Question

How do I find the particular solution to:

y''-2y'-3y=3-10sin(t)

I already know the complimentary solution, y(t) = c1e^-t+c2e^3t
I just need to know how to solve the particular using 3-10sin(t). Any help would be great assistance.

Answer 1

Hmmm, it's been a while since I've done these, but my guess is that the particular solution might be of the form \[
A\cos(t)+B\sin(t)+C
\]

Answer 2

Oh yeah, I forgot ot mention I got that, but I am having toruble after i find the derivatives. Because then i have cos and sin. how do I do it so that I get systems. like A+B = #
and so forth..

Answer 3

Well, find y' and y''.

Answer 4

y = At + Bsin(t) + Dcos(t)
y' = A + Bcos(t) - Dsin(t)
y' = -Bsin(t) - Dcos(t)

-Bsin(t)-Dcos(t)-2(A+Bcos(t)-Dsin(t)) - 3(At+Bsin(t)+Dcos(t)) = 3 - 10sin(t)
is that correct so far?

Answer 5

Yeah, sorry was distracted. I think you should try solving for A, B, and C.

Answer 6

somehow C became D though

Answer 7

Yeah, I used D instead of C. That's where i'm stuck on, how do I solve for A,B, and "C".

Answer 8

First thing you want to do is just simplify the thing.

Answer 9

You want it in the form of \[
(A + C + blah)\sin(t)+(blah_2)\cos(t)+blah_3
\] or something like that.

Answer 10

Oh!!! Ok. I see it now! Thanks for the help!! I was trying to separate it in some form but I didn't know how. Thanks!!