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### Messages - Jingze Wang

Pages: 1 [2]
16
##### Quiz-7 / Re: Q7 TUT 0501
« on: November 30, 2018, 04:29:02 PM »
To find critical points, let $x'=0, y'=0$
Thus, $-2x-y-x(x^2+y^2)=0, x-y+y(x^2+y^2)=0$
$(0,0)$ is one crtical point
Also, $(0.330757,1.09242), (0.330757,-1.09242)$ are also critical points

Let $f(x)=-2x-y-x(x^2+y^2), g(x)=x-y+y(x^2+y^2)$
$f_x=-2-3x^2-y^2, f_y=-1-2xy$
Also, $g_x=1+2xy, g_y=-1+x^2+3y^2$

$J=\begin{bmatrix} -2-3x^2-y^2&-1-2xy\\ 1+2xy&-1+x^2+3y^2\\ \end{bmatrix}$

Plug in the critical points to find eigenvalues of each linear system
For $(0,0)$,

We get $\begin{bmatrix} -2&--1\\ 1&-1\\ \end{bmatrix}$,
$r=\frac{-3\pm\sqrt{-3}i}{2}$
Therefore, it is stable at $(0,0)$

Similarly, for others critical points,
Plug in and eigenvalues are $-3.5092, 2.6771$
Thus, it is a saddle point therefore unstable

17
##### Term Test 2 / Re: TT2A-P4
« on: November 22, 2018, 03:46:14 PM »
Hello Samarth, I think your graph is not right, since the eigenvalues have no real parts, then graph should be center instead of spiral.

18
##### Term Test 2 / Re: TT2B-P4
« on: November 20, 2018, 03:37:20 PM »
\begin{align*}
Let A &= \begin{bmatrix}
-3 & -2 \\
5 & -5
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-3 - \lambda & -2 \\
5 & -5 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (5 + \lambda)(3 + \lambda) + 10\\
~\\
&= \lambda^{2} + 8\lambda +25\\
~\\
&= (\lambda +4)^{2} + 9 = 0\\
~\\
\lambda &= -4 \pm 3i \\
~\\
For \lambda = -4 + 3i, A - \lambda I &= \begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}\\
~\\
Since,\ nul(\begin{bmatrix}
1 - 3i & -2 \\
5 & 1 - 3i
\end{bmatrix}) &= span\{\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\  v &= \begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
Since \ e^{\lambda t}v &= e^{-4 + 3i}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}e^{3it}\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
&= e^{-4t}(cos3t + isin3t)\begin{bmatrix}
3i + 1\\
5 \\
\end{bmatrix}\\
~\\
&= e^{-4t}\begin{bmatrix}
3icos3t - 3sin3t + cos3t + isin3t\\
5cos3t +5isin3t \\
\end{bmatrix}\\
~\\
So, \ \phi_{1}(t) &= e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix}\\
~\\
\phi_{2}(t) &= e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
~\\
Thus, \ x(t) &= c_{1}\phi_{1}(t) + c_{2}\phi_{2}(t)\\
~\\
&= c_{1}e^{-4t}\begin{bmatrix}
- 3sin3t + cos3t\\
5cos3t\\
\end{bmatrix} + c_{2}e^{-4t}\begin{bmatrix}
3cos3t + sin3t\\
5sin3t \\
\end{bmatrix}\\
\end{align*}
(b) In the attachment.
Hi Yulin, I think you made a mistake in your graph, since c=5>0, your graph should be counterclockwise instead of clockwise

19
##### Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 02:35:45 PM »
here is my answer
Hi, my answer is same like yours except I got $C_1=-2ln2, C_2=0$
I am not confident about my answer, feel free to correct me

20
##### Term Test 2 / Re: TT2B-P2
« on: November 20, 2018, 11:17:19 AM »
a)
The coefficient of $y''$ is -1
then Wronskain is $Ce^{t}$
b)
Use homogeneous equation to find fundamental solutions
$y'''-y''+4y'-4y=0$
Then $r^3-r^2+4r-4=0$
Then $(r^2+4)(r-1)=0$
Then$r_1=-2i, r_2=2i, r_3=1$
Then the solution is $y=C_1\cos2t+C_2\sin2t+C_3e^t$
So
$W(y_1, y_2, y_3)(t) = \begin{bmatrix} \cos2t&\sin2t&e^{t}\\-2\sin2t&2\cos2t&e^{t}\\-4\cos2t&-4\sin2t&e^{t}\\ \end{bmatrix}=10e^{t}$
This is consistent with what we get in part (a)

c) Use undetermined coefficients method
Assume $y(t) = At\cos{2t}+Bt\sin{2t}$
$y'(t) = A\cos{2t}-2At\sin2t+B\sin2t+2Bt\cos2t$
$y''(t) = -4A\sin{2t}-4At\cos2t+8At\sin2t-8B\sin2t$
$y'''(t) = -12A\cos{2t}+8At\sin2t-12B\sin2t-8Bt\cos2t$

Plug into the equation,

We get $-8A-4B=8, 4A-8B=0$
we get $A=-\frac{4}{5}$
$B=-\frac{2}{5}$

Thus, $Y=C_1\cos2t+C_2\sin2t+C_3e^t-\frac{4}{5}t\cos2t-\frac{2}{5}t\sin2t$

21
##### Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 10:05:33 AM »
a)First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} 2&1\\ -3&-2\\ \end{bmatrix}$

