1. Second Order Differential Equation

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!

2. Solved, Thanks.