Recent Posts

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1
Term Test 1 / Re: Content on Term Test
« Last post by LilyBateman on October 19, 2019, 11:45:23 PM »
On Quercus, under modules then Test 1 it gives the sections that are on the test, the most recent section is 3.5
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Quiz-4 / Quiz 4 TUT0303
« Last post by ireneeewang on October 19, 2019, 11:09:59 PM »
 :)
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Term Test 1 / Content on Term Test
« Last post by amcgale on October 19, 2019, 11:07:23 PM »
Hi,
My TA told me the content on the midterm will be up to 3.3 but I feel like he might have made a mistake! Can someone please confirm/deny this info thankss
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Quiz-1 / TUT0301 Quiz-1
« Last post by Yuxin Liang on October 19, 2019, 06:35:29 PM »
Here is the question I received and my solution
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Quiz-4 / TUT0301 Quiz-4
« Last post by Yuxin Liang on October 19, 2019, 05:45:44 PM »
Here is the question I received and my solution
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Quiz-4 / MAT244 TUT0602
« Last post by dujiawe1 on October 19, 2019, 05:32:36 PM »
this is my quiz question and answer
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Quiz-4 / TUT0301 Quiz4
« Last post by yiqiwang on October 19, 2019, 05:18:14 PM »
Question: y′′ + 2y′ + y = 2e^−t
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Quiz-4 / TUT0602 Quiz4
« Last post by Yuchen Cong on October 19, 2019, 05:11:25 PM »
Question:
2y''+2y+1=0
Solution:
2r2+2r+1=0
r=[-2+-((-2)2-(4)(2)(1))(1/2)]/2(2)
r=[-2+-(-4)(1/2)]/4
r=-(1/2)+-(1/2)i
λ=-(1/2), µ=1/2
y(t)=C1e-(1/2)tcos(1/2)t+C2e-(1/2)tsin(1/2)t


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Quiz-4 / Quiz 4 TUT0202
« Last post by Mengyuan Wang on October 19, 2019, 04:46:15 PM »
4y'' + 12y' + 9y = 0
let y = e^rt
    y' = re^rt
    y''= r^2e^rt
   
4r^2 + 12r +9r = 0
    r = -(2/3)
thus
   y = c1e^(-(2/3))t + tc1e^(-(2/3))t
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Quiz-4 / Re: TUT0801
« Last post by Jason on October 19, 2019, 03:38:57 PM »
Which section does this question belong to?
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