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

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In summary, the conversation is about how to integrate a given equation involving velocity and time. The solution involves rewriting the left side using a substitution and then integrating to get (-m/k)ln abs(gm-kv)+C. Further simplification leads to the final solution of gm-kv=Ce^(-kt/m).
  • #1
Math10
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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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  • #2
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.
 
  • #3
What substitution?
 
  • #4
Never mind. I integrated and got (-m/k)ln abs(gm-kv)+C.
 
  • #5
Now, how to simplify (-m/k)ln abs(gm-kv)=t+C?
 
  • #6
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 [itex] \int {\frac{1}{g- \frac{kv}{m}} }dv [/itex] ?Could you do something like [itex] \int (3 -5v)^{-1} dv [/itex] ?
 
  • #7
No, now I got it. I simplified and got gm-kv=Ce^(-kt/m).
 

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