Air resistance Differential Equation Help

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Homework Help Overview

The discussion revolves around solving a differential equation related to air resistance, specifically the equation d²s/dt² + R ds/dt = g, where g represents gravity and R is a resistance constant. Participants are exploring the implications of this equation and its solutions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the form of the auxiliary equation and the general solution, questioning the correctness of the original equation and the proposed methods for solving it. There are attempts to linearize the equation and explore different methods, including substitution and separation of variables.

Discussion Status

The discussion is active, with participants providing feedback on each other's interpretations and methods. Some guidance has been offered regarding the linearization of the equation and the potential for different approaches, though there is no explicit consensus on the best method to proceed.

Contextual Notes

There are mentions of specific constants, such as g being constant at 10, and the distinction between homogeneous and non-homogeneous equations is being debated. Participants express uncertainty about the methods available to them based on their current knowledge.

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

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

Daniel.
 
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 ?
 
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.
 
So the general solution is wrong?
 
For the homogenous eq.is correct

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

Daniel.
 
well g is constant as it = to 10

so isn't the equation homogenous in essence

sid
 
No,a homogenous is when the constant is 0.

Daniel.
 
ok fair enough,

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

What do you definite for solving that equation

sid
 
  • #10
You tried one way.Try differently.Make a substitution

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

Daniel.
 
  • #11
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
 
  • #12
Separation of variables...?Write the new equation.You'll see that the variables would be separated.

Daniel.
 

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