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Mathematics Question on integral

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

A

v=ce−kmt−mgkv = ce^{-^{\frac{k}{m}t}} - \frac{mg}{k}

B

v=c−mgke−kmtv = c- \frac{mg}{k} e^{-^{\frac{k}{m}t}}

C

ve−kmt=c−mgkv e^{-^{\frac{k}{m}t}} = c- \frac{mg}{k}

D

vekmt=c−mgkv e^{^{\frac{k}{m}t}} = c- \frac{mg}{k}

Answer

v=ce−kmt−mgkv = ce^{-^{\frac{k}{m}t}} - \frac{mg}{k}

Explanation

Solution

dvdt+kmv=−g⇒dvdt=−km(v+mgk)\frac{dv}{dt}+\frac{k}{m}v = -g \Rightarrow \frac{dv}{dt} = -\frac{k}{m}\left(v+\frac{mg}{k}\right) ⇒dvv+mg/k=−kmdt⇒log(v+mgk)\Rightarrow \frac{dv}{v+mg/k} = -\frac{k}{m}dt \Rightarrow log \left(v+\frac{mg}{k}\right) =−kmt+log C= -\frac{k}{m}t+log\,C ⇒v+mgk=Ce−kt/m⇒v=Ce−kt/m−mgk\Rightarrow v+\frac{mg}{k}= Ce^{-kt/m} \Rightarrow v = Ce^{-kt/m} -\frac{mg}{k}