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### Messages - Victor Lam

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1
##### Final Exam / Re: FE-3
« on: April 17, 2013, 06:14:16 PM »
i did this

2
##### Final Exam / Re: FE-1
« on: April 17, 2013, 06:07:07 PM »
solution

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##### Final Exam / Re: FE-6
« on: April 17, 2013, 06:01:07 PM »
Solutions and phase portrait

4
##### Term Test 2 / Re: TT2 Question 1
« on: March 28, 2013, 04:35:02 PM »
It appears that I have left out a negative sign in my previous scan. My apologies. For completion, I have fixed the error in red in the re scan.

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##### Term Test 2 / Re: TT2 Question 1
« on: March 28, 2013, 01:01:44 AM »
my solution

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##### Term Test 2 / Re: TT2 Question 2
« on: March 28, 2013, 12:59:14 AM »
Solutions & sketches, etc

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##### Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 22, 2013, 06:11:22 PM »
Also, the line y = 0 is not included in the basin of attraction, as we look at the original system for dy/dt, there clearly is a y term there, which will just stay on the line y = 0. Hence, the basin of attraction for the two critical points can be represented by the 2 rectangles (one above y = 0, the other below y = 0).

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##### Quiz 4 / Re: Quiz 4--Problem (night sections)
« on: March 22, 2013, 04:09:30 PM »
We got saddle points (unstable) for both (0,0) and (-2,0). Also (-1,-1) and (2,2) are both asymptotically stable spirals of clockwise orientation. Solutions can be seen in the visual.
Note as t approaches infinity, all points satisfying x > -2 ultimately approach the particular critical points.

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##### MidTerm / Re: MT Problem 3
« on: March 07, 2013, 12:38:24 AM »
I basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this?

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##### MidTerm / Re: MT Problem 4
« on: March 07, 2013, 12:11:01 AM »
I'm a lil late , but here's a quick way to find Y(particular). A slower "coefficients" method can also verify that A = 1/13

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##### Quiz 3 / Re: Night Sections Problem 2
« on: February 27, 2013, 08:31:03 PM »
General solution is the summation of the homogeneous and particular solutions. See attachment.

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##### Quiz 2 / Re: Night Sections, 2.6 p 102, # 25
« on: January 31, 2013, 12:00:28 AM »
Scans attached.

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##### Quiz 2 / Re: Night sections, 3.2 p 156, # 17
« on: January 30, 2013, 11:41:58 PM »
Solution is attached.

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