Try:
d = 1;
Tf = 1;
DSolve[{G'[t] == a + b*G[t - d] - u*G[t], G[t /; t <= 0] == G0}, G[t], {t, 0, Tf}];
(* {{G[t] -> Piecewise[{{G0, t <= 0},
{(-a + a*E^(t*u) - b*G0 + b*E^(t*u)*G0 + G0*u)/(E^(t*u)*u),
Inequality[0, Less, t, LessEqual, 1]}}, Indeterminate]}}*)
DSolve
needs a value for d
(better exact) and Tf
to solve. If you don't gives then MMA can't solve or gives a Error
messages.
EDITED -27.04.2018.
Using Laplace transfrom to eq:
eq = G'[t] == a + b*G[t - d] - u*G[t];
eq2 = LaplaceTransform[eq, t, s]
(* -G[0] + s LaplaceTransform[G[t], t, s] ==
a/s - u LaplaceTransform[G[t], t, s] +
b LaplaceTransform[G[-d + t], t, s] *)
in the last expression MMA can't find transform of LaplaceTransform(G[t - d], t, s) see page 120 (it's
weakness of this function) and puttting initial condition G[0]=G0
eq3 = -G[0] + s LaplaceTransform[G[t], t, s] == a/s - u LaplaceTransform[G[t], t, s] +
b *Exp[-d*s]*LaplaceTransform[G[t], t, s] /. G[0] -> G0
sol = LaplaceTransform[G[t], t, s] /. Solve[eq3, LaplaceTransform[G[t], t, s]][[1]]
(* (E^(d s) (a + G0 s))/(s (-b + E^(d s) s + E^(d s) u)) *)
Inverse Laplace Transform:
Total@(InverseLaplaceTransform[#, s, t] & /@ {(sol // Apart)[[1]], (sol // Apart)[[2]]})
(* a/u + (E^(-t u) (-a + G0 u))/u +
b InverseLaplaceTransform[(a + G0 s)/(
s (s + u) (-b + E^(d s) s + E^(d s) u)), s, t]*)
MMA can only find for first experssion.