Finding the diferencial function of a moving particle

[Gbox]

Homework Statement

A particle is moving in a liquid ##a=-kv## when ##k## is a constant, if a constant force is being apply on the particle ##a=-kv+\frac{F}{m}##.
1. for the formula that's include the force, find the diferencial function where ##v## will appear in it explicitly, ad its derivites
2. define a new variable ##U=v-\frac{F}{mk}## and substitute in place of ##v## that is in the diferencial function.
3. guess a solution in the form of ##U=AE^{-Bt}## where A and B are contestants, find B that for it the equation is true
4. find A and when the velocity at ##t=0## is 0
5.find ##x(t)## when ##x(t=0)=0##
6. according to the result on 4, what will be the velocity of the particle when ##t\rightarrow \infty##

Homework Equations

##a=-kv##
##x+x_o+v_0t+\frac{at^2}{2}##

The Attempt at a Solution

1. I should be looking for the functions which its derivites is ##a=-kv+\frac{F}{m}## so ##v=-kvt+ frac{Ft}{m}##
And ##x=x_0+\frac{kvt^2}{2}+\frac{Ft^2}{2m}##?

2.##x=x_0+\frac{k*(v-\frac{F}{mk})*t^2}{2}+\frac{Ft^2}{2m}##

Are 1 and 2 are correct?

 

[Let'sthink]

what is a

 

[Gbox]

a Is the acceleration

 

[Let'sthink]

Statement of the question is in a clumsy language and confusing. Moment we say a particle is moving with a = -kv, means force is acting on it = ma = -mkv. and this is a velocity dependent force. Now if additional constant force is applied, we have new acceleration a' given by
ma' = F - kmv or
a' = d²x/dt² = (F/m) - k(dx/dt), because we want an expression with v and dv/dt, we have
dv/dt = (F/m) - kv
Your very first step is wrong. Integral of kv is not kvt because v is not a constant but function of t in general in both cases when F = or F is non-zero.. The rest thing you try using your mathematical skills related to integral calculus.