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How do differentials really work?

  1. Feb 23, 2012 #1
    I have quite some trouble thinking why we are allowed to manipulate differentials as we see fit when solving differential equations. I usually think of the derivative as the fundamental object upon which differentials are based. With this in mind I wince when I see derivatives appear separately, as in separation of variables when solving differential equations.

    For example, consider [itex] \frac {dy}{dx} = ky [/itex]
    We would "separate" the variables as follows: [itex] \frac {dy}{y} = k dx [/itex]
    And then we would integrate both sides... [itex] \int \frac {dy}{y} = \int k dx[/itex]

    What I don't understand is what allows us to separate the variables... since when are we allowed to multiply and divide both sides of an equation by a differential? I thought differentials were not like normal numbers, and you're not allowed to play with them unless some theorem specifically allows you to do so?
    Also, why are we allowed to integrate both sides of the equation with respect to different variables?

    I appreciate any help/advice. Thanks!

    BiP
     
  2. jcsd
  3. Feb 23, 2012 #2
    You're not really integrating the y side in terms of y, because y and dy are functions of x. You're using substitution. For example, if [tex]y=C e^{k x}[/tex], then [tex]\frac{dy}{y}=\frac{C k e^{k x} dx}{C e^{k x}}=k dx[/tex].

    As for the rest. I'm as lost as you are. I had a thread on this a long time ago. I'm looking for it now.
     
  4. Feb 23, 2012 #3

    HallsofIvy

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    Roughly speaking, the idea is this: the derivative is defined as
    [tex]\frac{df}{dx}(a)= lim_{x\to a}\frac{f(x)- f(a)}{x- a}[/tex]
    so that, while the derivative is NOT a fraction, it is the limit of a fraction. You can prove that the derivative "acts" like a fraction, has all the fraction properties, by going back before the limit, using the fraction property and then taking the limit again. That is why we define "dx" and "dy" symbolically and treat dy/dx as a fraction.
     
  5. Feb 23, 2012 #4
    There are, of course, also perfectly sound ways of treating differentials as actual quantities in nonstandard analysis. And they often have some pedagogical value, like Silvanus Thompson's wonderful book Calculus Made Easy, or if you prefer a more rigorous version Keisler's Elementary Calculus, available free online: http://www.math.wisc.edu/~keisler/calc.html
     
  6. Feb 23, 2012 #5

    Office_Shredder

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    Separation of variables can be explained by the chain rule. If your differential equation is [tex]\frac{dy}{dx} = f(x)/g(y) [/tex] (in your example, f(x)=k and g(y)=1/y) then we can write
    [tex] g(y) \frac{dy}{dx} dx = f(x)[/tex]
    These are both functions of x (because g(y) = g(y(x))). So integrate them both
    [tex] \int g(y) \frac{dy}{dx} dx = \int f(x)dx[/tex]
    and the result follows from noting that
    [tex] \int g(y(x)) \frac{dy}{dx} dx = \int g(y)dy[/tex]
    by doing integration by parts (with the substitution y=y(x)).
    Then you can consider the act of separating the variables as just a mnemonic shortcut to avoid a lot of notation
     
  7. Feb 24, 2012 #6
    As someone said here, dy and dx are limit cases of difference values. They are still differences after all and behaving as fraction. dy/dx just happened to be a famous notation that appears as it is in the books.
     
  8. Feb 24, 2012 #7

    lavinia

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    Right.

    So the equation is true with arbitrary accuracy/arbitrarily small error for very small values of delta x and delta y and there for it is correct to manipulate them as fractions and correct to equate the integrals. One just needs to remember that these equation are true with arbitrarily small error.
     
    Last edited: Feb 24, 2012
  9. Jun 7, 2012 #8
    This seems to casual to me. while dy/dx is a limit, dx and dy are not the limits of anything. I've seen dx and dy defined as real valued variables related by:

    dy = (dy/dx) dx
     
  10. Jun 7, 2012 #9

    HallsofIvy

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    I agree with pondhockey here.

    (Though not necessarily with resurecting a thread from three and a half months ago!)
     
  11. Jun 8, 2012 #10
    For which I apologize. If it helps the cause, here is a location (as newbie I can't post a link yet) that explains differentials in action, in the context of thermodynamic equations.

    //www.physicsforums.com/showthread.php?t=525071&highlight=differential
     
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