Integrating ODE: dv/dt=g-kv/m Step-by-Step Guide

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Homework Help Overview

The discussion revolves around integrating the ordinary differential equation (ODE) given by dv/dt = g - kv/m. Participants are exploring methods to integrate this equation and are sharing their attempts and questions regarding the integration process.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rewriting the left side of the equation for integration and consider potential substitutions. There are questions about specific integration techniques and simplifications of the resulting expressions.

Discussion Status

The discussion includes various attempts at integration, with some participants expressing uncertainty about the integration process. Others have made progress and shared their results, indicating a mix of understanding and confusion about the steps involved.

Contextual Notes

Some participants are questioning the assumptions behind the integration techniques and whether certain substitutions are appropriate. There is a focus on the integration of expressions that involve logarithmic functions and exponential forms.

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Homework Statement


How to integrate dv/dt=g-kv/m?

Homework Equations


This is ODE.

The Attempt at a Solution


dv/(g-(kv/m))=dt
Now what? I don't know how to integrate the left side.
 
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Math10 said:

Homework Statement


How to integrate dv/dt=g-kv/m?

Homework Equations


This is ODE.

The Attempt at a Solution


dv/(g-(kv/m))=dt
Now what? I don't know how to integrate the left side.
The left side can be rewritten as ##\frac{m \ dv}{mg - kv}##. There's a fairly obvious substitution you can use.
 
What substitution?
 
Never mind. I integrated and got (-m/k)ln abs(gm-kv)+C.
 
Now, how to simplify (-m/k)ln abs(gm-kv)=t+C?
 
Math10 said:
dv/(g-(kv/m))=dt
Now what? I don't know how to integrate the left side.

Are you saying you can't work \int {\frac{1}{g- \frac{kv}{m}} }dv ?Could you do something like \int (3 -5v)^{-1} dv ?
 
No, now I got it. I simplified and got gm-kv=Ce^(-kt/m).
 

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