Calculating Velocity of a Falling Water Drop Using Newton's Second Law

[Idoubt]

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=kt². 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 f_ydy}

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 = kt² 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_tdt = 2lnt = ln t²

integrating factor Q = e^{integral f_tdt} = e^lnt2 = t²multplying the DE by the int factor

t²dv + 2vtdt - gt²dt = 0

the first two terms combine to form a perfect integral

dU where U = vt²

so dU - gt² = 0

integrating

U - gt³/3 = C , where C is a constant

ie vt² - gt³/3 = 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.

 

[Idoubt]