Difference between the dimensions $years^{-1}$ and, for instance, $\dfrac{animals}{years}$ in mathematical modelling using differential equations?

Question

I am told that an animal population has a natural growth rate of $k \ years^{-1}$ and is harvested at a rate of $a \ \dfrac{animals}{years}$.

Note that the constants $k$ and $a$ are both given as rates but, from a dimensional analysis perspective, they are different. I do not understand why one rate is given in the units $years^{-1}$ and the other is given in $\dfrac{animals}{years}$; indeed, I would have expected the growth rate to also be given in $\dfrac{animals}{years}$, since it is, after all, the growth rate of animals per year?

Obviously, this difference matters, since our units must make sense when modelling using differential equations. But I'm having trouble reasoning about this difference; it seems to make no sense when attempting to reason about it intuitively.

Why are the units $years^{-1}$ instead of $\dfrac{animals}{years}$, as they are for the other rate? Doesn't it make sense to have them both as $\dfrac{animals}{years}$, since they are rates?

I would greatly appreciate it if people could please take the time to help me understand this.

EDIT:

If readers are interested in the specific problem, it is from this MIT differential equations problem set:

Answer 1

One way to see the difference is to think about the differential equation involved. If $N$ is the number of animals, then the differential equation will be in terms of $\frac{dN}{dt}$, which has units of animals per year. This means that every term has to have units of animals per year. The first term will have $k$, but $k$ by itself has the incorrect units; it has to be multiplied by $N$ to have the correct units. The second term will simply be $a$ since $a$ already has the same units as the differential term. For the first term, the rate depends on the population; for the second, the rate is a constant. So the final differential equation is $$\frac{dN}{dt}=kN-a$$

Answer 2

I think it is the same reason that angular speed is sometimes measured in $\mathrm{time}^{-1}$ (such as $\mathrm{s}^{-1}$) rather than $\frac{\mathrm{angle}}{\mathrm{time}}$ (such as $\frac{\mathrm{rad}}{\mathrm{s}}$).

It is because you consider some variables (angles, animals...) as dimensionless variables. This is because they are not physical quantities and they can be reduced to a proportionality constant (you measure the quantity 'animals' in the unit 'number of animals'). $$\frac{\mathrm{animals}}{\mathrm{years}}=\frac{1}{\text{total number of animals}}\frac{1 \text{ animal}}{\mathrm{years}}\simeq k \,\mathrm{years}^{-1}$$

It may seem reasonable to show the results in $\frac{\mathrm{animals}}{\mathrm{years}}$, specially is you want to talk about several ratios (like $\frac{\mathrm{predators}}{\mathrm{years}}$ or $\frac{\mathrm{preys}}{\mathrm{years}}$), but notice that all these ratios are comparable between each other (for example, 'predator population grow at a slower rate than prey population') because they have the same physical unit: $\mathrm{years}^{-1}$.