Solving Linear Differential Equations: A Beginner's Guide

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Discussion Overview

The discussion focuses on understanding the process of solving linear differential equations, specifically the reasoning behind transforming the equation into a particular form for integration. Participants explore the implications of different approaches to integration and the effects of coefficients on the solution process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant questions why the equation is transformed into the form \(\frac{dp}{p-900}=\frac{1}{2} dt\) instead of \(\frac{dp}{0.5p-450}=1 dt\), seeking clarification on the reasoning behind this choice.
  • Another participant suggests that both forms are equivalent, noting that using \(0.5\) simplifies the integration process by making the coefficient in front of \(p\) equal to \(1\).
  • A later reply acknowledges the importance of including \(dt\) in the integration process and reflects on the integration steps that lead to different solutions if not handled correctly.
  • One participant realizes that forgetting to account for the factor of \(2\) during integration can lead to confusion and incorrect solutions, indicating a learning moment in understanding the chain rule.

Areas of Agreement / Disagreement

Participants generally agree that the two forms of the equation are equivalent, but there is some uncertainty regarding the integration process and how to properly account for coefficients. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

Participants express uncertainty about the integration steps and the impact of coefficients on the resulting solutions. There are mentions of missing factors and the importance of careful integration, but no consensus is reached on a definitive method.

ImAnEngineer
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Hey guys.

I've recently started studying differential equations. There is one thing I don't understand and of which I simply can't find an explanation.

I'm trying to solve some linear differential equations without using standard solutions.
Say we have the equation:
[tex]\frac{dp}{dt}=0.5p - 450[/tex]

The next step is (according to my book):

[tex](1) \frac{dp}{p-900}=\frac{1}{2} dt[/tex]

All of the next steps that lead to the solution are clear to me. They use the chain rule to integrate, exponentiate, and get: [itex]p=900+ce^\frac{t}{2}[/itex].
But what I don't understand, is why they first write it in the form of eq.(1), and not as, say:
[tex](2) \frac{dp}{.5p-450}=1 dt[/tex] ?

Possibly it's a silly question, but nevertheless, please help me out :) .
 
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It's equivalent, it's just that using .5 for the right hand side means the coefficient infront of p is simply 1 and the integration is slightly easier. If you used (2), a 2 would simply come out of the left hand side integration as well which makes up for the 1/2 on the right side.

Also, you do realize there is the 'dt' on the right side correct? It can't just disappear :).
 
Pengwuino said:
It's equivalent, it's just that using .5 for the right hand side means the coefficient infront of p is simply 1 and the integration is slightly easier. If you used (2), a 2 would simply come out of the left hand side integration as well which makes up for the 1/2 on the right side.

Also, you do realize there is the 'dt' on the right side correct? It can't just disappear :).
I thought it should be equivalent, but when I integrate it I get a different solution. I'm trying to integrate it again now to see how I should get the 2 on the left hand side. Thanks for your answer.

PS: I realized that I forgot the dt's when I had turned of the computer and went to bed :D
 
Ah, now I see why the 2 comes in. If you integrate it without the 2, then you forget to compensate for the chain rule factor (or whatever you call it). I don't know if that sentence makes sense, but I get it now :) .
 

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