First, rewrite your equation as
$$
\frac{dy(t)}{dt}=b+f(t)(y(t))^n, \qquad \qquad \qquad (*)
$$
where $f(t)=1/a(t)$.

For $n=1$ you have a linear inhomogeneous ODE which is easily solved.

For $n=2$ you get a special case of the so-called [general Riccati equation][1]
$$
\frac{dy(t)}{dt}=b+f(t)(y(t))^2, 
$$
solving which is equivalent to solving a second-order linear ODE.
Indeed, introduce a new independent variable $\tau(t)=\int a(t) dt$ 
you end up with the "standard" Riccati equation
$$
\frac{dy(\tau)}{d\tau}=h(\tau)+(y(\tau))^2,
$$
where the function $h$ is defined so that $h(\tau(t))=b a(t)$.
and putting $y(\tau)=-(1/z(\tau))dz(\tau)/d\tau$ yields
$$
d^2z(\tau)/d\tau^2+h(\tau)z(\tau)=0.
$$
However, this linear equation in general is not necessarily solvable by quadratures.

For $n=3$ (*) is a special case of the Abel differential equation of the first kind,
see e.g. [here][2] and references therein for details.

For $n\neq 1,2,3$ your equation in full generality is probably not solvable by quadratures 
but some of its special cases may be, see e.g. the book (in German)

E. KAMKE: Differentialgleichungen, Lösungen und Lösungsmethoden, Band I, Leipzig, 1951, 

and [this list][3]. 




  [1]: http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf
  [2]: http://eom.springer.de/A/a010110.htm
  [3]: http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc1.htm