Y''+4y'+4y = 6.5e^(-2t), I guessed y = At^2*e^(-2t) but still wrong any ideas?

[mr_coffee]

Hello everyone!
I'm stuck on this problem! as usual.

Find a particular solution to
y'' + 4 y' + 4 y = 6.5 e^{-2 t}.

Here is my work:
y''+4y'+4y = 6.5e^(-2t)
r^2+4r+4 = 0;
r = -2, repeated root
So i made the guess:

y = A*t^2*e^(-2t);
y' = A(2t*e^(-2t) + t^2(-2)*e^(-2t))
y'' = A(2*e^(-2t)+t^2*4*e^(-2t))So i equated co-efficents of e^(-2t) and got:
2A +4A = 6.5
6A = 6.5
A = 13/12

y = c1*e^(-2t)+c2*t*e^(-2t)+(13/12)*t^2*e^(-2t);
So as the answer i submitted was:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/42/a68981ceebd32318c1f96a3eee5ab31.png
but it was wrong any ideas why?
Thanks!

 

[arildno]

Your second derivative is wrong; besides, try an expansion y=(At^2+Bt)e^-2t

 

[mr_coffee]

Thanks for the help arildno, but I'm still stuck now. I don't seem to know what to do about equating co-efficents, here is my work:
http://suprfile.com/src/1/1suf6p/lastscan.jpg

 

[mr_coffee]

N/m, i got it right now, seems like you only equate e^(-2t) for A's because verything else must go to Zero! i got
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/f2/6fbe99cb135fd7a93100695be3cdab1.png
Thanks for the help. Quick question though, how did you know to guess:
y=(At^2+Bt)e^-2t ? and not just y = At^2*e^(-2t)?