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Simple question - Derivation for obtaining Intgrating Factor

  1. Mar 8, 2014 #1
    I was wondering if somebody could clear up some confusion I have regarding this.

    I've been going over the derivation for obtaining the integrating factor again in my book and there is one step I don't understand.

    There's no point going through the whole thing from scratch, but I've got to the point where we need to multiply the whole DE by a function [tex]\mu(t)[/tex] such that the LHS of the DE is recognizable as the derivative of some function.


    Need to choose [tex]\mu(t)[/tex] to satisfy:

    [tex]\frac{d\mu(t)}{dt}=2[/tex] for this particular example.

    [tex]\frac{d\mu(t)/dt}{\mu(t)}=2[/tex]


    But I don't see how the next step follows from the previous one:

    [tex]\frac{d}{dt}ln|\mu(t)|=2[/tex]

    In particular, I don't see where the [tex]ln|\mu(t)|[/tex] has come from.

    Can anyone tell me how this works?
     
  2. jcsd
  3. Mar 8, 2014 #2

    Fredrik

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    You need to use the chain rule and that ##\frac{d}{dx}\ln x=\frac 1 x##.

    For all t such that ##\mu(t)>0##,
    $$\frac{d}{dt}\ln \mu(t)=(\ln\circ\mu)'(t)=\ln'(\mu(t))\mu'(t)=\frac{1}{\mu(t)}\mu'(t).$$ For all t such that ##\mu(t)<0##,
    $$\frac{d}{dt}\ln(-\mu(t))=(\ln\circ(-\mu))'(t)=\ln'((-\mu)(t))(-\mu)'(t)=\frac{1}{-\mu(t)}(-\mu'(t))=\frac{1}{\mu(t)}\mu'(t).$$
     
  4. Mar 8, 2014 #3
    Thanks for explaining that Frederik... I think I can see it now.
     
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