Solving Differential Equation: ln |9/64| = k

Click For Summary

Homework Help Overview

The discussion revolves around solving a differential equation involving logarithmic expressions, specifically focusing on the term k = ln |9/64|. Participants are exploring the implications of using logarithms in the context of the equation and the steps that follow.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive the next steps after establishing k = ln |9/64|. Questions are raised about the validity of using ln 0 and the transition from one form of the equation to another. There are also inquiries about the disappearance of terms and the correctness of the integration process.

Discussion Status

The discussion is ongoing, with participants sharing their work and questioning each other's reasoning. Some guidance is provided regarding the integration of functions, but there is no clear consensus on the next steps or the correctness of the approaches taken.

Contextual Notes

Participants are navigating the complexities of logarithmic properties and differential equations, with some expressing uncertainty about the steps taken and the assumptions made in their calculations.

kyu
Messages
12
Reaction score
0


i got k = ln |9/64|

then how can the next step using ln 0 doesn't make sense. what should i do?
 

Attachments

  • ma2.jpg
    ma2.jpg
    16.1 KB · Views: 478
Physics news on Phys.org
kyu said:
i got k = ln |9/64|

then how can the next step using ln 0 doesn't make sense. what should i do?

Please show us your work.
 
Mark44 said:
Please show us your work.

should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
ln 9 - ln 64 = k
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)
 
kyu said:
should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
Where did the dh go?
kyu said:
ln 9 - ln 64 = k
How did you get this (above)?
kyu said:
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)
 
kyu said:
should be wrong but here goes

dh/dt = -k h^(1/2)
1/h^(1/2) = -k dt
ln 9 - ln 64 = k
ln |9/64| = k

ln 0 - ln 64 = ln |9/64| (t-0)

The integral of [itex]h^{-1/2}[/itex] is given by the usual rule for powers, [itex]\int x^\alpha\,dx = \frac{x^{\alpha + 1}}{\alpha + 1} + C[/itex], not the exception [itex]\int x^{-1}\,dx = \ln |x| + C[/itex].
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K