16
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.
Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Pages: 1 [2]
17
Term Test 2 / Re: TT2B-P4
« on: November 20, 2018, 03:37:20 PM »\begin{align*}Hi Yulin, I think you made a mistake in your graph, since c=5>0, your graph should be counterclockwise instead of clockwise
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.
18
Term Test 2 / Re: TT2-P3
« on: November 20, 2018, 02:35:45 PM »here is my answerHi, 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
19
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, $Y=C_1\cos2t+C_2\sin2t+C_3e^t-\frac{4}{5}t\cos2t-\frac{2}{5}t\sin2t$
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, $Y=C_1\cos2t+C_2\sin2t+C_3e^t-\frac{4}{5}t\cos2t-\frac{2}{5}t\sin2t$
20
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 $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$
$A=\begin{bmatrix}
2&1\\
-3&-2\\
\end{bmatrix}$
$det(A-rI)=(2-r)/(-2-r)+3=0$
$r^2-1=0$
$r=\pm1$
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 $$y=c_1\begin{pmatrix}\hphantom{-} {1 }\\{-3}\end{pmatrix}e^{-t}+c_2\begin{pmatrix}\hphantom{-} {1 }\\{-1}\end{pmatrix}e^{t}$$
21
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
22
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
$A=\begin{bmatrix}
5&5\\
-5&-1\\
\end{bmatrix}$
$det(A-rI)=(5-r)(-1-r)+25=0$
$r^2-4r+20=0$
$r=\frac{4\pm\sqrt{-64}}{2}$
$r=2\pm4i$
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, 03:10:49 PM »
I am sorry, but what's the difference? I cannot find it.
25
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
$A=\begin{bmatrix}
2&-5\\
\alpha&-2\\
\end{bmatrix}$
$det(A-rI)=(2-r)(-2-r)+5\alpha=0$
$r^2-4+5\alpha=0$
$r=\frac{\pm\sqrt{16-20\alpha}}{2}$
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
26
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
$A=\begin{bmatrix}
\alpha&1\\
-1&\alpha\\
\end{bmatrix}$
$det(A-rI)=(\alpha-r)(\alpha-r)+1=0$
$r^2-2{\alpha}r+\alpha^2+1=0$
$r=\frac{2\alpha\pm\sqrt{-4}}{2}$
$r=\alpha\pm2i$ $\color{red}{r_\pm =\alpha \pm i\; V.I.}$
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
27
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
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
Pages: 1 [2]