Motion with mass decreasing: A rocket launch

Question:
A rocket is launched in a constant gravity, 9.80 m/s$^2$. The initial velocity is 400 m/s, and the burn time is 100 seconds. If the exhaust velocity is 2000 m/s, and the mass decreases by a factor of three; namely, the current mass divided by the original mass is equal to $\frac{1}{3}$, find the final velocity of the rocket.

Answer:
The equation of motion is given as
\begin{equation}
m\frac{dv}{dt}=-mg-u \frac{dm}{dt}
\end{equation}
where $u$ is the exhaust velocity. Divide both sides
by the mass, $m$.
\begin{equation}
\frac{dv}{dt}=-g-u \frac{dm}{dt}\frac{1}{m}
\end{equation}
Multiply $dt$ by both sides. (This manipulation is not for mathematicians.)
\begin{equation}
dv=-gdt - u \frac{dm}{m}
\end{equation}
Integrate this.
\begin{equation}
\int^{v}_{v_0} dv=\int^{t}_0 -gdt - u \int^{m}_{m_0}\frac{dm}{m}
\end{equation}
Therefore,
\begin{equation}
v-v_0 = -gt - u \ln \frac{m}{m_0}
\end{equation}
Plug in the numbers.
\begin{equation}
v = 400 - 9.80 \cdot 100 - 2000 \ln \frac{1}{3}=1620 \ \mathrm{m/s}
\end{equation}

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