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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
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
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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