Question

The solution of $\frac{dv}{dt}+\frac{k}{m}v=-g$ is

• A. $v=c{e}^{-\frac{k}{m}t}-\frac{mg}{k}$
• B. $v=c-\frac{mg}{k}e-\frac{k}{m}t$
• C. $ve-\frac{k}{m}t=c-\frac{mg}{k}$
• D. $ve\frac{k}{m}t=c-\frac{mg}{k}$

Answer 1

The correct option is A $v=c{e}^{-\frac{k}{m}t}-\frac{mg}{k}$

$\frac{dv}{dt}+\frac{k}{m}v=-g$
Integrating factor (I.F.) = $e^{\int \frac{k}{m}dt}=\frac{k}{em}t$
$\therefore v\frac{k}{em}t=-\int g{e}^{\frac{kt}{m}+c}v\frac{k}{em}t=\frac{-gm}{k}\frac{k}{em}t+cv=C.\frac{-k}{em}t-\frac{mg}{k}$