Does this ODE have an exact or well-established approximate analytical solution? The equation looks like this:
$$\frac{\mathrm{d}y}{\mathrm{d}t} = A + B\sin\omega t - C y^n,$$
where $A$, $B$, $C$ are positive constants, and $n\ge1$ is an integer.  Actually I am mainly concerned with the $n=4$ case.
The $n=1$ case is trivial.  Otherwise the only method I can think of is kind of an iterative approximation, in which a "clean" expression seems not easy to obtain.
Just in case it is too "easy" for the experts, we may generalize it to the form
$$\frac{\mathrm{d}y}{\mathrm{d}t} = f(t) - g(y).$$

If $\frac{dy}{dt} = F(t,y)$ has $F$ and $\partial_2F$ continuous on a rectangle then there exists a unique local solution at any particular point within the rectangle. This is in the basic texts. For $F(t,y) = f(t)-g(y)$ continuity requires continuity of $f$ and $g$ whereas continuity of the partial derivative of $F$ with respect to $y$ require continuity of $g'$. Thus continuity of $f$ and continuous differentiability of $g$ suffice to indicate the existence of a local solution. Of course, this is not an analytic expression just yet. That said, the proof of this theorem provides an iteratively generated approximation.
Alternatively, you could argue by Pfaff's theorem there exists an integrating factor which recasts the problem as an exact equation. So, you can rewrite $dy = [f(t)-g(y)]dt$ as
$Idy+I[f(t)-g(y)]dt=dG$ for some function $G$. Then the solution is simply $G(t,y)=k$ which locally provides functions which solve your given DEqn. Of course, the devil is in the detail of how to calculate $I$. The answer for most of us is magic.
Probably a better answer to your problem is to look at the Bernoulli problem. There a substitution is made which handles problems which look an awful lot like the one you state.
