Initial Value Problem for (DE)

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SUMMARY

The discussion focuses on solving the initial value problem defined by the differential equation dv/dt = 9.8 - (v/5) with the initial condition v(0) = 0. Participants clarify the necessity of rewriting the equation in the form (dv/dt)/(9.8 - (v/5)) = 1dt for integration purposes. The solution involves integrating both sides, leading to the expression -5ln(9.8 - (v/5)) = t + C, where the integral of 1/x is identified as ln(x). This understanding is crucial for determining the time to reach 98% of the limiting velocity and the distance fallen during that time.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with integration techniques, specifically natural logarithms
  • Knowledge of initial value problems in calculus
  • Basic physics concepts related to velocity and acceleration
NEXT STEPS
  • Study the method of separation of variables in differential equations
  • Learn about integrating factors for solving linear differential equations
  • Explore applications of exponential decay in physics
  • Review the concept of limiting velocity in motion problems
USEFUL FOR

Students studying calculus, particularly those tackling differential equations, as well as educators looking for clear explanations of integration techniques and initial value problems.

Vanessa Avila
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Homework Statement


dv/dt = 9.8 - (v/5) , v(0) = 0

(a) The time it must elapse for the objet to reach 98% of its limiting velocity
(b) How far does the object fall in the time found in part (a)?

Homework Equations


(dv/dt)/(9.8-(v/5))

The Attempt at a Solution


I'm a little overwhelmed by this class and I think the problem I have is I'm not catching on as to why the next answer is the way it is, so it would be nice if someone could explain to me why.
As I read on my textbook, I'm supposed to rewrite the form of the eqn first in which I attempted to do:

(dv/dt)/(9.8-(v/5)) = 1dt
but why do we have to rewrite the eqn in this form? I checked the solution for this, and it said it was right. I know afterwards you then integrate both sides, but I saw that they got
−5ln(9.8−(v/5))=t+C
I understand where the t+C comes from but not where the -5ln(9.8-(v/5)) comes from. Can someone explain to me how to get to that?
 
Physics news on Phys.org
What is the integral of 1/x?
 
Orodruin said:
What is the integral of 1/x?
ln(x)! Ah I missed that. Thanks!
 

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