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Integrating Factor

  1. Oct 20, 2007 #1
    Hi,

    I'm learning differential equations, and although I understand the methods I have learned thus far, I often have trouble seeing what is the reasoning behind them.

    Take for example, the use of the integrating factor when solving first order linear ODE's. I understand how to use it, but i'm not sure where it came from. In the resources that i'm using there's really not explanation, instead the discussion is limited to: assume there's a function by which we can multiply our differential equation to make it integrable.

    Can anybody explain why we use this approach?

    thanks,

    Erik
     
  2. jcsd
  3. Oct 20, 2007 #2

    arildno

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    Diff.eqs are subtle, complex and extremely varied creatures.
    What this entails is that it is exceedingly difficult to generate theorems and general truths about them that at the same time is readily applicable.

    Following this, is that whenever we DO have some powerful, easy technique that can be used on a sub-class of diff.eqs, the rationale behind that technique might be little else than "it works".

    That is, our equipment for handling diff.eqs might be described as just a bag of tricks; isolated techniques that work only in a few special cases.

    That is one of the reasons why many professional mathematicians shy away from working directly with the diff.eqs themselves, because it is difficult to find some interesting, general results there. The work they DO make, might well have an impact on generating new techniques for handling diff.eqs, as a side benefit.
     
  4. Oct 20, 2007 #3
    mathworld has a detailed derivation/proof/whatchamacallit.

    just google first order linear differential equation.
     
  5. Oct 21, 2007 #4

    HallsofIvy

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    It's really based on the simple observation that d(p(t)y)dt=p(t)dy/dt+ p'(t)y. (The product rule.)

    If you have a differential equation of the form dy/dt+ g(t)y= f(t), and you multiply by any function p(t), you have p(t)dy/dt+ p(t)g(t)y= p(t)q(t). Now compare the two forms. It should be obvious that the left side of your new equation is "exact" ( equals d(p(t)y)/dt) if and on p'(t)= p(t)f(t). That's separable equation, easy to solve for p(t).
     
  6. Oct 21, 2007 #5
    Thanks everyone for your responses! It's very much appreciated.

    HallsofIvy, that's pretty much what I was looking to hear. The book i'm using my studies is "advanced engineering mathematics" by zill & cullen. Personally I think it's not worth reading, having or even using as a coaster. I looked at some other books, and one of them gave me a similar explanation to what you gave me.

    I feel the need to understand the "why" behind ideas like that. It is not satisfying and somehow an obstruction to understanding to just know that it works. I suppose that's what you get in current engineering curricula...

    Thanks again,

    Erik
     
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