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Simple diff.eq

  1. Dec 5, 2005 #1
    d/dx(y*x^2)=x^2/(1+x^2)

    I guess I can't write this as:
    dyx^2=x^2dx/(1+x^2)
    dy=dx/(1+x^2)
    beacuse I don't get the right answer...

    So what do I do?
     
  2. jcsd
  3. Dec 5, 2005 #2

    Tide

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    You can't just divide out the [itex]x^2[/itex] in the differential. I suggest direct integration of the DE with respect to x.
     
  4. Dec 5, 2005 #3
    well, it's the d/dx(y*x^2) that confused me.
    we have only learned to solv the types:
    dy/dx=f(x)g(y)

    what is direct integration?

    Thanks
     
  5. Dec 5, 2005 #4

    Tide

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    Direct integration means this:
    [tex]\int_{x_0}^{x} \frac {d}{dx'}\left(x'^2 y\right) dx' = \int_{x_0}^{x} \frac {x'^2}{1+x'^2} dx'[/tex]
     
  6. Dec 6, 2005 #5
    Then I don't think we have learned about direct integration...
     
  7. Dec 6, 2005 #6

    dextercioby

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    Well, Tide said that, simply

    [tex] \int du = u +\mathcal{C} [/tex]

    You must understand what i've written, else why attempt to solve diff. eqns. ?

    Daniel.
     
  8. Dec 6, 2005 #7
    okey, I understand what you just wrote :P

    but I still don't know how to solve my problem...
     
  9. Dec 6, 2005 #8

    HallsofIvy

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    Tide's point was just that you need to divide both sides by x2 to get simply [itex]dy= \frac{dx}{1+x^2}[/itex].

    "Direct integration" just meant regular integration!

    [tex]y= \int dy= \int \frac{dx}{1+ x^2}[/itex]
     
    Last edited: Dec 6, 2005
  10. Dec 6, 2005 #9
    Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses.

    When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be:
    (pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2

    (the pi/4 comes from the start value...)
     
  11. Dec 6, 2005 #10

    HallsofIvy

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    Oops! Sorry about that. I misread the original equation!

    It's not that the yx2 was "encapsulated by parentheses" but that here, specifically, the parentheses mean that the entire function yx2 is being differentiated. I read it as [itex]x^2\frac{dy}{dx}[/itex] rather than [itex]\frac{dx^2y}{dx}[/itex].

    Given that, Tide meant that
    [tex]x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}[/tex].

    That right hand side is a little more complicated. First divide it out:
    [tex]\frac{x^2}{x^2+1}= 1- \frac{1}{x^2+1}[/itex]
    and it becomes easy to integrate.
     
    Last edited: Dec 6, 2005
  12. Dec 6, 2005 #11
    Oh, thanks! Now I see it :)
     
  13. Dec 6, 2005 #12
    Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :(
     
  14. Dec 6, 2005 #13

    HallsofIvy

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    Fundamental Theorem of Calculus!

    [tex] \int d(Anything)= Anything+ C[/tex]

    [tex]\int d(x^2y)= x^2y + C[/tex]
     
  15. Dec 6, 2005 #14
    Oh! sorry! I see it! bah

    hehe, the yx^2 messed up my head :P
     
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