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Differential Equations - Linear Factor Proof

  1. Feb 9, 2005 #1
    My class has recently done an intro to differential equations, and although I understand the method of solving simple equations, I want to know why the method of Linear Factors works. Unfortunately my book hasn't provided a proof for it. :grumpy:

    Also in the final step where you integrate both sides of the equation:

    [tex]\frac{d}{dx}[uy]=uq(x)[/tex]
    the book says to integrate each side in respect to the variable in them

    So I would have [tex]\int\frac{d}{dx}[uy] dy= \int uq(x)dx[/tex]
    This doesn't make sense, considering each side has been multiplied by different differentials.
     
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  3. Feb 9, 2005 #2

    dextercioby

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    No,you've missed something and the final expression is totally wrong.
    [tex] \frac{d}{dx}(uy)=uq(x) \Rightarrow d(uy)=uq(x) dx[/tex]

    And now integrate.

    Daniel.
     
  4. Feb 10, 2005 #3

    HallsofIvy

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    That is what I would call using an "integrating factor" to solve a linear equation.

    You have a differential equation of the form
    [tex]\frac{dy}{dx}+ p(x) y= q(x)[/tex]
    and multiply by some function u(x) such that
    [tex]u(x)\frac{dy}{dx}+ u(x)p(x) y= \frac{d(u(x)y)}{dx}[/tex]

    It's easy to show that u(x) must satisfy u'= u(x)p(x), a separable equation.

    I doubt that what you gave is, in fact, exactly what is in your book. (If it is it's a typo!)
    What is correct is that
    [tex]\int d[uy]= \int u(x)q(x)dx[/tex]

    Of course, the integral on the left is just u(x)y.
     
  5. Feb 10, 2005 #4
    But the left side is missing the dy...

    It seems suspect that you can multiply the dx as a fraction like that, then d[uy] really doesn't mean anything.

    It's like performing math on notation... I'm sure it's just shorthand =/
     
  6. Feb 10, 2005 #5

    dextercioby

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    Yes.

    Only to the ones which are unfamilar to methods involving differentials.


    Of course ot does.It's the differential of the product u(x)y.

    Daniel.
     
  7. Feb 10, 2005 #6
    Correct me if I'm wrong, but [tex]\frac{d}{dx}[uy][/tex] is just notation for the derivative of uy, the dx isn't really a differential at all.

    If it was, then in reality you would be dividing by dx... Are you telling me that d[uy] = dy?
     
  8. Feb 10, 2005 #7

    dextercioby

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    That "dx" BEARS the name "differential of the independent variable".It's not a differential "stricto sensu".

    As for the last,i'm not that naive to claim such thing...They obviously represent different objects.

    Daniel.
     
  9. Feb 10, 2005 #8
    So where does that leave us? Also, don't you need a dy to integrate the left side?
     
  10. Feb 10, 2005 #9

    dextercioby

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    You have a "genuine" differential in the LHS which u can integrate without any problem...,ain't it so...?

    Daniel.
     
  11. Feb 10, 2005 #10
    I thought that you needed to know in respect to which variable. I've just been mechanically integrating both sides, but I don't know why the left side doesn't need a dy.

    As far as I'm concerned, the LHS is just an equation which happens to be the differential. It would still need a dy (it reminds me of the fundamental theorem of calculus for some reason...?)
     
  12. Feb 10, 2005 #11

    dextercioby

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    Well,what does the FTC say and why do you think it would apply here directly...?

    Daniel.
     
  13. Feb 11, 2005 #12
    dexter, where did you learn all of your math from?
     
  14. Feb 11, 2005 #13

    dextercioby

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    About 99% from school,the rest individual study.Why?

    Daniel.
     
  15. Feb 11, 2005 #14

    HallsofIvy

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    The left side: d[uy] doesn't need a left side because it is a differential all by itself:

    [itex]\int dx= x[/itex], [itex]\int dy= y[/itex], [itex]\int d(uy)= uy[/itex], etc.

    You could, if you like, say [itex]\int\frac{d(uy)}{dx}dx= uy[/itex] but most people prefer not to have that 'double' dx.
     
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