Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating Factor (Derivation)

  1. Sep 11, 2010 #1
    Hello all! :smile: I through a section in my text (by https://www.amazon.com/Advanced-Engineering-Mathematics-Michael-Greenberg/dp/0133214311") on 1st Order linear ODEs. I am understanding the derivation of the integrating factor method pretty well; however, there are some aspects of the mathematics that I am getting hung up on. I have an "engineering" background in calculus and hence I get slowed down by the details sometimes.

    In order to find the solution to an ODE of the form:

    [tex]y' + p(x)y = q(x)\qquad(1)[/tex]

    we first consider the homogeneous case; i.e.,
    [tex]y' + p(x)y = 0\qquad(2)[/tex]

    After some hootenanny, we arrive at the solution to (2) given by
    [tex]y(x) = Ae^{-\int p(x)\,dx}\qquad(3)[/tex]

    Now, in order to solve for the constant 'A' given the initial condition y(a) = b, it says that

    Question 1:

    I am not sure why he even brought this up? Why would changing the dummy variables cause the two integrals to differ *at all* ?


    Moving on to the homogeneous case:

    Basically we wish to multiply (2) by some function of x, say [itex]\mu(x)[/itex] giving

    [tex]\mu(x) y' + \mu(x)p(x)y = \mu(x)q(x)\qquad(5)[/tex]

    so that the left-hand-side of (5) is a derivative. After much more hootenanny, we arrive at the solution

    [tex]y(x) = e^{-\int p(x)\,dx}\left(\int e^{\int p(x)\,dx}q(x)\,dx + C\right)\qquad(6)[/tex]

    Now again, if we wish to solve for the constant 'C' given the initial condition y(a) = b, we are again advised to use a different expression of (6), namely

    [tex]y(x) = e^{-\int_a^x p(\xi)\,d\xi}\left(\int_a^x e^{\int_a^\xi p(\zeta)\,d\zeta}q(\xi)\,d\xi + C\right)\qquad(7)[/tex]

    This is just too much changing of dummy variables for my feeble mind to handle :redface: .

    Question(s) 2:

    There are 3 dummy variables being used here:

    [tex]e^{-\int_a^x p(\xi)\,d\xi}\qquad(8)[/tex]

    [tex] e^{\int_a^\xi p(\zeta)\,d\zeta}\qquad(9)[/tex]

    [tex]q(\xi)\,d\xi\qquad(10).[/tex]

    I am not really sure what is happening here. They are using [itex]\xi[/itex] in (8) and (10) as dummy variables and in (9) as an endpoint of the integral.

    I don't expect anyone to answer all of this at once, but i would love to start a discussion about it so that I can solidify my understanding of the derivation. It's not enough anymore for me to just be able *to use* the integrating factor, but instead to *understand it.*

    Thanks,
    Casey
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 14, 2010 #2
    Anyone have any thoughts or comments? Thanks :smile:
     
  4. Sep 14, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Changing the dummy variables doesn't change anything and that's not what they are saying. The important change is the limits of integration.
    [tex]\int p(x)dx[/tex]
    without any limits of integration , is the "indefinite integral". If F(x) is an anti-derivative of p(x) then [itex]\int p(x)dx= F(x)+ C[/itex] where C is an arbitrary constant.
    [tex]\int_a^x p(\xi)d\xi[/tex]
    is an definite integral just with the variable x as an upper limit: with F(x) an anti-dervative as befor, it would be equal to F(x)- F(a). In other words, the "C" is now specifically -F(a).


    One good reason for doing that is that a person seeing
    [tex]e^{-\int p(x) dx}\int e^{\int p(x)dx}[/tex]
    might think, mistakenly, that the two exponentials cancel. They don't of course- you cannot take a function of x "inside" the integral like that.

    No, that is not the case. In
    [tex]e^{-\int_a^x p(\xi)d\xi[/tex]
    [itex]\xi[/itex] is a dummy variable- the integral will be a function of x.

    In
    [tex]e^{\int_a^\xi p(\zeta)d\zeta[/tex]
    [itex]\zeta[/itex] is a dummy variable but [itex]\xi[/itex] is not- this integral will actually be a function of [itex]\xi[/itex] so that it can be integrated with respect to [itex]\xi[/itex] in the outer integral:
    [tex]\int_a^x E(\xi)q(\xi) d\xi[/tex]
    where "[itex]E(\xi)[/itex]" is the result of the previous calculation,
    [tex]e^{\int_a^\xi p(\zeta)d\zeta[/tex]

    In that last integral, yes, [itex]\xi[/itex] is a dummy variable- the integration is a function of x.

     
    Last edited by a moderator: May 4, 2017
  5. Sep 15, 2010 #4
    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,by (Erik006)
     
  6. Sep 15, 2010 #5
    I think the key is product rule. namely
    (my)'=m'y+my'
    Let's say you have a DE
    y'+py=q
    you want to make it look like product rule by multiplying both sides such that:
    my'+mpy=mq
    No quite product yet, but if mp=m' then
    (my)'=mq
    now that y can be solved by integrating both sides. m is very easy to solve given mp=m'.
     
    Last edited: Sep 16, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrating Factor (Derivation)
  1. Integration Factors (Replies: 3)

  2. Integrating Factors (Replies: 1)

  3. Integrating factor (Replies: 1)

  4. Integrating Factor (Replies: 4)

  5. Integrating factor (Replies: 1)

Loading...