How to solve $y′′′+y'=2-\sin(x)$ I have tried to solve this but with no luck.
So far I just get,
$$y_p(x) = A \sin x + B \cos x \\  
  y'_p(x) = A \cos x - B \sin x  \\
 y''_p(x) =-A \sin x - B \cos x  \\
y'''_p(x) =-A \cos x + B \sin x $$
$$y'''+ y' =  -A \cos x + B \sin x + A \cos x - B \sin x \\ \\ 
         =  \cos x (-A+A) + \sin x (B-B) = 2 - \sin x $$
I would really appreciate help in solving this

here is a way to why $x$'s are added. we will find a particular solution to $$y'' + y = \sin x \tag 1$$ the equation is forced by a solution of the homogeneous problem(called  the resonant case). first you look at a more general problem of the solution of 
$$y'' + y = \sin kx$$ a particular solution is $$y_p = \frac{\sin kx}{1-k^2}$$ so is $$y_p(x) = \frac{\sin kx - \sin x}{1-k^2}= -\frac{1}{1+k}\frac{\sin kx - \sin x}{k - 1} $$ in the limit as $k \to 1, $ we have 
$$y_p = -\frac{1}{2}\left( \frac{d}{dk}\sin kx \right)|_{k = 1} 
=-\frac{1}{2}x\cos x $$
you can see that a particular solution of $(1)$ is $$y_p = -\frac{1}{2}x\cos x. $$
