### Author Topic: TT2 # 3  (Read 5927 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
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##### TT2 # 3
« on: November 19, 2014, 08:46:44 PM »
Find the general solution, sketch the phase portrait and determine the type of behavior near the origin of the the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t= -x - 4y\ , \\
&y'_t=\ \ x -\ y .
\end{aligned}\right.
\end{equation*}

#### Yuan Bian

• Jr. Member
• Posts: 14
• Karma: 4
##### Re: TT2 # 3
« Reply #1 on: November 19, 2014, 10:33:18 PM »
xâ€²=(âˆ’1 âˆ’4)(x) .
y'=(1    -1) (y)

find eigenvalues

det(Aâˆ’rI)=r2+2r+5=0
r1=-1+2i
r2=-1-2i
then, find eigenvectors

(-2i -4)   v1=(2i)
(1  -2i)         (1)

(2i -4)   v2=(2i)
(1  2i)         (-1)
x(t)=C1e(âˆ’1+2i)t(2i)+C2eâˆ’(-1-2i)t(2i)
(1)                      (-1)

stable spiral point
-4<0, counterclockwise
« Last Edit: November 19, 2014, 11:59:15 PM by Yuan Bian »

#### Chang Peng (Eddie) Liu

• Full Member
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##### Re: TT2 # 3
« Reply #2 on: November 19, 2014, 10:37:09 PM »
#3

#### Chang Peng (Eddie) Liu

• Full Member
• Posts: 19
• Karma: 7
##### Re: TT2 # 3
« Reply #3 on: November 19, 2014, 10:37:31 PM »
Picture

#### Tao Hu

• Jr. Member
• Posts: 6
• Karma: 3
##### Re: TT2 # 3
« Reply #4 on: November 19, 2014, 10:38:32 PM »
first write the equation in matrix form:

\begin{equation*}\textbf{x}'=\begin{pmatrix}\hphantom{-}-1 & -4\\\hphantom{-}1 &-1\end{pmatrix}\textbf{x}\ . \end{equation*}

find eigenvalues:

\begin{equation*} r^2 - trace(A) + (ad - bc)=  r^2+ 2r + 5 = 0\implies r_1= -1 + 2i,   r_2=-1 -2i\end{equation*}

then, find eigenvectors, which are conjugated

\begin{equation*} \begin{pmatrix} -1 - r & \hphantom{-}-4\\  \hphantom{-}1 &-1 -r\end{pmatrix}\begin{pmatrix}\mathbf{\xi}_1\\\mathbf{\xi}_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

find the two conjugate eigenvectors

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

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

therefore

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

the attachment is the phase portrait generated by PPLANE
(spiral point, stable)
« Last Edit: November 20, 2014, 07:15:34 PM by Tao Hu »

#### Victor Ivrii

Despite some glitches (one should write \sin (t) resulting in upright $\sin$ rather than sin (t) in LaTeX) the last solution is the only one I can read