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##### 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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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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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

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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Here is the question I received and my solution

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Here is the question I received and my solution

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this is my quiz question and answer

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Question: y′′ + 2y′ + y = 2e^−t

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Question:

2y''+2y+1=0

Solution:

2r^{2}+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)t}cos(1/2)t+C2e^{-(1/2)t}sin(1/2)t

2y''+2y+1=0

Solution:

2r

r=[-2+-((-2)

r=[-2+-(-4)

r=-(1/2)+-(1/2)i

λ=-(1/2), µ=1/2

y(t)=C1e

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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

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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Which section does this question belong to?