Solved, Thanks.
y''' + y' = 2 + sin(x) y(0)=0 y'(0)=1 y''(0) = -1
I have my homogeneous solution of:
y(h) = c1 + c2* cos(t) + c3* sin(t)
When finding my particular solution, I have:
y(p) = A + B*cos(t) + C*sin(t)
When taking the y(p)''' + y(p)', I get Zero.
I am hoping for any hints in finding the Particular solution so I get
y = y(h) + y(p) to solve for my initial conditions.
Thanks!