### Author Topic: Quiz 2 Problem 2 (night sections)  (Read 2450 times)

#### Chang Peng (Eddie) Liu

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##### 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*}
« Last Edit: October 02, 2014, 03:56:44 AM by Victor Ivrii »

#### Michael Boutros

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##### Re: Quiz 2 3.4 #14
« Reply #1 on: October 01, 2014, 11:21:32 PM »
Hello,

With all due respect, this question should not have been on the quiz. The course website states the following: "Each consecutive quiz covers homework from the previous quiz up to and including the week preceding the week this quiz is administered."

According to Dr. Ivrii's post last Wednesday, the coverage for last week's lecture was upto 3.2. I did the questions upto that section and was prepared for the other question on the quiz. The questions for 3.4 were never assigned and therefore I believe they should not have been on today's quiz. I'm sure my classmates will agree that although the question wasn't super difficult, we were not told to prepare for it and therefore did not prepare.

Thank you,
Michael Boutros

#### Yeming Wen

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##### Re: Quiz 2 3.4 #14
« Reply #2 on: October 01, 2014, 11:31:36 PM »
I guess the graph looks like this.

#### Victor Ivrii

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##### Re: Quiz 2 3.4 #14
« Reply #3 on: October 02, 2014, 01:38:07 AM »
Michael Boutros,
I wrote:
Quote
I plan to give intro to 2nd order equations and then to cover sections 3.1, 3.2 and may be start 3.3.
It was a plan, not the actual content. Instead, in L5101 we covered 3.1, 3.2 and 3.4 because 3.4 is much more simple and shorter,  than 3.3 and there was no point to start 3.3 after Quiz 1.

All other sections did the same. When students asked me to write what I am planning to do I warned that plans are always approximate.

« Last Edit: October 02, 2014, 01:39:42 AM by Victor Ivrii »

#### Victor Ivrii

A bit simpler: the general solution is $y=\bigl(C_1 + C_2 (t+1)\bigr) e^{-2 (t+1)}$: it is equivalent (with the different constants to $\bigl(C_1 + C_2 t\bigr) e^{-2 t}$ but we simply moved the "time" to start from $0$ rather than from $-1$: $t_{\text{new}}=t+1$.
Then $y(-1)=C_1=2$ and $y'(-1)= -2 C_1+ C_2=1$. Then $C_1=2$, $C_2=5$ and $y=(7+5t)e^{-2t+2}$. Indeed graph is similar to one sketched by Yeming Wen: it tends to $+0$ as $t\to +\infty$, crosses $y=0$ as $t=-7/5$ and fast tends to $-\infty$ as $t\to -\infty$.