Theoretical question on cyclic function

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Homework Help Overview

The discussion revolves around a theoretical question regarding a cyclic function defined by a continuous function \( a(x) \) with period \( T \) and its relationship to the differential equation \( y' = a(x)y \). Participants are exploring the implications of the integral \( \lambda = \int_{0}^{T} a(x) dx \) on the behavior of solutions \( u(x) \).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the separability of the equation and the integration process over one period. There are questions about how to properly integrate and what steps to take following the integration.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the integration process and its implications for the solution \( u(x) \). Some guidance has been offered regarding the integration of the equation, but there is no explicit consensus on the claims presented by the original poster.

Contextual Notes

There is a noted lack of theoretical basis from the original poster to fully address the claims regarding the behavior of \( u(x) \) based on the value of \( \lambda \). Participants are also questioning the assumptions underlying the integration process.

nhrock3
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a(x) is continues on R with cycle T ,a(x+T)=a(x)
u(x) is non trivial soluion of y'=a(x)y
[tex]\lambda=\int_{0}^{T}a(x)dx[/tex]

which of the following claims is correct:

A. if [tex]\lambda>0[/tex] then [tex]\lim_{x\rightarrow\infty}u(x)=\infty[/tex]
B. if [tex]\lambda=0[/tex] then u(x) is a cyclic function

i don't have the theorectical basis to solve it
 
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The equation is separable. Write it as

[tex]\frac{y'}{y} = a(x)[/tex]

and integrate both sides over one period.
 
how to integrate over a cycle
?
what to do after i integrate over a cycle
[tex]\int\frac{dy}{y}=\int a(x)dx[/tex]

[tex]\ln y=\inta(x)dx[/tex]
 
nhrock3 said:
how to integrate over a cycle
?
what to do after i integrate over a cycle
[tex]\int\frac{dy}{y}=\int a(x)dx[/tex]

[tex]\ln y=\inta(x)dx[/tex]

Try

[tex]\int_{u(0)}^{u(T)}\frac{dy}{y}=\int_0^T a(x)dx[/tex]
 

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