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

1. Feb 23, 2006

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:

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) ?

Last edited: Feb 23, 2006
2. Feb 23, 2006

arildno

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

3. Feb 23, 2006

mr_coffee

Thanks,
Okay i did it this time, but still got it wrong:

The answwer i submitted was:

i just let c1 and c2 = 1, they give no inital condition! so what am i suppose to do with that c1 and c2?

4. Feb 23, 2006

arildno

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

5. Feb 23, 2006

mr_coffee

Thanks a ton arildno, worked great!! :D