Evaluate $\dot{v}=-\frac{\gamma}{m}v+g$

Question

Solve: $$\dot{v}=-\frac{\gamma}{m}v+g$$

Where $\gamma,m,g$ are constants

I have to to separate the equation as $\dot{v}=\frac{dv}{dt}$ and got to:

$$\frac{dv-g}{v}=-\frac{\gamma}{m}dt$$

Answer 1

Your rearrangement is incorrect; what you can condlude is that $\frac{dv/dt-g}{v}=-\frac{\gamma}{m}$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.

Our original equation is $\dot{v}+Pv=Q$ with $P:=\frac{\gamma}{m},\,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.

Write $R:=\exp\int Pdt$ so $R^\prime=RP,\,(Rv)^\prime=RQ,\,v=R^{-1}\int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.

In the case at hand $v=e^{-\gamma t/m}\int ge^{\gamma t/m} dt=Ce^{-\gamma t/m}+\frac{mg}{\gamma}$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-\frac{mg}{\gamma})e^{-\gamma t/m}+\frac{mg}{\gamma}$.

Edit: as for separation, we get $t-t_0=-\frac{m}{\gamma}\ln |g-\frac{\gamma v}{m}|\implies v=g\mp\frac{m}{\gamma}\exp\frac{\gamma (t-t_0)}{m}$, which can be manipulated in the same way.