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Messages - Yueran Hu

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Quiz-5 / Re: LEC5101 Quiz5
« on: November 01, 2019, 02:13:30 PM »
Sure, it can be the answer as well. But if you see my answer carefully, you will find that (c-1) is a constant as well, which could be written as C1.

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Quiz-5 / LEC5101 Quiz5
« on: October 31, 2019, 07:58:03 PM »
Find the general solution of the given equation y'' + 4y' + 4y = t-2e-2t, t>0

First, solve y'' + 4y' + 4y = 0.

So r2 + 4r + 4 = 0,    r = -2 and r = -2

We have Yc(t) = ce-2t +c2te-2t

y1 = e-2t
y2 = te-2t

Determine the Wronskian as follows:

W(y1, y2)(t) = e-4t

Since Y(t) = u1(t)y1(t) +  u2(t)y2(t)

u1 = -ln t

u2 = -t-1

Hence Y(t) = u1(t)y1(t) +  u2(t)y2(t)
                = -ln te-2t + -t-1te-2t
                = -e-2tlnt-e-2t

So the general solution is y = yc + Y(t) = c1e-2t + c2te-2t-e-2tlnt
with c-1 = c1

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Quiz-4 / Quiz4 TUT0303
« on: October 18, 2019, 08:12:58 PM »
Find the general solution of the given equation.

y" + 2y' + y = 2e-t

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Quiz-3 / TUT0303 QUIZ3
« on: October 11, 2019, 10:13:14 PM »
Find the Wronskian of two solution of the given equation.

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Quiz-2 / TUT0303 QUIZ2
« on: October 04, 2019, 08:58:13 PM »
Here is quiz2's answer with typed.

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