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?
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 $v=0$.