Solving momentum and mass equation 
A rocket has velocity $v$. Burnt fuel of mass $\Delta m$ leaves at
  velocity $v-7$. Total momentum is constant:
  $$mv=(m-\Delta m)(v+\Delta v) + \Delta m(v-7).$$
  What differential equation connects $m$ to $v$? Solve for $v(m)$ not $v(t)$, starting from $v_0 = 20$ and $m_0 = 4$.

Simplifying the equation for momentum, we have $m\Delta v-\Delta m\Delta v=7\Delta m$. Dividing by $\Delta m$, we get $m\frac{\Delta v}{\Delta m}-\Delta v=7$. As $\Delta m\to0$, $m\frac{dv}{dm}=7$ or $\frac{dv}{dm}=\frac7m$. Then, $v=7\ln m+C$.
At this point, I am getting a feeling that something is not right, because this is a section about exponentials, and logarithms are in the next section. Even if I ignore this, I do not know how to proceed with $v_0$ and $m_0$, since that seems to involve $t=0$. Could you provide some tips on this problem?
 A: I assume you meant:
$$mv=(m-\Delta m)(v+\Delta v) + \Delta m(v-7)$$
So that:
$$0=m\Delta v-\Delta m \Delta v - 7\Delta m$$
To the first order, we have that: $$\frac{dm}{dv} = \frac{1}{7}m(v)$$
An exponential is indeed the solution for this differential equation:
$$m(v) =  m_0e^{v/7}$$
Note $m_0$ which is the mass when $v=0$.
A: To formulate the equation correctly, you should be considering the mass of the rocket at time $ t+\delta t$ as $ m\color{red}{+}\delta m$ (and velocity as $v+\delta v$). By conservation of mass, the particle of ejected mass is$\color{red}{-}\delta m$.
All such variable mass equations should be set up this way. This is in spite of the fact that we know in this case that the mass is actually decreasing in time. If you don't do this, you end up with a solution to the differential equation which defies common sense.
We therefore have $$mv=(m+\delta m)(v+\delta v)-\delta m(v-7)$$ and this leads to $$m\frac{dv}{dm}=-7$$
Solving this and applying the initial conditions gives the solution as$$v=20+7\ln\frac{4}{m}$$
Now you have the expected result that velocity increases as mass decreases, which is not what you had before.
