Understanding Homogeneous Differential Equation Simplification

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Homework Help Overview

The discussion revolves around the simplification of a homogeneous differential equation, specifically focusing on the transformation of the equation from its original form to a new expression involving a substitution.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand the simplification process from the differential equation \(\frac{dy}{dx} = \frac{y-x}{y+x}\) to the expression \(\frac{v-1}{v+1}\) using the substitution \(v = \frac{y}{x}\). Some participants discuss the substitution method and its implications for the simplification.

Discussion Status

The discussion includes an explanation of the substitution method, with one participant successfully grasping the concept after receiving clarification. There appears to be a productive exchange of ideas, although no consensus is explicitly stated.

Contextual Notes

Participants are working within the constraints of understanding homogeneous differential equations and the specific substitution method involved. The original poster expresses confusion regarding the simplification process, indicating a need for further clarification on the underlying principles.

snowJT
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This is a basic simplification, but I'm going to post this here because it becomes homogeneous, and I know [tex]v = \frac{y}{x}[/tex] but I don't see this simplification, I don't understand how it gets from this...

[tex]\frac{dy}{dx} = \frac{y-x}{y+x}[/tex]

To THIS:

[tex]= \frac{v-1}{v+1}[/tex] (I'm just only showing the RHS here)

If someone wouldn't mind explaining it, that would be great because I'm lost, unless this is a special rule.
 
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Well, let y=vx, then sub into the RHS and you get [tex]\frac{vx-x}{vx+x}=\frac{x(v-1)}{x(v+1)}=\frac{v-1}{v+1}[/tex]
 
oh wow, that's incredible, I get it, thanks cristo
 
snowJT said:
oh wow, that's incredible, I get it, thanks cristo
You're welcome!
 

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