homework and exercises - Hamiltonian equations: can I divide a solution of motion for a constant?

I'm solving an exercise about Hamiltonian equations. I have followed the proceeding below. The results given by the book are different to mine because its first result is the half of mine (and the second one linked to the first one is different to mine). I think that my proceeding is correct and so I can't understand...

Given these two Hamiltonian equations:

$$\tag{1} \dot p ~=~ - \alpha pq,$$ $$\tag{2} \dot q ~=~\frac{1}{2} \alpha q^2.$$

Find $q(t)$ and p$(t)$, considering initial conditions $p_0$ and $q_0$.

I have integrated the second equation and obtained:

$$\tag{3} q(t)~=~\frac{2q_0}{2-q_0 \alpha (t-T_0)}$$

Then I have pugged this, in the second canonical eq, and I have obtained:

$$\tag{4} p(t)~=~p_0(2-q_0 \alpha (t-t_0))^2.$$

The solutions given by the book are:

$$\tag{5} q(t)~=~\frac{q_0}{1- \frac{1}{2} \alpha q_0 (t-t_0)},$$ $$\tag{6} p(t)~=~p_0[1-\frac{1}{2} \alpha q_0 (t-t_0)]^2.$$

I can obtain the solutions of the book if I divide numerator and denominator of $q$ for 2.. but.. can I do it?

Is my proceeding correct?