What does it mea for an equation to be homogeneous?

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Discussion Overview

The discussion revolves around the concept of homogeneity in equations, specifically focusing on its meaning in the context of differential equations. Participants explore different interpretations and applications of the term in both first-order and higher-order equations.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant notes that the term "homogeneous" has two distinct meanings in differential equations: one for first-order equations and another for higher-order linear equations.
  • For first-order equations, a function f(x,y) is considered homogeneous if f(ax, ay) = f(x, y) for any number a, implying that f can be expressed as a function of y/x only.
  • For higher-order linear equations, an equation is homogeneous if the non-homogeneous term f(x) equals zero for all x.
  • Another participant expresses gratitude for clarification, indicating a specific interest in second-order differential equations.

Areas of Agreement / Disagreement

Participants appear to agree on the definitions provided for homogeneity in different contexts, but there is no consensus on a singular interpretation, as the term is used in multiple ways.

Contextual Notes

The discussion does not resolve the potential confusion arising from the different uses of the term "homogeneous" in various contexts of differential equations.

bmed90
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As the title says, what does it mean for an equation to be homogenous?
 
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Unfortunately, there are two quite different uses of the term "homogeneous" in differential equations.

1) As applied to first order equations, an equation of the form y/dx= f(x,y) is "homogeneous" if and only if f(ax, ay)= f(x, y) for any number a. That is the same as saying that f can be thought of as a function of y/x only.

2) As applied to a linear equation of order higher than 1, the equation a_n(x)d^n/x^n+ a_{n-1}(x)y/dx^{n-1}+ \cdot\cdot\cdot+ a_1(x)dy/dx+ a_0(x)y= f(x) is homogeneous if and only if f(x)= 0 for all x.
 
Hey thanks for your replies. I guess I should have been more specific in regards to the fact that I was inquiring about 2nd order diff eqs. @ HallsofIvy YOu answered my question perfectly. Simple and straight to the point. Thanks
 

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