### Author Topic: TT2 #2  (Read 4940 times)

#### Victor Ivrii

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##### TT2 #2
« on: November 19, 2014, 08:46:06 PM »
(a) Determine the type of behavior (phase portrait) near the origin of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t=y\ , \\
&y'_t=2x + y .
\end{aligned}\right.
\end{equation*}

(b) Solve for the system of ODEs from \textbf{2a} the initial value problem with $\ x(0)=2 \ ,\ y(0)=1\$.

#### Yuan Bian

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##### Re: TT2 #2
« Reply #1 on: November 19, 2014, 10:19:36 PM »
x'= (0  1)
y'= (2  1)
r2-r-2=0
(r-2)(r+1)=0
r1=2, r2=-1
r1>0> r2
unstable saddle
(x)=c1e2t(1)+c2e-t(1)
(y)          (2)          (-1)
2b) c1+c2=2
2c1-c2=1
so c1=1,c2=1
(x)=e2t(1)+e-t(1)
(y)       (2)     (-1)
« Last Edit: November 19, 2014, 11:56:24 PM by Yuan Bian »

#### Chang Peng (Eddie) Liu

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##### Re: TT2 #2
« Reply #2 on: November 19, 2014, 10:20:52 PM »
This is what I did... Please feel free to correct me if I'm wrong! :p

#### Chang Peng (Eddie) Liu

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##### Re: TT2 #2
« Reply #3 on: November 19, 2014, 10:21:18 PM »
2b

#### Yuan Bian

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##### Re: TT2 #2
« Reply #4 on: November 19, 2014, 10:31:33 PM »
different c1 and c2...
and c2 is still in your final answer.
« Last Edit: November 19, 2014, 11:31:25 PM by Yuan Bian »

#### Tao Hu

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##### Re: TT2 #2
« Reply #5 on: November 20, 2014, 05:09:36 PM »
A typed solution may be more helpful
2(a):
\begin{equation*} \textbf{x}'=\begin{pmatrix}\hphantom{-}0 & 1\\\hphantom{-}2 &1 \end{pmatrix}\textbf{x}\ . \end{equation*}

find eigenvalues

\begin{equation*} r^2 - trace(A) + (ad- bc) =  r^2 -r - 2 = 0\implies r_1= 2, r_2=-1\end{equation*}

then, find eigenvectors, when r = 2

\begin{equation*} \begin{pmatrix} 0 - 2 & \hphantom{-}1\\  \hphantom{-}2 &1 -2\end{pmatrix}\begin{pmatrix}\mathbf{\xi}{^1}{_1}\\\mathbf{\xi}{^1}{_2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

then

\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}1\\2\end{pmatrix}\end{equation*}

when r = -1

\begin{equation*} \begin{pmatrix} 0  + 1 & \hphantom{-}1\\  \hphantom{-}2 &1 +1\end{pmatrix}\begin{pmatrix}\mathbf{\xi}{^2}{_1}\\\mathbf{\xi}{^2}{_2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

then

\begin{equation*}\mathbf{\xi}^2=\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}

Therefore

\begin{equation*}\mathbf{x}(t)= C_1e^{2t}\begin{pmatrix}1\\2\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}

Real eigenvalues with distinct signs, the type of origin is a saddle point.

2(b):

\begin{equation*} \mathbf{x}(0)=C_1+ C_2=2\\\mathbf{y}(0)=2C_1- C_2=1 \end{equation*}
Easy Calculation:
\begin{equation*} C_1 = 1, C_2 = 1 \end{equation*}
final answer:
\begin{equation*}\mathbf{x}(t)= e^{2t}\begin{pmatrix}1\\2\end{pmatrix}+ e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}
« Last Edit: November 20, 2014, 05:35:46 PM by Tao Hu »

#### Victor Ivrii

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##### Re: TT2 #2
« Reply #6 on: November 20, 2014, 06:35:16 PM »

#### Yuan Bian

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##### Re: TT2 #2
« Reply #7 on: November 20, 2014, 08:58:13 PM »
I agree with you that typed version is better. As math students, I think many students wants to get bonus by answering online question and using typed version. but situation now, it's few of us can type math formula using latex(? or something else....),
As a math course, I think we should focus on math idea and correctness.
and I think copy my answer above to get bonus is not fair, and even not sure I get bonus or not.
« Last Edit: November 20, 2014, 09:04:42 PM by Yuan Bian »

#### Victor Ivrii

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##### Re: TT2 #2
« Reply #8 on: November 21, 2014, 01:05:50 AM »
You got bonus. But bonus in this case is for the usefulness to community (for math part everyone gets much more as a regular mark for a test).  In this case typed solution which could be edited by the author or by a moderator is much more useful