# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Yueran Hu on October 31, 2019, 07:58:03 PM

Title: LEC5101 Quiz5
Post by: Yueran Hu 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
Title: Re: LEC5101 Quiz5
Post by: Lan Cheng on October 31, 2019, 09:54:55 PM
Hi, I think your general solution should be
$y(t)=Y(t)+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$

Title: Re: LEC5101 Quiz5
Post by: Yueran Hu 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.