$\m$ is an arbitrary sign. This equation can be easily integrated once, and the integration constant can be fixed from the constraint . This leads to the first integral $$E=|{\bf c}|^2\left(\frac{1}{4}\dot{\f}^2+\z^2e^{-2\m\f}\right)$$ and the solution $$\f = \m\ln\cosh (y)
      +\frac{\m}{2}\ln\left(\frac{|{\bf c}|^2\z^2}{E}\right)$$ for the breathing mode $\f$, where $$y=\frac{2\sqrt{E}}{|{\bf c}|^2}(\t -\t_1)$$ is a rescaled time variable and $\t_1$ is a constant. The seven bulk moduli are proportional to $\f$ and can be obtained by inserting this result into Eq. . From Eq.  one can then, in turn, obtain the solutions for the blow-up moduli. After another integration one finds $$z_i = \frac{d_i}{2}\left(\tanh (y)+1\right)+z_{0i}\;$$ where $$d_i = \frac{\sqrt{E}\z_i}{\z^2}$$ and $z_{0i}$ are independent integration constants. As advertised earlier, the blow-up moduli $z_i$ indeed approach constants asymptotically. More precisely, at early time, $y\rightarrow -\infty$, we have $z_i\rightarrow z_{0i}$ and at late time, $y\rightarrow\infty$, we have $z_i\rightarrow z_{1i}\equiv z_{0i}+d_i$, that is, $d_i$ is the distance by which the blow-up modulus $z_i$ moves. We also note that the non-trivial evolution of $z_i$ happens around the time $y\simeq 0$.

Let us now interpret this solution. From Eq.  the orbifold volume is directly measured by the breathing mode $\f$ via $$\cV_0 = \frac{1}{8}\exp\left(\frac{16}{7}\f\right)\; .$$ Further, from Eq. , the 