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MAT244-2014F => MAT244 Math--Tests => TT2 => Topic started by: Victor Ivrii on November 19, 2014, 08:46:06 PM

Title: TT2 #2
Post by: Victor Ivrii 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\$.
Title: Re: TT2 #2
Post by: Yuan Bian 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
(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)
Title: Re: TT2 #2
Post by: Chang Peng (Eddie) Liu on November 19, 2014, 10:20:52 PM
This is what I did... Please feel free to correct me if I'm wrong! :p
Title: Re: TT2 #2
Post by: Chang Peng (Eddie) Liu on November 19, 2014, 10:21:18 PM
2b
Title: Re: TT2 #2
Post by: Yuan Bian on November 19, 2014, 10:31:33 PM
different c1 and c2... ::)
Title: Re: TT2 #2
Post by: Tao Hu 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*}