$det(A-rI)=(2-r)/(-2-r)+3=0$

$r^2-1=0$

$r=\pm1$

When r=-1, $3x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}$ is the corresponding eigenvector
When r=1,  $x_1+x_2=0$
We get $\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}$ is the corresponding eigenvector
Then the general solution is $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$

22
##### Term Test 2 / Re: TT2-P4
« on: November 20, 2018, 08:53:07 AM »
Also, I am wondering can I just say it is spiral instead of focus

23
##### Term Test 2 / Re: TT2-P4
« on: November 20, 2018, 08:38:46 AM »
First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} 5&5\\ -5&-1\\ \end{bmatrix}$

$det(A-rI)=(5-r)(-1-r)+25=0$

$r^2-4r+20=0$

$r=\frac{4\pm\sqrt{-64}}{2}$

$r=2\pm4i$

Since they are complex conjugates
Then just use one of the eigenvector to find real solution
Use eigenvalue $r=3+4i$ to find its corresponding eigenvector

\begin{bmatrix}
3-4i&5\\
-5&-3-4i\\
\end{bmatrix}

The eigenvector is
$\begin{bmatrix} 5\\ 4i-3 \end{bmatrix}$

$X=e^{2+4i}\begin{bmatrix}5\\4i-3\end{bmatrix}\cos4t+i\sin4t$

Rearrange this, we get $U=e^{2t}\begin{bmatrix} \cos4t\\ -3cos4t-4\sin4t \end{bmatrix}$

Also, $V=e^{2t}\begin{bmatrix} 5\sin4t\\ 4\cos4t-3\sin4t \end{bmatrix}$

And they are real valued solutions

Since -5<0, it is clockwise

Also real parts is 2>0, it is unstable spiral

24
##### Quiz-6 / Re: Q6 TUT 0401
« on: November 18, 2018, 07:39:14 PM »
Our graphs are all clockwise

25
##### Quiz-6 / Re: Q6 TUT 0401
« on: November 18, 2018, 03:10:49 PM »
I am sorry, but what's the difference? I cannot find it.

26
##### Quiz-6 / Re: Q6 TUT 5101
« on: November 17, 2018, 04:00:25 PM »
First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} 2&-5\\ \alpha&-2\\ \end{bmatrix}$

$det(A-rI)=(2-r)(-2-r)+5\alpha=0$

$r^2-4+5\alpha=0$

$r=\frac{\pm\sqrt{16-20\alpha}}{2}$

Notice that $-4+5\alpha$ determines the type of roots, so $\alpha=4/5$ is the critical value

Case 1

When $-4+5\alpha=0, \alpha=0$, there is a repeated eigenvalue 0 with one eigenvector

Case 2

When $-4+5\alpha>0, \alpha>4/5$, there are two distinct complex eigenvalues without real parts

Case 3

When $-4+5\alpha<0, \alpha<4/5$, there are two distinct real eigenvalues with different signs

27
##### Quiz-6 / Re: Q6 TUT 0401
« on: November 17, 2018, 03:56:11 PM »
First, try to find the eigenvalues with respect to the parameter

$A=\begin{bmatrix} \alpha&1\\ -1&\alpha\\ \end{bmatrix}$

$det(A-rI)=(\alpha-r)(\alpha-r)+1=0$

$r^2-2{\alpha}r+\alpha^2+1=0$

$r=\frac{2\alpha\pm\sqrt{-4}}{2}$

$r=\alpha\pm2i$       $\color{red}{r_\pm =\alpha \pm i\; V.I.}$

Notice there are always complex eigenvalues, and $\alpha=0$ is critical value since $\alpha=0, \alpha>0, \alpha<0$ have different phase portraits

When $\alpha=0$ , real parts of eigenvalues are 0

When value of $\alpha$ is slightly below 0
Then $\alpha<0$ , real parts of eigenvalues are negative

When value of $\alpha$ is slightly above 0
Then $\alpha>0$ , real parts of eigenvalues are positive

28
##### MAT244--Misc / How to post the typed solutions properly
« on: November 16, 2018, 05:20:32 PM »
When I tried to post typed solutions in Latex, they just lost the format and looked like bad. Anyone could help me with this? Thanks.

29
##### Term Test 1 / Re: TT1 Problem 4 (morning)
« on: October 16, 2018, 08:29:02 AM »
At first, find complement solution
Let y''(t)+4y'(t)+5y(t)=0
we get r^2+4r+5=0
r=-2+i and -2-i
Yc=e^(-2t)(C1 cost+C2 sint)
Then, find particular solution
Let y''(t)+4y'(t)+5y(t)=e^(-2t)
Let Yp1=Ae^(-2t)
Yp1'=-2Ae^(-2t)
Yp1''=4Ae^(-2t)
Plug in we get A=1
Let y''(t)+4y'(t)+5y(t)=8sint
Let Yp2=Bcost+Csint
Yp2'=-Bsint+Ccost
Yp2''=-Bcost-Csint
Plug in, we get B=-1, C=1
Finally, the general solution is Y=e^(-2t)(C1 cost+C2 sint)+e^(-2t)-cost+sint

30
##### MAT244--Lectures & Home Assignments / Graphs about differential equations
« on: October 12, 2018, 08:55:19 AM »
Can we possibly be tested to draw graphs of solutions to differential equations?

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