Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz5 => Topic started by: Yueran Hu on October 31, 2019, 07:58:03 PM

Find the general solution of the given equation y'' + 4y' + 4y = t^{2}e^{2t}, t>0
First, solve y'' + 4y' + 4y = 0.
So r^{2} + 4r + 4 = 0, r = 2 and r = 2
We have Y_{c}(t) = ce^{2t} +c_{2}te^{2t}
y_{1} = e^{2t}
y_{2} = te^{2t}
Determine the Wronskian as follows:
W(y_{1}, y_{2})(t) = e^{4t}
Since Y(t) = u_{1}(t)y_{1}(t) + u_{2}(t)y_{2}(t)
u_{1} = ln t
u_{2} = t^{1}
Hence Y(t) = u_{1}(t)y_{1}(t) + u_{2}(t)y_{2}(t)
= ln te^{2t} + t^{1}te^{2t}
= e^{2t}lnte^{2t}
So the general solution is y = y_{c} + Y(t) = c_{1}e^{2t} + c_{2}te^{2t}e^{2t}lnt
with c1 = c_{1}

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

Sure, it can be the answer as well. But if you see my answer carefully, you will find that (c1) is a constant as well, which could be written as C_{1}.