Air resistance Differential Equation Help

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SUMMARY

The forum discussion centers on solving the differential equation for air resistance, represented as d²s/dt² + R ds/dt = g, where g is the gravitational constant (10 m/s²) and R is defined as k/m (resistance constant). Participants confirm the general solution as s(t) = A + Be^(-Rt) with R being positive. The discussion emphasizes the need to linearize the equation to determine the resistance constant, suggesting that plotting ln(S) against T will yield a gradient of -R. The original equation is validated as correct, and alternative methods such as separation of variables are recommended for solving it.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with concepts of air resistance and gravitational force.
  • Knowledge of linearization techniques in mathematical modeling.
  • Experience with plotting logarithmic graphs for data analysis.
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  • Learn about linearization techniques for differential equations.
  • Study the method of separation of variables for solving differential equations.
  • Explore the application of resistance constants in physics and engineering.
  • Investigate the use of auxiliary equations in solving linear differential equations.
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Students and professionals in physics, engineering, and applied mathematics who are interested in solving differential equations related to air resistance and gravitational forces.

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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 ...e^{-Rt} ("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

s(t)=A+Be^{-Rt},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

\frac{ds(t)}{dt}=:u(t)

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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