Y''+9y = 6sin(3t); Diff EQ, weee where did i go wrong?

[mr_coffee]

Ello ello!
I'm looking at an example problem my professor did which is very similar to the one that i am doing, the only difference is his problem is:
y''+y = sin(t)
mine is:
y''+9y = 6sin(3t);

THe problem says:
Find a particular solution to
y'' + 9 y = 6sin(3 t) .

y_{p} =
so here is what i did:
http://img239.imageshack.us/img239/6219/lastscan8oo.jpg

I basically copied what my professor did, but instead of equating sin(t) to 1, i equated it to 6.

Any ideas on what I'm doing wrong? THanks!

Looking at it again should i have started out with the guess of:
y =A*t*sin(3t) + B*t*cos(3t) ?

 

[arildno]

You have forgotten that when differentiating the term Btcos(3t), a 3 jumps out every time you differentiate the cos-factor.

 

[mr_coffee]

Thanks,
Okay i did it this time, but still got it wrong:
http://img159.imageshack.us/img159/9669/lastscan0bk.jpg
The answwer i submitted was:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/5c/fbfb91a1714cd260613bc82810dfbe1.png
i just let c1 and c2 = 1, they give no inital condition! so what am i suppose to do with that c1 and c2?

 

[arildno]

You must have -tcos(3t) as your particular solution, not -tcos(t).

 

[mr_coffee]

Thanks a ton arildno, worked great! :D