Can an arbitrary constant in the solution of a differential equation really take on any value? Consider the first order differential equation $y' = -2y^{\frac{3}{2}}$.
It has $y = \dfrac{1}{(x+c)^2}$ as the solution. 
Now, if I divide both the numerator and denominator by $c^2$ (assuming $c \neq 0$), we get $y = \dfrac{C^2}{(Cx+1)^2}$, where $C = \dfrac{1}{c}$. Since $c$ is arbitrary, $C$ is also arbitrary.

Can $C$ take on the value of $0$?

It seems simple enough to argue that since $C = \dfrac{1}{c}$, $C \neq 0$. 
But $y = \dfrac{C^2}{(Cx+1)^2}$ satisfies the differential equation, which implies that it's a solution for all possible values of $C$. Moreover, the function obtained when $C = 0$ is $y=0$, which satisfies the differential equation. 
So where am I going wrong in my reasoning?
 A: Yes, you are right, $C$ can take the value $0$.
The solution to the original DE involves dividing both sides by $y^{3/2}$, before integrating.  They should consider the special case when $y=0$ at that point.
The limit of $1/(x+c)^2$ as $c\to\infty$ is 0 at every point $x$, so the new formula is a limit of old formulas.
A: Your reasoning is not valid, though your conclusion is correct. It is true that $y=\frac{1}{(x+c)^2}$ is a solution for all $c$, and since if $C=\frac{1}c$ you get $y=\frac{C^2}{(Cx+1)^2}$, that must also be a solution. However, this only covers cases where $C\neq 0$, since as you note, $0$ cannot be written as $\frac{1}{\text{something}}$. There is no rule that says that any constant that appears in a differential equation can be replaced by any other one and there is not likely any proof in this vein.
Rather, to establish the $C=0$ case, you can either argue by continuity (e.g. as $C$ approaches $0$, the function $(x,C)\mapsto \frac{C^2}{(Cx+1)^2}$ is continuous and has a continuous derivative with $x$, and thus since every $C$ is any punctured neighborhood of $0$ is a solution, so $C=0$ is a solution), or you can check it directly.
