is\;
(\begin{array}{cc} -1 \\ 1 \end{array})\\
when\; \lambda=2\\
\begin{pmatrix}
-1&2\\
1&-2\\
\end{pmatrix}->
\begin{pmatrix}
-1&2\\
0&0\\
\end{pmatrix}\\
so\; -x_1+2x_2=0\\
let\; x_2=t\;\;\;then\;x_1=2t\\
so\;the\;corresponding\;eigenvector\;is\;
(\begin{array}{cc} 2 \\ 1 \end{array})\\
so\;the\;general\;solution\;is\;x=c_1e^{-t}(\begin{array}{cc} -1 \\ 1 \end{array})+c_2e^{2t}(\begin{array}{cc} 2 \\ 1 \end{array})\\
$$

Jingjing Cui

  • Full Member
  • ***
  • Posts: 16
  • Karma: 5
    • View Profile
Re: Problem 3 (noon)
« Reply #4 on: November 19, 2019, 07:17:44 AM »
b)
$$
\phi(t) U'(t)=g(t)\\
\begin{pmatrix}
-e^{-t}&2e^{2t}\\
e^{-t}&e^{2t}\\
\end{pmatrix}(\begin{array}{cc} U_1' \\ U_2' \end{array})=(\begin{array}{cc} 0 \\\frac{6e^{3t}}{e^{2t}+1} \end{array})\\
-U_1'e^{-t}+2U_2'e^{2t}=0\\
U_1'e^{-t}+U_2'e^{2t}=\frac{6e^{3t}}{e^{2t}+1}\\
U_1'=\frac{4e^{4t}}{e^{2t}+1}\\
so\;U_1=2e^{2t}-2ln|e^{2t}+1|+C_1\\
U_2'=\frac{2e^{t}}{e^{2t}+1}\\
so\;U_2=2arctan(e^t)+C_2\\
x=\phi(t)U(t)\\
so\;the\;solution\;is\;:\\
x=(2e^{2t}-2ln|e^{2t}+1|+C_1)(\begin{array}{cc} -e^{-t} \\ e^{-t} \end{array})+(2arctan(e^t)+C_2)(\begin{array}{cc} 2e^{2t} \\ e^{2t} \end{array})\\
$$

OK, except LaTeX sucks:

1) text should not be a part of math formulae or included like \text{blah blah}
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
« Last Edit: November 24, 2019, 09:59:17 AM by Victor Ivrii »

NANAC

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 2
    • View Profile
Re: Problem 3 (noon)
« Reply #5 on: November 19, 2019, 09:15:02 AM »
Please see the attachment for the answer

anntara khan

  • Jr. Member
  • **
  • Posts: 6
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #6 on: November 19, 2019, 02:20:39 PM »
Please see attached file, Thanks!

baixiaox

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #7 on: November 19, 2019, 05:41:10 PM »
answer for tt2 question3 a)

baixiaox

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #8 on: November 19, 2019, 05:42:18 PM »
answer for tt2 question3 b)

Mingdi Xie

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #9 on: November 20, 2019, 02:45:02 PM »
this is my solution for part a

Mingdi Xie

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 0
    • View Profile
Re: Problem 3 (noon)
« Reply #10 on: November 20, 2019, 02:54:05 PM »
Solution for part b

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2550
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 3 (noon)
« Reply #11 on: November 24, 2019, 10:00:42 AM »
What everybody is missing
In problem got lost "classify point $(0,0)$"

saddle
« Last Edit: November 24, 2019, 10:02:33 AM by Victor Ivrii »