There are more basic conceptual issues with what the OP wrote than not using FLRW.
I think it's fine to use Newtonian motion in one dimension to help make sense of what's going on. Ryden's textbook does it in the same chapter the OP refers to (so I'm not sure how helpful it is to tell the OP it's wrong and should learn GR instead - IMHO).
We have a ball and we throw it up in a central gravitational field, and that's meant to be analogous to a galaxy receding away. There are limitations to this approach, but they don't show up here.
Still, it has to be done properly. Here, for example:
jeremyfiennes said:
A cosmological constant to balance this tendency would need to have the form lambda = const/r^2.
This assumes for some reason that the cosmological constant should balance gravity. If it did that, it would nett steady expansion with constant recession velocities. Same as we would have V=const if we canceled out gravity for our ball. We could have some large constant in front of our ##1/r^2## force, such that lambda>Fg. But that would nett acceleration for all time. This is not what is observed or claimed.
Whereas to get outward acceleration >at some point in the evolution of the system<, all we need is the extra force to scale with r slower than gravity's ##r^{-2}##, i.e. anything with the exponent higher than -2. So, e.g. for a force that scales as r, we have something of a form: ##\ddot x=-a/r^2+br##. Which is the same form as the last equation George posted above.
As long as the ball (galaxy) has escape velocity, the second term will start to dominate at some point in the evolution of the system. I.e., we'll have a period of deceleration followed by a period of acceleration.
Further:
jeremyfiennes said:
Hubble's law gives a radial expansion rate dr/dt = H0r, Fig.b. Differentiating gives an acceleration d^2r/dt^2 = H0^2*r. Requiring an outward expansionist force Fh proportional to r, Fig.c. And not proportional to the 1/r^2 of a cosmological constant.
This in turn show misunderstanding of the Hubble law that leads to comparing apples and oranges.
Linear Hubble law with constant ##H_0## is not an equation of motion. It does not mean that if you have an object today at some distance, receding with some velocity, it'll recede with twice velocity once it gets to twice the distance (having thus accelerated). It shows the spread of velocities of >different objects at one point in time<. Only in this sense can H be treated as a constant. In general, it's time-dependent and monotonically decreasing.
In the very far future, when matter dilutes, the Hubble parameter will approach a constant value, and it will be approximately valid to use ##H_\infty## as a constant in the Hubble law. At that point the expansion will approximate exponential, and it will be true that a single object moving to twice the distance accelerates to twice the velocity.
This is equivalent to the first term in:
##\vec{g} = \left(-\frac{GM}{r^2} + \frac{c^2 \Lambda}{3} r \right) \hat{r}##
vanishing for large r.
So there's no disagreement between a) and c) in the OP.