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Simple differential equation - please help!

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    dy/dx = 2x / (y+1)

    2. Relevant equations

    n/a

    3. The attempt at a solution

    I have to solve this - and i know it's separation of variables, but I always get y on both sides when I try to solve and end up with:

    y = 2(x^2) / (y+1)

    which is clearly wrong.

    Where am I messing up? :S
     
  2. jcsd
  3. Oct 20, 2008 #2
    Sorry the latex doesn't seem to be working, but since it's seperable, go...


    integral[ (y+1) dy] = integral[ 2x dx]

    and when you integrate that you should be on your way.
     
  4. Oct 20, 2008 #3
    that's what i did - but that came out as the solution i gave in my original post - which isn't right is it?
     
  5. Oct 20, 2008 #4

    tiny-tim

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    Hi mikehibbert! :smile:
    No, it doesn't … try again! :smile:
     
  6. Oct 20, 2008 #5
    argh i've literally been thinking about this for like two hours, it's ridiculous.

    integrating (y+1) wrt y gives: (y^2)/2 + y

    and integrating 2x wrt x gives: x^2

    giving (y^2)/2 + y = x^2

    and rearranged gives y = 2x / (y+1)

    i still get it no matter how many times i think of it :(

    arrgghhh
     
  7. Oct 20, 2008 #6

    tiny-tim

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    I assume you mean y = 2x2 / (y+1) …

    but shouldn't there be a 2y in there somewhere?

    and what about a constant of integration? :redface:
     
  8. Oct 20, 2008 #7
    yes i did mean y = 2(x^2) / (y+1) sorry.

    and there should be a +c on the end, granted.

    but my solution is still wrong isn't it?

    you can't have a y on both sides?
     
  9. Oct 20, 2008 #8
    oh and i can't see where a 2y would come from??

    the LHS integral is (y^2) / 2 + y

    which factorizes to 0.5y(y+1) ???

    then re-arranges to what i said earlier.
     
  10. Oct 20, 2008 #9

    tiny-tim

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    Nope … 0.5y(y+2) :redface:
     
  11. Oct 20, 2008 #10
    grrr i'm literally an idiot...
     
  12. Oct 20, 2008 #11
    i swear i'm going to kill the person than wrote this question...

    now i'm getting y = 2(x^2) + x^2 + c

    which doesnt really look right to me :S
     
  13. Oct 20, 2008 #12

    tiny-tim

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    D'oh! :rolleyes:

    Get some sleep … :zzz:
     
  14. Oct 20, 2008 #13
    i'm gunna have to.
    gotta be handed in at 10am tomo grrr.
    i give in :P
    the easiest question on the sheet has beaten me!
     
  15. Oct 20, 2008 #14

    tiny-tim

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    It'll all be clear at breakfast! :smile:
     
  16. Oct 20, 2008 #15
    Just rewrite it as y^2+y-2x^2-C=0, and you want to factor it. So use the quadratic formula and you'll be able to get the two solutions for the differential equation.
     
  17. Oct 21, 2008 #16

    HallsofIvy

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    I don't think that is the right person to kill!

    You have, correctly, (1/2)y2+ y= x2+ C and you want to solve for y. That is a quadratic equation:
    y2+ y- (x2+ C)= 0. Use the quadratic formula.
     
  18. Oct 21, 2008 #17

    tiny-tim

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    Hey guys!

    Why do you keep writing the wrong equations? :redface:
     
  19. Oct 21, 2008 #18
    Hi!
    I'd like to venture a guess (I'm currently in the middle of calculus 1 and haven't really done any differential equations)

    For the integral of (y+1) I have [(y^2)/2] + y

    For the integral of 2x I have x^2

    then I set them equal and solved for x to get

    x = the +-squareroot of {[y^2+2y]/2} I thought it looked nicer to give them a common denominator.

    Anyway since originator of the post has already turned in the assignment I didn't see what it would hurt to post a possible answer, but anyway could someone let me know if I am anywhere close or if not post the right answer.

    Thank!
     
  20. Oct 21, 2008 #19

    tiny-tim

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    Hi Geekchick! :smile:

    i] You must include a constant of integration.

    ii] Don't use √s … as HallsofIvy said, this is a quadratic equation … so leave it as such … it tells you what shape the curve is. :smile:
     
  21. Oct 21, 2008 #20
    alright...Best I can come up with is y=-1+-square root of (2x^2+C)
     
    Last edited: Oct 21, 2008
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