Find the only periodic solution of an ODE

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Discussion Overview

The discussion revolves around finding the only periodic solution of the ordinary differential equation (ODE) given by 𝑦′+𝑦=𝑏(π‘₯), where 𝑏 is a periodic function. The focus is on the conditions for periodicity and continuity of the solution.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant presents the general solution to the ODE as 𝑦(π‘₯)=exp(βˆ’π‘₯)⋅𝑐+1 and 𝑦(π‘₯)=exp(βˆ’π‘₯)β‹…π‘βˆ’1, questioning how to determine the constant 𝑐 for periodicity.
  • Another participant notes that for 𝑐 = 0, the solution would not be continuous, implying a need for a different value of 𝑐.
  • A subsequent reply questions whether the solution jumps from -1 to 1, indicating a concern about continuity.
  • One participant acknowledges a misunderstanding and retracts their previous comments, suggesting a lack of clarity in the discussion.
  • Another participant suggests that the two given values provide ranges for 𝑐, and emphasizes that continuity will influence the prefactors in the solution.
  • There is confusion regarding the notation used, specifically why π‘₯ depends on two parameters, indicating a need for clarification on the formulation.

Areas of Agreement / Disagreement

Participants express differing views on the continuity of the solution and the implications of the constant 𝑐. There is no consensus on how to proceed with finding the periodic solution.

Contextual Notes

The discussion highlights potential limitations in understanding the notation and the assumptions regarding continuity and periodicity of the solution.

michii15
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TL;DR
Find the only periodic solution for 𝑦′+𝑦=𝑏(π‘₯) with 𝑏:ℝ→ℝ has a period of 2𝑇 and is 1 for π‘₯(0,𝑇) and βˆ’1 for π‘₯(βˆ’π‘‡,0).
Find the only periodic solution for 𝑦′+𝑦=𝑏(π‘₯) with 𝑏:ℝ→ℝ has a period of 2𝑇 and is 1 for π‘₯(0,𝑇) and βˆ’1 for π‘₯(βˆ’π‘‡,0).

The ODE is easy to solve: 𝑦(π‘₯)=exp(βˆ’π‘₯)⋅𝑐+1 and 𝑦(π‘₯)=exp(βˆ’π‘₯)β‹…π‘βˆ’1. But how can I find the 𝑐 such that the solution is periodic with a period of 2𝑇?

The solution is: https://i.stack.imgur.com/S21Ze.png

Thanks for your help.
 
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sorry forgot to mention the function should be continuous.
 
but for c = 0 you won't have a continuous solution
 
it jumps frome -1 to 1?
 
Sorry I misread your question. You are right. Deleting all my previous comments...
 
Well, you have two values given, that gives you ranges to play with, continuity will determine the prefactors.

I'm not sure if I understand your notation, however. Why does x depend on two parameters?
 

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