# Homework Help: Question about differential equations?(nonhomogeneous)

1. Feb 28, 2009

### hard_assteel

For this problem I want to find the particular solution;
but i am not sure if it is right setup.

y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

Thank You

2. Feb 28, 2009

### Staff: Mentor

No. Here's your particular solution: yp = Asin(9t)+Bcos(9t)
This would be your particular solution for a right hand side of any of the following:
c1sin(9t)
c2cos(9t)
c1sin(9t) + c2cos(9t)

3. Feb 28, 2009

### Tom Mattson

Staff Emeritus
That's true, but it's of no use in finding the particular solution.

To be honest, I can't figure out what you're thinking. But no, what you have written is not right. All you need is $y_P=A\sin(9t)+B\cos(9t)$. Plug it into your differential equation and solve for $A$ and $B$.