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

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SUMMARY

The discussion focuses on integrating the ordinary differential equation (ODE) represented by dv/dt = g - kv/m. The integration process involves rewriting the equation as dv/(g - (kv/m)) = dt and applying the substitution method. The final solution derived is gm - kv = Ce^(-kt/m), where C is the constant of integration. This step-by-step guide clarifies the integration process and highlights the importance of substitution in solving ODEs.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with integration techniques
  • Knowledge of substitution methods in calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study advanced integration techniques for ODEs
  • Learn about the method of separation of variables
  • Explore the application of exponential functions in differential equations
  • Investigate the use of initial conditions in solving ODEs
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Students studying calculus, mathematicians focusing on differential equations, and educators teaching integration techniques will benefit from this discussion.

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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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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