Simple problem in Mechanics, weird differential equation

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Homework Help Overview

The discussion revolves around a mechanics problem involving a particle of mass m subjected to a retarding force described by a differential equation. The original poster seeks to solve the equation for the velocity function v(t) given an initial condition.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to formulate the differential equation and expresses uncertainty about solving it. Some participants question the method of separating variables and suggest rewriting the equation for integration. Others provide guidance on integrating the equation and imply a relationship involving logarithms.

Discussion Status

The discussion is active, with participants offering insights on how to approach the problem. Some guidance has been provided regarding integration techniques, although there is no explicit consensus on the final solution or method.

Contextual Notes

Participants note the nature of the retarding force and its implications for the differential equation, indicating that the setup involves specific assumptions about the forces acting on the particle.

dujardin
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There is a problem I couldn't figure out , it says :

it says that a particle of mass m moves along a straight line and is acted on by a retarding force (one always directed against the motion) F=b*exp(a*v(t)),
b, a are constants and v is the velocity.

At t=0 it is moving with velocity V

and I am aked to solve the differential equation that results from this to get a function of v(t).

I found that the differential equation that has to be solved is :

dv/dt = (b/m)*exp[a*v]

so this is like solving a non-linear differential equation of the form y'=exp(y)

How do you do that??
 
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Do you know how to separate variables ?
 
You have simply to re-write the equation in terms of differentials and you will be able to integrate:

[tex]\frac{du}{dt} = f(u)[/tex]
becomes:
[tex]du = f(u) dt[/tex]
and then
[tex]\frac{du}{f(u)} = dt[/tex]
you can then integrate:
[tex]\int \frac{du}{f(u)} = t + C[/tex]
which gives you [tex]u(t)[/tex] implicitly.

In your case you should get [tex]v(t)[/tex] in terms of a logarithm of t.
 
jambaugh said:
You have simply to re-write the equation in terms of differentials and you will be able to integrate:

[tex]\frac{du}{dt} = f(u)[/tex]
becomes:
[tex]du = f(u) dt[/tex]
and then
[tex]\frac{du}{f(u)} = dt[/tex]
you can then integrate:
[tex]\int \frac{du}{f(u)} = t + C[/tex]
which gives you [tex]u(t)[/tex] implicitly.

In your case you should get [tex]v(t)[/tex] in terms of a logarithm of t.


I understood everything. Thank you so much
 
Just a note! :smile:

Since you have a retarding force, the ODE should be

[tex]\frac{d\,v}{d\,t}=-\frac{b}{m}\,e^{a\,v}[/tex]
 

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