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### Messages - Chang Peng (Eddie) Liu

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1
##### MT / Re: MT problem 3
« on: December 05, 2014, 11:57:12 PM »
can someone please tell me what section of the textbook this question was from?

2
##### TT2 / Re: TT2 # 4
« on: November 20, 2014, 12:09:28 AM »
Graph of #4

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##### TT2 / Re: TT2 # 4
« on: November 20, 2014, 12:08:35 AM »
I just showed the steps for finding the second eigenvector..

After that, you would proceed to plug it into the standard general equation for a repeated eigenvalue question.

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##### TT2 / Re: TT2 # 3
« on: November 19, 2014, 10:37:31 PM »
Picture

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##### TT2 / Re: TT2 # 3
« on: November 19, 2014, 10:37:09 PM »
#3

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

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##### TT2 / Re: TT2 #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

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##### Quiz 3 / Q3 problem1 (night sections)
« on: October 22, 2014, 09:38:38 PM »
4.3 #2

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##### Quiz 2 / Quiz 2 Problem 2 (night sections)
« on: October 01, 2014, 10:41:33 PM »
3.4 #14
Solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing $t$.
\begin{equation*}
y''+ 4y'+ 4y = 0, \qquad y(-1) = 2,\quad y'(-1) = 1.
\end{equation*}

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##### Quiz 2 / Quiz 2 Problem 1 (night sections)
« on: October 01, 2014, 10:39:30 PM »
3.2 #17

If the Wronskian $W$ of $f$ and $g$ is $3e^{4t}$ , and if $f (t) = e^{2t}$ , find $g(t)$.

Can anyone type solution? - V.I.

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##### Quiz 1 / Re: Q1 problem 1 (L5101)
« on: September 25, 2014, 12:40:57 PM »
Hi Prof. Ivrii,

It took me close to an hour to come up with that in MSW because I'm not used to typing out equations; so by the time I finished, Roro already posted it! But thank you!

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##### Quiz 1 / Re: Q1 problem 1 (L5101)
« on: September 25, 2014, 01:04:58 AM »
I'm having trouble typing out equations with proper format in this forum, so I did it in MSW and screenshot the work.. Apologies in advanced!

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##### Quiz 1 / Re: Q1 problem 2 (Night sections)
« on: September 24, 2014, 11:04:15 PM »
Firstly, I'd like to apologize for the really ugly notations and equations.. there is the solution! I rewrote it--V.I.

Find the solution of the given initial value problem.
\begin{gather}
y' - 2y = e^{2t},\label{A}\\
y(0) = 2.\label{B}
\end{gather}

First, we need to find the integrating factor, which is $I = e^{\int -2\,dt}= e^{-2t}$

so we multiple the entire equation by I, thus, giving us
$e^{-2t} y' - 2e^{-2t} y = 1$.
We see that the left side of the equation can be rewritten as $[e^{-2t}y]'$
and we see that the right side of the equation is in fact 1

therefore: $[e^{-2t} y]' = 1$

we now take the integral of both sides, giving us: $e^{-2t} y = t + C$   where $C$ is a constant. To find  $C$, we substitute $y = 2$ when $t$ = 0 (this information is given in the question). We find out that $C = 2$.

rearranging the formula, we come to the solution as
\begin{equation*}
y = (t+2)  e^{2t}.
\end{equation*}

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