(2D-3E)cos(t) + (-2E-3D)sin(t)+2F=5sin(t)
now we match coefficients.
on the right side there are no cos(t) terms so we assume (2D-3E)=0
on the right side we have 5*sin(t) so it must be that (-2E-3D)=5
on the right we have no constants so 2F=0
which allows us to solve for D, E and F.
2D=3E
D=3E/2
-2E-3*3E/2=5
-13E/2=5
E=-10/13
D=(3/2)*(-10/13)
D=-30/26=-15/13
F=0
so xp= (-15*sin(t)-10*cos(t) ) /13
the final solution is
x=xh+xp
where
xh=Ae^-0.5t + Be^-2t + C // homogeneous part of the solution
xp=(-15*sin(t)-10*cos(t) ) /13 // particular solution
or together:
x=Ae^-0.5t + Be^-2t + C -(15*sin(t)+10*cos(t) ) /13
we still have A,B and C that are not known (need three conditions to solve for those)