Solving 1st Order ODE for Laminar Flow: A Simple Guide | Jerome, Engineer

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Discussion Overview

The discussion revolves around solving a first-order ordinary differential equation (ODE) related to laminar flow, specifically the equation dp/dx + A/p - B = 0. Participants explore various methods to approach this problem, including separation of variables and the implications of different values for the constant B.

Discussion Character

  • Technical explanation, Mathematical reasoning, Homework-related

Main Points Raised

  • Jerome expresses difficulty in solving the ODE and seeks assistance.
  • One participant suggests that for B = 0, the solution can be expressed as p^2 = -2Ax + c.
  • Another participant notes that for B not equal to 0, Mathematica provides a solution involving a ProductLog, indicating that a simple solution may not exist except for specific values of A and B.
  • A different participant presents an implicit solution derived from separation of variables: p + (A/B)ln(Bp - A) = Bx + constant.
  • Jerome requests clarification on the separation of variables method due to confusion regarding the constant term B.
  • Another participant responds by stating that separation can be achieved, leading to the equation p/(Bp - A) dp = dx.
  • Jerome expresses gratitude for the clarification and indicates that it has been helpful.

Areas of Agreement / Disagreement

The discussion contains multiple approaches to solving the ODE, with no consensus on a single method or solution. Participants present differing views on the solvability of the equation depending on the value of B.

Contextual Notes

Participants have not fully resolved the implications of the constant term B in the context of separation of variables, and there are unresolved mathematical steps in the proposed solutions.

plasticpigeon
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Hello people

I am an engineer and therefore not a great mathematician.

To solve a problem involving laminar flow i need to solve the equation with a general form

dp/dx + A/p - B = 0 which I don't know how to do.

Can anyone shed any light on how to solve this simple looking problem.

Many thanks

Jerome
 
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For B = 0 the solution can be written as
[tex]p^2 = - 2 A x + c[/tex]

For B not equal to 0, Mathematica gives something with a ProductLog, so there is probably no nice solution, except for special values of A and B.
 
separation of variables gives you the implicit solution

p+(A/B)ln(Bp-A)=Bx + constant.
 
Dear Defunc

Many thanks for your reply. I'd be very grateful if you could explain to me how you got the solution. I could not see how to separate variables because of the constant term B.

Many thanks

Jerome
 
You can separate it to obtain te following:


p/(Bp-A) dp=dx.
 
thanks, that has helped me a lot!
 

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