Air resistance Differential Equation Help

[$id]

Hi guys

I need help solving the following differential equation for air resistance

d^2s/dt^2 + R ds/dt = g


Where g = gravity I presume and R = k/m where k = resistance constant

Using my limited knowledge I think that the auxiliary equation for this is

Y^2+ RY = 0

Y = 0 or R

General Solution

S = A + Be^Rt

Which log graph would i need to plot to work out resistance constant.

thanks a lot

sid

 

[dextercioby]

You meant "-R" and [itex]...e^{-Rt} [/itex] ("R" positive,i presume).

Linearize the equation.U'll find R from the negative slope...

Daniel.

 

[$id]

Is the actualy equation correct though?

Or am i missing something

sid

BTW by linearise i think you mean

ln(S)=LnA + LnB -RT

So a plot of Kn(S) against T for some values should yeild the gradinet at -R

yeah ?

 

[dextercioby]

Yes,for the second part.The initial equation may be correct,i don't know how u've gotten it.But the solution you had found was incorrect.

Daniel.

 

[$id]

So the general solution is wrong?

 

[dextercioby]

For the homogenous eq.is correct

[tex] s(t)=A+Be^{-Rt} [/tex],with R>0...

Daniel.

 

[$id]

well g is constant as it = to 10

so isn't the equation homogenous in essence

sid

 

[dextercioby]

No,a homogenous is when the constant is 0.

Daniel.

 

[$id]

ok fair enough,

The orginal differential equation is definate correct, it was taken from a book

What do you reccomend for solving that equation

sid

 

[dextercioby]

You tried one way.Try differently.Make a substitution

[tex] \frac{ds(t)}{dt}=:u(t) [/tex]

Daniel.

 

[$id]

Unfortunately i am not aware of that method

the only ones i know are auxiliary equation, separation of variables and intergrating factors

Do you which one of those will work.

sid

 

[dextercioby]

Separation of variables...?Write the new equation.You'll see that the variables would be separated.

Daniel.