How can I verify the separability of a differential equation with constants?

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Homework Help Overview

The discussion revolves around verifying the separability of a differential equation of the form dy/dx = f(ax + by + c), where a, b, and c are constants. Participants are exploring how to manipulate the equation to demonstrate that it can be expressed in a separable form.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express the equation in terms of functions g and h to achieve separability. They consider substituting u = ax + by + c and explore the implications of this substitution. Other participants suggest rearranging terms to isolate variables and provide hints on how to proceed with the manipulation.

Discussion Status

The discussion is active, with participants providing guidance and suggestions for approaching the problem. The original poster expresses confusion but receives tips that may help clarify the path forward. There is no explicit consensus, but productive dialogue is ongoing.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to work with constants and the requirement to demonstrate separability without providing a complete solution.

fluidistic
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Homework Statement


Verify that the following ODE can be reduced to an ODE of separable variables.
[tex]\frac{dy}{dx} =f(ax+by+c)[/tex] where a, b and c are constants.2. The attempt at a solution
I think I must show that there exist functions g and h such that [tex]g(y)dy=h(x)dx[/tex].
I have that [tex]dy=f(ax+by+c) dx[/tex]. I was at a loss. So I talked to a friend and he told me to write [tex]u=ax+by+c[/tex].
So I get [tex]dy=f(u)dx \Rightarrow y= \int f(u)dx=\frac{u-ax-c}{b}[/tex], [tex]y'=\frac{u'-a}{b}[/tex], [tex]y''=u''[/tex]. I want to write [tex]f(u)[/tex] as [tex]\phi _1 (x) \phi _2 (y)[/tex] but I'm totally stuck.
I'd love a tip.
Thank you.
 
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You have figured out that y' = (u'-a)/b and if you plug that in for y' you get

(u'-a)/b = f(u)

Write u' as du/dx and, remembering that a and b are constants, see if you can't get the u terms and the x terms on opposite sides.
 
ya lckurtz sums it up

also try looking here:http://www.tutorvista.com/math/separable-differential-equation"
 
Last edited by a moderator:
Thank you guys, I got it.
 

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