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### Messages - Yangming Cai

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##### Quiz 5 / Re: Problem 2, night sections
« on: November 21, 2013, 06:59:10 PM »
I used mathtype in word, is there anything wrong with this s.w?

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##### Quiz 5 / Re: Problem 1, night sections
« on: November 21, 2013, 12:21:49 PM »
\begin{array}{l}\det (A - rI) = \left( {\begin{array}{*{20}{c}}{3 - r}&{ - 2}\\2&{ - 2 - r}\end{array}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {r^2} - 1 - 2 = 0\\r = 2,\;r =  - 1\\{\rm{when}}\;r = 2\\1{\xi _1} = 2{\xi _2}{\rm{and}}{\xi ^1} = \left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\\{\rm{when}}\;r =  - 1\\2{\xi _1} = 1{\xi _2}{\rm{and}}{\xi ^2} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\\x = {c_1}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right){e^{ - t}}\end{array}
\begin{array}{l}{\rm{for }}{c_1} = 0,as\;t \to  + \infty ,x \to 0.as\;t \to  - \infty ,x \to  + \infty \\{\rm{for }}{c_1} \ne 0,as\;t \to  + \infty ,{\rm{the first term dominates, so }}x \to  + \infty .\\{\rm{               }}as\;t \to  - \infty ,{\rm{the second term dominates}},sox \to  + \infty \end{array}

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##### Quiz 5 / Re: Problem 2, night sections
« on: November 21, 2013, 12:18:29 PM »
\begin{array}{l}\det (A - rI) = \left( {\begin{array}{*{20}{c}}{2 - r}&{ - 5}\\1&{ - 2 - r}\end{array}} \right) = {r^2} + 1 = 0\\r =  \pm i\\{\rm{when }} r = i\\(2 - i){\xi _1} = 5{\xi _2}{\rm{   and }}{\xi ^1} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right)\\{x^1} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right){e^{it}} = \left( {\begin{array}{*{20}{c}}5\\{2 - i}\end{array}} \right)(\cos t + i\sin t)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\begin{array}{*{20}{c}}{5\cos t}\\{2\cos t + \sin t}\end{array}} \right) + i\left( {\begin{array}{*{20}{c}}{5\sin t}\\{2\sin t - \cos t}\end{array}} \right)\\x = {c_1}\left( {\begin{array}{*{20}{c}}{5\cos t}\\{2\cos t + \sin t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{5\sin t}\\{2\sin t - \cos t}\end{array}} \right)\end{array}

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##### Quiz 5 / Re: Problem 2, night sections
« on: November 20, 2013, 09:30:32 PM »

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##### Quiz 5 / Re: Problem 1, night sections
« on: November 20, 2013, 09:23:47 PM »

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##### Quiz 4 / Re: Problem 1 Night Sections
« on: November 13, 2013, 09:34:32 PM »

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##### Quiz 4 / Re: Problem 2 Night Sections
« on: November 13, 2013, 09:09:54 PM »

(a). $W(\mathbf{x}^{(1)},\mathbf{x}^{(2)})=t\cdot 2t-1cdot t^2=t^2$;

2. When $t=0$ we have $W=0$;  then $\mathbf{x}^{(1)}(0)$ and $\mathbf{x}^{(2)}(0)$  are linearly dependent, so $\mathbf{x}^{(1)}$ and $\mathbf{x}^{(2)}$ are linearly independant on intervals where $t\ne 0$;

3. The coefficients of the ODE are discontinous at x=0. If $\mathbf{x}$ satisfies this system $\mathbf{x}'+A\mathbf{x}=0$ then $A$ must be singular at $t=0$.

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##### Quiz 3 / Re: Problem 2 (night sections)
« on: November 06, 2013, 09:08:39 PM »
I am so impressed with your speed.

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##### Quiz 2 / Re: Problem 1, Night sections
« on: October 30, 2013, 08:56:05 PM »
if $y_1$ and $y_2$ are zero at the same point in $I$ï¼Œthen its Wronskian , which is $y_1y_2'-y_2y_1'=0$   and then $y_1$ and $y_2$ are not linearly independent, indicating that they cannot form a fundamental solution on that interval

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##### Quiz 2 / Re: Problem 2, night sections
« on: October 30, 2013, 08:51:01 PM »