A simple differential equation

[John O' Meara]

A body is falling in the Earth's atmosphere. The forces on it are the Earth's gravitational attraction and air resistance, which is proportional to its speed v. Its acceleration a is given by a=g-kv. Expressing a in the form dv/dt, obtain an expression for v. Show that whatever its initial speed, as t becomes very large the final speed of the body tends to g/k. I seem to go wrong somewhere in this equation. And what meaning has the air resistance to this question?
\frac{1}{g-kv}\frac{dv}{dt}=1 \\ Therefore \int \frac{1}{g-kv} dv = \int dt \\ Let u = g-k*v => du/dt=-k therefore dv=-du/k. Therefore the integral = \frac{-1}{k}\int \frac{1}{u}du = -ln(u)/k, which gives \frac{-\ln(g-kv)}{k}= t+c \\ Therefore \ln{(g-kv)} = -kt+kc Therefore -kv=-g + \exp^{-kt+kc} The answer is actually: v=\frac{g}{k}+c\exp^{-kt}

 

[Dick]

Divide both sides of your solution by -k. Write exp(-kt+kc)/(-k)=exp(-kt)*exp(kc)/(-k). Now define a new constant C=exp(kc)/(-k). Voila. After all, exp(kc)/(-k) is just 'some constant'.