Mastering Equation Solving: m(dv/dt)=mg-bvn | Step-by-Step Guide

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SUMMARY

The discussion centers on the differential equation m(dv/dt) = mg - bvn, which is questioned for its dimensional consistency and clarity. Participants clarify that the equation likely represents a falling body with a resistance force proportional to velocity raised to the power of n. The correct formulation is suggested to be m(dv/dt) = -mg - bvn, indicating a need for proper integration techniques to solve it. The conversation emphasizes the importance of understanding the physical meanings of the symbols involved in the equation.

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  • Understanding of differential equations
  • Familiarity with Newton's laws of motion
  • Knowledge of dimensional analysis
  • Basic integration techniques
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  • Study the derivation of differential equations in physics
  • Learn about the concept of drag force in fluid dynamics
  • Explore integration methods for solving ordinary differential equations
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Students of physics, mathematicians, and engineers interested in mastering differential equations and their applications in motion analysis.

Opressor
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m (dv / dt) = mg-bvn
 
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Opressor said:
m(d2v/dt2)=-mg-bvn
This equation doesn't make sense if the symbols have their usual meanings. Did you mean m(dv/dt)=-mg-bvn ?
 
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What does it represent? It is a differential equation (second derivative of velocity?) equals -(mass)(gravity?) - (some number b) times (velocity?)^n
If I am even close, it does not look like the dimensions match.
 
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is the equation of a falling body having as resistance force bv ^ n
 
It looks like you are asking about solutions to dx/dt = 1 - axn? For a and n = constants you can just integrate it, so I guess there is more to your equation than we understand?
 
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