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First Differential Equation Ever

  1. Dec 3, 2011 #1
    [tex]\frac{dy}{dx}=15x^2-\frac{2xy}{1+x^2}[/tex]

    This is the first differential equation problem I have ever done and I am kind of confused so let me know where I went wrong.

    [tex]\frac{1}{y}dy=15x^2-\frac{2x}{1+x^2}[/tex]

    [tex]ln(y)=5x^3-ln(1+x^2)+C[/tex]

    Then I raised each side to the e and got an answer that was not correct :(

    Thanks for any help.
     
  2. jcsd
  3. Dec 3, 2011 #2

    Simon Bridge

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    In the first step,
    - you divided both sides by y, you forgot the first term on the RHS needs to be divided also;
    - you multiplied the LHS by dx, but forgot to do the same to the RHS.

    This is a bit rough for your first DE, you'd normally start with homogeneous ones and build your skills from there.
     
  4. Dec 3, 2011 #3
    Just to clarify, this problem was assigned by my teacher, I'm not trying to self-study or anything.

    Can you give me an additional hint on how to get the y's on one side and x's on the other so that I can integrate both sides. I know this is asking for more help then you are supposed to on this forum, but I have only had 50 minutes of lecture on DE.
     
  5. Dec 3, 2011 #4

    Simon Bridge

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    You want the equation to look like:

    [tex]\frac{d}{dx}y+a(x)y(x)=b(x)[/tex]
    ... where a and b are known functions of x, y is the one you want to find.

    Then you use an integrating factor [itex]\sigma(x)[/itex]:

    [tex]\frac{d}{dx}\sigma y = \sigma \frac{d}{dx} y + \sigma a y[/tex]

    ... find that (hint: compare with the power rule) and you can multiply through by [itex]\sigma[/itex] and simplify the DE into something you can solve.

    Since this is supplied by your teacher, you must have some course notes or a text book which covers this. lynchpin: "integrating factor".

    -----------------------
    http://www.physics.ohio-state.edu/~physedu/mapletutorial/tutorials/diff_eqs/non_homo.htm [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Dec 3, 2011 #5

    vela

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    The method you're referring to is called separation of variables. It's where you try to get all the x's on one side and all the y's on the other. If it works, great, but typically, like in this problem, you can't achieve the necessary separation. You'll need to try a different method.

    As Simon has wisely counseled, look up "integrating factor." It's a standard method of attacking first-order differential equations when separation fails.
     
  7. Dec 4, 2011 #6
    Thank you for the help on that problem!! For:
    [tex]e^{2x+3y}\frac{dy}{dx}=1[/tex] I can't put it in the form for finding the integrating factor (or am I missing something?). Is there an easier way to solve this where you don't have to find I?
     
  8. Dec 4, 2011 #7

    SammyS

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    [itex]\displaystyle e^{2x+3y}=e^{2x}e^{3y}[/itex]

    So this is separable. [itex]\displaystyle e^{3y}\,dy=e^{-2x}\,dx[/itex]
     
  9. Dec 4, 2011 #8
    Thank you so much.
     
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