Though upper and lower case, I suspect that probably the two v's are intended to be the same, in which case, separating the variables,
\(\displaystyle \frac{dv}{dt}=9.8-2v\),
\(\displaystyle \frac{dv}{9.8-2v}=dt\),
\(\displaystyle \int\frac{1}{9.8-2v}dv=\int dt\),
etc..
Alternatively, you could write the equation as
\(\displaystyle \frac{dv}{dt}+2v=9.8\)
and use the integrating factor technique,
(multiply throughout by exp(2t)).
Tiggsy.