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-$$\frac{bv}{m}$$
[tex]\frac{dv}{g-$$\frac{bv}{m}$$}[/tex] = 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).