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## Homework Statement

A drop of water of mass m is falling vertically towards the ground. Due to moisture, the mass of the drop is increasing as given by m=kt

^{2}. The equation of motion is

mdv/dt + vdm/dt = mg

find v after 1 second.

## Homework Equations

The integrating factor of an imperfect integral

Adx + Bdy = 0

is given by Q = e

^{integral fydy }

where f

_{y}= 1/A[ dB/dx - dA/dy]

## The Attempt at a Solution

rearranging the equation of motion

dv/dt + (v/m)dm/dt -g = 0

substituting m = kt

^{2}and dm/dt = 2kt

dv/dt + 2v/t - g =0

i.e dv + (2v/t -g )dt = 0 , now it is the form Adv + Bdt, where A=1, B=(2v/t-g)

here f

_{t}= 1/1[ 2/t-0] = 2/t, integral f

_{t}dt = 2lnt = ln t

^{2}

integrating factor Q = e

^{integral ftdt}= e

^{lnt2}= t

^{2}

multplying the DE by the int factor

t

^{2}dv + 2vtdt - gt

^{2}dt = 0

the first two terms combine to form a perfect integral

dU where U = vt

^{2}

so dU - gt

^{2}= 0

integrating

U - gt

^{3}/3 = C , where C is a constant

ie vt

^{2}- gt

^{3}/3 = C

when t=0, C=0 so

v = gt/3

I can't find anything wrong with my work, but the answer choices to the question are

1, 0.25 g

2, 0.5 g

3, 0.75 g

4, g

Can some1 tell me where I messed up? thanks in advance.