Answer
$v(t)=v(0)e^{-\frac cmt}+\frac{mg}c(1-e^{-\frac cmt})$
Work Step by Step
Equation 1.9:
$\frac{dv}{dt}=g-\frac cmv$
Multiply by $e^{\frac cmt}$
$v'e^{\frac cmt}+\frac cmve^{\frac cm t}=ge^{\frac cmt}$
Product rule
$(ve^{\frac cm t})'=ge^{\frac cmt}$
$v(t)e^{\frac cmt}-v(0)=g\int\limits_0^te^{\frac cm\bar{t}}d\bar{t}$
Note: bar over t is to differentiate the integral variable from the time at an instant.
$v(t)=v(0)e^{-\frac cmt}+\frac{mg}c(1-e^{-\frac cmt})$
