Derivation of Drag Force in air(D=bv)

[shayaan_musta]

I have stuck at a step during derivation.
Here it is.
D=bv
mg-bv=ma
a=g-\frac{bv}{m}
\frac{dv}{dt}=g-\frac{bv}{m}
\frac{dv}{g-\frac{bv}{m}} = dt
or,
\frac{dv}{\frac{mg-bv}{m}} = dt
\frac{dv}{mg-bv} = \frac{dt}{m}
let, u=mg-bv
\frac{du}{dv}=0-b
\frac{du}{dv}=-b
multiplying both sides by, "-b"
\frac{-bdv}{mg-bv}=\frac{-bdt}{m}
\frac{du}{u}=\frac{-bdt}{m}
integrating L.H.S. from 0 to u and Integrating R.H.S. from 0 to t
\int^{u}_{0}\frac{du}{u}=\frac{-b}{m}\int^{t}_{0}dt
ln(u)|^{u}_{0}=\frac{-bt}{m}
ln(u)=\frac{-bt}{m}
ln(mg-bv)=\frac{-bt}{m}
mg-bv=e^{\frac{-bt}{m}}

Now, I am stuck
L.H.S. must be \frac{mg-bv}{mg}
but I am unable to get mg in denominator.

Where I am wrong?

 

[torquil]

I haven't checked everything, but one thing is that you cannot integrate u from 0, since that integral doesn't converge. So you cannot choose to have u=0 when t=0 as you have done by your choice of integration limits.

How about selecting v=0 at t=0 as your initial condition, i.e. you'll do the u-integral with the lower limit 'mg' instead of 0.

What do you think?

 

[AlephZero]

The place you went wrong was

integrating L.H.S. from 0 to u and Integrating R.H.S. from 0 to t

If v = 0 when t = 0, what does u equal when t = 0? (Hint, the answer is not 0).