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I Quick Differential Form Question

  1. Jun 14, 2017 #1

    Drakkith

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    Staff: Mentor

    I've been going through my book learning about differential equations of multiple variables and I have a quick question about differential forms.

    If you are working a problem and get to the point where you're left with a differential form like ##(y)dx##, does that mean that the change in the function as ##x## changes is zero, such that ##(y)dx=0##?

    For example, one of the example problems in my book is:$$(2x^2+y)dx+(x^2y-x)dy=0$$
    It says that if you multiply the function by the integrating factor ##u(x)=x^{-2}## you get $$(2+yx^{-2})dx+(y-x^{-1})dy=0$$ and you end up losing the solution ##x≡0##.

    I understand that plugging 0 into the original equation yields ##(y)dx+(0)dy=0##, but I'm not sure why 0 is a solution.
     
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  3. Jun 14, 2017 #2

    fresh_42

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    ##ydx## isn't automatically zero, but the system ( ##x\equiv 0 \wedge ydx=0## ) has a solution which will get lost if we divide by ##x^{2}##. So as always in these cases, consider this system first and in the next case ##x \not\equiv 0\,##.
     
  4. Jun 14, 2017 #3

    Drakkith

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    I'm sorry but I don't know what that bit in the parentheses means.

    What do you mean by "consider this system first"?
     
  5. Jun 14, 2017 #4

    fresh_42

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    I meant ##x \equiv 0## is a possibility at first. In this case we conclude ##ydx=0## which can be integrated and leads to a solution with ##x \equiv 0 ## and ##y = \text{ anything }##. From this point on we can assume ##x \not\equiv 0## and consider this remaining case.
     
  6. Jun 14, 2017 #5

    Drakkith

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    Staff: Mentor

    Ah okay. Thanks fresh.
     
  7. Jun 14, 2017 #6

    WWGD

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    As a general comment not every operation on your equation/system will preserve the solution set. Remember, e.g., when you have a system of linear equations that anything other than switching rows, non-zero scaling and adding a multiple of one row to another, will change the solution .
     
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