Question

The solution of $\frac{dv}{dt}+\frac{k}{m}v=-g,$ where $m,g$ and $k$ are constants and $c$ is the constant of integration, is

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

Answer 1

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

Given,
$\frac{dv}{dt}+\frac{k}{m}v=-g$
$\Rightarrow \frac{dv}{dt}=-\frac{k}{m}(v+\frac{mg}{k})$
$\Rightarrow \frac{dv}{v+\frac{mg}{k}}=-\frac{k}{m}dt$
$\Rightarrow \ln |v+\frac{mg}{k}|=-\frac{k}{m}t+c$
$\Rightarrow |v+\frac{mg}{k}|=e^{-\frac{kt}{m}+c}$