NO!!!
YOURSsolution is NOT a solution of ODE, even if calculated by hand, and it does NOT fulfill YOURSsolution[ 0 ] == 0.
In[150]:= YOURSsolution /. t -> 0
% // FullSimplify
Out[150]= 1/2 B r (1/(R - I (w - w0)) + 1/(R + I (w - w0)))
Out[151]= (B r R)/(R^2 + (w - w0)^2)
You may drop the condition y[ 0 ] == 0 in your ODE to get
ysol1 = y /. DSolve[{y'[t] == (I*w0 - R)*y[t] + B*r/2*(Exp[I*w*t] + Exp[-I*w*t])}, y, t][[1, 1]]
Then drop, as you wanted to, C[1]
ysol2 = ysol1 /. C[1] -> 0
But
In[171]:= ysol2[0]
Out[171]= 1/2 B r (1/(R + I (w - w0)) + 1/(R - I (w + w0)))
That is NOT what you wanted.
ysol2 is a solution of ODE
In[172]:=
ysol2'[t] == (I*w0 - R)*ysol2[t] + B*r/2*(Exp[I*w*t] + Exp[-I*w*t]) // FullSimplify
Out[172]= True
ysol2 looks very similar to YOURSsolution (but remember: it does not fulfill your initial condition)
In[181]:= (B r)/2 (2/(B r) ysol2[t] // Simplify // Expand)
Out[181]= 1/2 B r (E^(I t w)/(R + I (w - w0)) + E^(-I t w)/( R - I (w + w0)))
but it is not the same
In[173]:= ysol2[t] - YOURSsolution // FullSimplify
Out[173]= (I B E^(-I t w) r w0)/((R - I w)^2 + w0^2)
Look at the signs in the w / w0 terms , I have the suspicion that there is a typo in that dissertation.