Using the inverse hyperbolic tangent function to solve ODE

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Homework Help Overview

The discussion revolves around solving the differential equation \(\frac{dv}{dt} = g(1 - \frac{\rho}{g}v^2)\), with a focus on the use of the inverse hyperbolic tangent function in the solution process.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to separate variables for integration but expresses uncertainty about integrating \(\frac{dv}{1-\frac{\rho}{g}v^2}\) and considers a substitution. Another participant suggests using partial fractions for the expression. A different approach involving the hyperbolic tangent function is also mentioned.

Discussion Status

Participants are exploring various methods to approach the problem, including separation of variables and substitution techniques. Some guidance has been provided, particularly regarding the use of hyperbolic functions and partial fractions, but no consensus has been reached on a single method.

Contextual Notes

The discussion indicates a need for clarity on integration techniques and the application of hyperbolic functions, with no explicit resolution of the problem yet.

bitrex
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Homework Statement


Hi all. I have to solve the differential equation [tex]\frac{dv}{dt} = g(1 - \frac{\rho}{g}v^2)[/tex].

The Attempt at a Solution



Apparently the solution should involve the inverse hyperbolic tangent function - with the equation in this form it should just be separable, correct? However, when separating variables I have to integrate the function [tex]\frac{dv}{1-\frac{\rho}{g}v^2}[/tex] which I am not sure how to go about. I think a substitution of some kind? Any tips would be appreciated.
 
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Partial fractions: [itex]1-\frac{\rho}{g}v^2= (1- \sqrt{\frac{\rho}{g}}v)(1+ \sqrt{\frac{\rho}{g}}v)[/itex].
 
Or v=sqrt(g/rho)*tanh(u), if you want to stick with the hyperbolic function approach. You'll get the same answer, though it will look different.
 
I see it now. Thanks guys!
 

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