Pain in the Neck Question - Just want confirmation of method?

Question

Wondering if anyone could help me answer this swine of a question I have been trying to do for ages.
I am trying to solve a difference equation
u[n+2] + 3u[n+1] + 2u[n]=cos([3x(Pi)xn]/4)
Anyway I found the complementary difference equation easily enough but can't get the particular solution
I put y[p]=Acos([3x(Pi)xn]/4) + Bsin([3x(Pi)xn]/4)
Then I found u[n +1] etc using the compound angle formulaes and subbed into the original equation.
I then compared coefficients of sin and cos.
However my simultaneous equations dont solve to the right answers.
I don't expect anyone to actually do this question but could someone tell me if I am doing the right method?
Thanks.

Answer 1

Yes, this is the right method.
I'll do some of the steps below - you can compare your answers and perhaps find the mistake.

Incidentally, I'm assuming your x symbols above mean multiplication, i.e. RHS is cos(3πn/4).

With y[p] as above you should get
u[n+1] = A cos (3πn/4 + 3π/4) + B sin (3πn/4 + 3π/4)
= A[cos 3πn/4 cos 3π/4 - sin 3πn/4 sin 3π/4] + B[sin 3πn/4 cos 3π/4 + cos 3πn/4 sin 3π/4]
= cos 3πn/4 (B-A)/√2 + sin 3πn/4 (-A-B)/√2

u[n+2] = A cos (3πn/4 + 3π/2) + B sin (3πn/4 + 3π/2)
= A sin 3πn/4 - B cos 3πn/4

Putting these back into the equation gives
cos 3πn/4 [-B + 3(-A+B)/√2 + 2A] + sin 3πn/4 [A - 3(A+B)/√2 + 2B] = cos 3πn/4
and hence
A (2-3/√2) + B (-1+3/√2) = 1
A (1-3/√2) + B (2-3/√2) = 0

From the second equation we get A(3-√2) = B(2√2 - 3) and hence B = A.(3-√2)/(2√2-3). We can substitute back into the first equation and solve for A and hence B. I wind up with A = (4-3√2)/(28-18√2) and B = (18-13√2)/(110√2-156), but I haven't checked these values.