ODE Integrating factor

Discussion in 'Math' started by kokkie_d, Oct 12, 2010.

  1. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
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    0
    Hi,

    Could someone explain to me why the following step is allowed?
    <br />
e^{x}\frac{dy}{dx} + e^{x}y = e^{x}x<br />
    <br />
\frac{d}{dx}[ e^{x}y] = e^{x}x<br />

    cheers,
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    The chain rule.
     
  3. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    I think the chain rule has nothing to do here. It's just the expansion of the term
    (e^x \cdot y)'=(e^x)'\cdot y+e^x \cdot y'=e^x \cdot y+e^x \cdot y'
     
  4. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
    72
    0
    @georacer:

    Risking sounding silly here: What do you mean "expaning of the term"?
    I mean I am familiar with finding zeros by expanding term as in:
    <br />
x^{2}-9 = (x+3)\cdot(x-3)<br />
    but I am not sure how this applies here.

    Please elaborate.
     
  5. bradstormer

    Member

    Aug 6, 2010
    65
    1
    i think the reason it is allowed is because its the same as multiplying fractions, for example

    2/3*6 = (2*6)/3 = 12/3 = 4

    think of the "d" part of the equation as just a function, just as if the equation was

    -2/3*6 = -(2*6)/3 = -12/3 = -4

    hope this was of some use :)
     
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Nonononono! Nothing of the sort!

    There is an identity in differential analysis that says that given two functions f and g which are differentiable, the followin is valid:
    \frac{d}{dx}(f(x) \cdot g(x))=\frac{df(x)}{dx} \cdot g(x) + f(x) \cdot \frac{dg(x)}{dx}

    Is that clear?
     
    kokkie_d likes this.
  7. bradstormer

    Member

    Aug 6, 2010
    65
    1
    well ya helped me anyway, thank you!
     
  8. kokkie_d

    Thread Starter Active Member

    Jan 12, 2009
    72
    0
    Yes, thank you.

    I have found where it came from (had to revert to manual for dummies) lol

    If you take the derivative of: e^{x}y
    i.e. \frac{d}{dx}[e^{x}y]

    you get:
    e^{x}\frac{dy}{dx}+e^{x}y

    I was looking way to far out.
     
  9. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    Georacer, I was going to respond angrily -- then I realized that my feeble brain mixed up the product rule with the chain rule. Thanks for correcting me. My only excuse is that I took calculus before most of you were born, so I can claim feeble-mindedness and forgetfulness. :p Of course, my wife would just claim it's overall incompetence. :D
     
  10. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Well, if your wife competes with you in math, then you have much more serious issues that mathematical mixups...

    No harm done, be sure of that. The time will come when you will correct me in electronics, I 'm sure of that.

    Have a nice evening!
     
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