The function a[t] is at the denominator of the equation. This makes the initial datum a[0] == 0 singular. However, removing the datum does not solve the equation, except in the special case

DSolve[{eq[1, t, 1] == 0}, a[t], t]

Taking the inspiration from the solution formula I tried this change of independent variable:

DSolve[Numerator@Simplify[eq[x, t, k] /.
      {a -> Function[z, b[Exp[k (z - x)]]]} /.
     (Solve[E^(k (t - x)) == z, t][[1]] /. C[1] -> 0)] == 0,
 b[z], z]

which gives a solution formula in version 13.3.