Classifying a Diff Eq: Linear vs. Non-Linear Techniques

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Discussion Overview

The discussion revolves around classifying a differential equation of the form x(dy/dx) = y*e^(x/y) - x. Participants explore various classification techniques, including separability, linearity, homogeneity, Bernoulli, and exactness, while expressing challenges in breaking down the equation due to the presence of the exponential term.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests the equation might be linear but expresses uncertainty about how to analyze it due to the e^(x/y) term.
  • Another participant asserts that the equation is not linear and proposes dividing by x to reformulate the right-hand side as a function of g(y/x).
  • A different participant attempts to rearrange the equation into a separable form but notes that the integral remains complex due to the exponential term.
  • Another participant introduces a substitution v(x) = y(x)/x, leading to a reformulated equation that they claim is separable, although they advise against attempting to solve it.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the classification of the differential equation, with differing opinions on its linearity and separability remaining unresolved.

Contextual Notes

Participants express uncertainty regarding the integration of the exponential term and the implications of their proposed transformations, indicating potential limitations in their approaches.

formulajoe
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x(dy/dx) = y*e^(x/y) - x

its either separable, linear, homogeneous, bernoulli or exact. only thing i can figure is that its linear.

how do i break it down to figure it out? the e^x/y is what's throwing me off.
 
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It's not linear!
Divide by x.
Note that your right-hand side can now be written as some function g(y/x).
 
if i break it apart i get dy/dx = (y*e^(x/y) / x) - 1. i can see how it could be separable, but the e^x/y would still be there when i integrate. and that integral would be fairly impossible. i don't see what else it can be.
 
Introduce the variable:
[tex]v(x)=\frac{y(x)}{x}[/tex]
We have then:
[tex]v'(x)=\frac{y'(x)}{x}-\frac{v}{x}[/tex]
Or:
[tex]y'(x)=xv'(x)+v(x)[/tex]
Hence, inserting this into your diff. eq., you have:
[tex]xv'(x)=ve^{\frac{1}{v}}-(1+v)[/tex]
This is a separable equation (I wouldn't try solving it, though..)
 

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