Integrating the separable, first-order ordinary differential equation $m \frac{dv}{dt} = mg - av$

Question

I can't solve the very first problem from Slater & Frank's book, and have no one to help me (I'm self-studying it in these vacations):

1. A particle moves in a vertical line under the action of gravity and a viscous force $(-av)$ where $v$ is its velocity. Show that the velocity at any time is given by $$v = \left(v_0 + \frac{mg}{a}\right) e^{-\frac{a}{m}t} - \frac{mg}{a}.$$ Show that this solution reduces [...].

What I did is:

The resultant force is $mg - av$, and so by Newton's Second Law, $$m \frac{dv}{dt} = mg - av .$$

Since $m$, $a$, and $g$ are constants in this case, this is a first-order ODE, which also happens to be a separable one.

We rewrite it as a relation between differentials: $$m \frac{dv}{mg - av} = dt .$$

But I have no idea how to integrate the left side relative to $v$. Despite the fact that the exercise already gave the solution, $v(t)$.

Am I doing it right? Or maybe there is a simpler way using methods specials from mathematical mechanics?

Answer 1

Hint The denominator of l.h.s. of the differential equation $$\frac{m \,dv}{m g - a v} = dt$$ is linear in $v$, so this can be readily integrated. To see things a little more clearly, make the (linear) change of variables $u = m g - a v$, $du = -a \,dv$.

This gives $$\frac{m \,dv}{m g - a v} = -\frac{m}{a} \frac{du}{u} .$$ As you probably recall, $\int \frac{du}{u} = \log |u| + C$.