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1st order linear differential equations - integrating factors

  1. Sep 27, 2009 #1
    Hi guys, this is my first post but have read the forums for a long time - a quick search didnt bring up anything that could help me. So i was wondering if someone could please explain something to me.

    I have a differential equation
    (excuse the pictures, I dont know how to use the equation writer in this?)
    and basically I have to get it into a seperable form and simplify like I have done in the picture.

    0m9fW.jpg

    In the picture I explain what I want to know, basically
    how come the dy/dx e^(8x) + 8ye^(8x) can be written as d[e^(8x)]/dx ... where do the other terms go? I have been doing a few of these problems by following examples, and its not explained and I have been taking it for granted that this happens ... but I want to know why?
     
  2. jcsd
  3. Sep 27, 2009 #2

    lurflurf

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    That is due to the product rule for differentiation
    u'v+uv'=(uv)'
    y' (e^(8x)) + y(8e^(8x))=y' (e^(8x)) + y(e^(8x))'=[y(e^(8x)) ]'
     
  4. Sep 27, 2009 #3

    CompuChip

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    Science Advisor
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    It's simply the product rule for differentiation:

    [tex]\frac{d[f(x) g(x)]}{dx} = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx}[/tex]


    Here follows a somewhat technical explanation of why you do this, hopefully you can follow.

    So you start out with a differential equation of the form
    dy/dx + p(x) y = f(x). ​
    If you have some function I(x) which is nowhere zero, you can multiply the whole equation by I(x) to get
    dy/dx I(x) + p(x) I(x) y = f(x) I(x). ​
    The trick is now to choose I(x) such that the left hand side is precisely a derivative, so you can directly integrate both sides. Well, by the product rule
    d[y I(x)]/dx = dy/dx I(x) + y dI/dx. ​
    So you are happy when you can get I(x) to satisfy dI(x)/dx = I(x) p(x). This equation can be solved; for example, using separation:
    (dI(x)/dx) / I(x) = p(x),​
    d(ln I(x))/dx = p(x)​
    because I(x) was nowhere zero. Integrate both sides:
    ln I(x) = ∫ p(x) dx​
    and solve for I(x):
    I(x) = exp[ ∫ p(x) dx ] ​

    Thus, you have obtained the equation
    d[y I(x)]/dx = I(x) p(x)​
    and you can now solve it by integrating:
    y I(x) = ∫ I(x) p(x) dx,​
    provided you can do the integral on the right.

    Hope that was not too advanced for you? :)
     
  5. Sep 27, 2009 #4
    =D
    The light bulb is starting to brighten now! The above steps were not clear in my examples, but by seeing exactly where
    I(x) = exp[ ∫ p(x) dx ] ​

    came from made it alot easier to understand. Thankyou very much for the help!
     
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