If you use the numeric initial conditions, you get the trivial solution only, that being 0.
If you want to see the full expression (you can substitute in for the initial conditions later), this woirks:
syms y(t) y0 Dy0
Dy = diff(y,t);
D2y = diff(y,t,2);
fun = 0.001*D2y == -((1050)*Dy+(1/0.0047)*y);
cond1 = y(0) == y0;
cond2 = Dy(0) == Dy0;
conds = [cond1 cond2];
ySol(t) = dsolve(fun,conds);
%ySol(t) = dsolve(fun);
ySol = simplify(ySol, 'Steps',20)
disp(ySol(t))
producing:
(608855155^(1/2)*exp(t*((1000*608855155^(1/2))/47 - 525000))*(47*Dy0 + 24675000*y0 + 1000*608855155^(1/2)*y0))/1217710310000 - exp(-t*((1000*608855155^(1/2))/47 + 525000))*((608855155^(1/2)*Dy0)/25908730000 - y0/2 + (105*608855155^(1/2)*y0)/5181746)
