Integrating Factor, how do you get this?

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Homework Help Overview

The discussion revolves around the concept of integrating factors in differential equations, specifically focusing on the manipulation of terms in the context of integration. Participants are examining how certain variables are treated during the integration process.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the reasoning behind the treatment of the integrating factor, questioning whether it was dropped or solved for during integration. There are attempts to clarify the steps involved in deriving the integrating factor and its implications.

Discussion Status

The discussion is ongoing, with some participants providing clarifications about the integration process and the treatment of the integrating factor. There is a recognition of differing interpretations regarding the steps taken in the original problem, and no consensus has been reached yet.

Contextual Notes

Participants are grappling with the clarity of the steps shown in the original problem, and there are indications of varying levels of familiarity with the concepts involved in integration and differential equations.

flyingpig
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Homework Statement



My book shows that

[PLAIN]http://img845.imageshack.us/img845/4875/unleduot.jpg

and then they arrived at the results

[PLAIN]http://img202.imageshack.us/img202/5442/unleddq.jpg

So my problem is that, how exactly did they drop the mu in the first picture?
 
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I don't understand your question. I can see nowhere that the "dropped" the mu. In the "first picture", they wound up solving for mu.
 
[PLAIN]http://img703.imageshack.us/img703/2056/unledid.jpg
 
Last edited by a moderator:
As I said, they didn't "drop" mu, they solved for it!
If
[tex]\frac{du}{dx}= F(x)u[/tex]
then
[tex]\frac{du}{u}= F(x)dx[/tex]
and, integrating both sides,
[tex]ln(u)= \int F(x)dx[/tex]

Now take the exponential of both sides.
 
Ohhh okay that wasn't clear to me at first. Is there any good way to memorize it? It takes too long to derive it lol
 
The fact that
[tex]\frac{dy}{dt}= f(x)g(y)[/tex]
gives
[tex]\frac{dy}{g(y)}= f(x)dx[/tex]
should be immediate from Calculus.
 
How do know that mu(x) is linear?
 

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