Hyperbolastic rate equation of type II already has its initial condition in it?

I’m modelling some real-world gene expression data with various growth models including linear, exponential, and Verhurst growth but not all of the genes are showing these forms of time-dependence. Thus I am exploring other models, and that has led me to the hyperbolastic functions.

According to wikipedia, the type II hyperbolastic rate equation is given by

$$frac{dP(x)}{dx} = frac{alpha delta P^2(x)x^{gamma-1}}{M} tanh left( frac{M – P(x)}{alpha P(x)} right)$$

which has the solution

$$P(x) = frac{M}{1 + alpha sinh^{-1} left(e^{-delta x gamma} right)}$$

that doesn’t explicitly have the initial conditions in it. Then wikipedia says I can substitute

$$alpha := frac{M – P_0}{P_0 sinh^{-1}left(e^{-delta x_0 gamma} right)}$$

which has the initial condition $x_0$ in it. Since $alpha$ appears in both the rate equation and its solution, it seems that the initial condition doesn’t fall out of the derivation of the solution as I often find in solving differential equations. I don’t see why one can’t have arbitrary constants in the rate equation, but I would like to confirm that I’ve understood correctly that the above assignment for $alpha$ applies to both the rate equation and its solution. If so, I should be able to rewrite the rate equation as

$$frac{dP(x)}{dx} = frac{M – P_0}{P_0 sinh^{-1}left(e^{-delta x_0 gamma} right)} frac{delta P^2(x)x^{gamma-1}}{M} tanh left(frac{P_0 sinh^{-1}left(e^{-delta x_0 gamma} right)}{M – P_0} frac{M – P(x)}{P(x)} right)$$