Since D and N are pre-defined Mathematica function names, do some variable substitutions. Since the resulting system seems to take a long time and since some of your functions are simple linear combinations of others, do some function substitutions.
In[1]:= Simplify[{a[t] + c[t] - f[t] == 0, d[t] + e[t] + f[t] - i[t] == 0, 2 g[t] + h[t] == 0,
a[t] - e[t] == 0, 2 d[t] + e[t] - h[t] == 0, a'[t]/3 + c'[t] == capN*2 a[t],
d'[t]/3+e'[t]/2+f'[t] == capD*(d[t]+e[t]) - e[t], g'[t]/3+h'[t]/2+i'[t] == capM*i[t]-capA} //.
{a[t] -> e[t], a'[t] -> e'[t],
h[t] -> -2 g[t], h'[t] -> -2 g'[t],
f[t] -> a[t] + c[t], f'[t] -> a'[t] + c'[t],
i[t] -> d[t] + 2 a[t] + c[t], i'[t] -> d'[t] + 2 a'[t] + c'[t],
g[t] -> - d[t] - e[t]/2, g'[t] -> -d'[t] - e'[t]/2}]
Out[1]= {True, True, True, True, True, 6 capN e[t] == 3 c'[t] + e'[t],
6 capD d[t] + 6 (-1 + capD) e[t] == 6 c'[t] + 2 d'[t] + 9 e'[t],
capA + c'[t] + (5 d[t])/3 + (7 e'[t])/3 == capM (c[t] + d[t] + 2 e[t])}
Then this
In[2]:= DSolve[{6 capN e[t] == 3 c'[t]+e'[t], 6 capD d[t]+6 (-1+capD) e[t] == 6 c'[t]+2 d'[t]+9 e'[t],
capA + c[t] + (5 d'[t])/3 + (7 e'[t])/3 == capM (c[t] + d[t] + 2 e[t])}, {c[t], d[t], e[t]}, t]
solves in a couple of seconds. Unfortunately the result is large and complicated, but with some study you might be able to see how to simplify this greatly. Then you can use that result to determine each of the functions that were eliminated.