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Nonlinear First Order Differential Equations

  1. May 28, 2006 #1
    Hello. I am taking a self study diff e course, and I have run into a problem with no one to ask for help. Here is the problem:
    [tex]y\prime=1+x+y^2+xy^2[/tex]

    The question asks to find the general solution. I simply don't understand how to solve this problem. Here is the direction I am going in:
    [tex]dy=(1+x+y^2+xy^2)dx \Rightarrow
    \int dy = \int{dx} \ + \ \int{xdx} \ + \ y^2*\int{dx} \ + \ y^2*\int{xdx} \Rightarrow
    y = x + \frac{x^2}{2} + xy^2 + \frac{y^2 x^2}{2} + C[/tex]

    Where the heck do I go from here? I can't sepperate the equation, so how do I solve it?
     
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    arildno

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    This is totally wrong. Do you understand what separation of variables is about?
    To help you along a bit, note that your right-hand side may easily be transformed:
    [itex]1+x+y^{2}+xy^{2}=(1+x)y^{2}[/itex]
    Thus, your diff. eq. can be given in the form:
    [tex]y'=(1+x)y^{2}[/tex]
     
  4. May 28, 2006 #3
    First I would like to say that I wrote the problem very sloppily (i am still learning how to write in the math tex), I think I have fixed it if you want to look at it again.
     
  5. May 28, 2006 #4
    I don't understand what you did here.?
     
  6. May 28, 2006 #5
    Shouldn't [itex](1+x)y^{2} = y^{2}+xy^{2}[/itex]?

    Do you mean: [itex](1+x)+(1+x)y^{2} = 1+x+y^{2}+xy^{2}[/itex]
     
    Last edited: May 28, 2006
  7. May 28, 2006 #6

    arildno

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    Oh, dear, you're right.
    The correct identity is:
    [tex]1+x+y^{2}+xy^{2}=(1+x)(1+y^{2})[/tex]
    Sorry about that.
     
  8. May 28, 2006 #7
    The RHS should be transformed into (1+x)(1+y^2).
    That's probably what arildno meant to say.

    Arun

    edit: He's quick to correct himself.
     
  9. May 28, 2006 #8
    Alright. And then it should go:
    [tex](1+y^{2})dy=(1+x)dx[/tex]?
     
  10. May 28, 2006 #9

    arildno

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    Again:
    Do you understand what separation of variables is about?
     
  11. May 28, 2006 #10
    What the heck am I doing wrong with the LaTex that I wrote in the first post? Why are the equations all on the same line?
     
  12. May 28, 2006 #11
    Didn't I just seperate variables?
     
  13. May 28, 2006 #12

    arildno

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    Not correctly, anyway.
     
  14. May 28, 2006 #13
    Oh wait, I should have the reciprocal of (1+y^2) on the left, right?
     
  15. May 28, 2006 #14

    arildno

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    The double slash option for separating lines in Latex is not available here
     
  16. May 28, 2006 #15
    [tex]\frac{1}{1+y^{2}}dy=(1+x)dx[/tex]

    Is this the correct sepperation of variables?
     
  17. May 28, 2006 #16

    arildno

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    That's right. :smile:
     
  18. May 28, 2006 #17
    [tex]\tan^{-1}(y) = x + \frac{x^{2}}{2} + C[/tex]

    That is supposed to say arctan(y) on the right, I don't know what I did wrong.
     
    Last edited: May 28, 2006
  19. May 28, 2006 #18

    arildno

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    Try:
    [tex]\tan^{-1}(y)=...[/tex]
     
  20. May 28, 2006 #19
    Awesome, I prefer to write it that way anyway.
     
  21. May 28, 2006 #20
    Hey, I have a question about the LaTex. When I go to edit the LaTex, and then resubmit it to post the edit, the edit doesn't show up. It just shows the same thing as before the edit. I have to restart my computer to see the corrections I make. Do you know what I am doing wrong?
     
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