LEC5101 Quiz5

[Yueran Hu]

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 Y_c(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 = y_c + Y(t) = c₁e^-2t + c₂te^-2t-e^-2tlnt
with c-1 = c₁

[Lan Cheng]

Hi, I think your general solution should be
$y(t)=Y(t)+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$

[Yueran Hu]

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.