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Homework Help: Homogeneous equation; Initial Value

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Given,

    (y+2)dx + y(x+4)dy = 0, y(-3) = -1


    2. Relevant equations
    v=y/x


    3. The attempt at a solution

    I've been REALLY struggling with homogeneous equations for some reason...I just don't understand them all.

    so far i've tried two things.

    (1)dx -(y)dy
    ----- -------
    (x+4) = (y+2)

    Not even sure what else to do from here. I tried getting dy/dx and got

    y+2 dy
    ---- = -----
    y(x+4) dx



    Thanks in advanced for any assistance... I've been looking at this problem for hours and have other hw to get to.
     
  2. jcsd
  3. Sep 15, 2008 #2
    I think the idea here is to use separable equations. There should be a section in your math book on this concept.
     
  4. Sep 15, 2008 #3

    Dick

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    Take your first form and integrate both sides. You've 'separated' the variables.
     
  5. Sep 15, 2008 #4
    thanks guys.

    Dick,
    Glad to know i was on the right track. Just realized it isn't homogeneous.

    As far as integrating:

    dx/(x+4), I get lx|x+4|

    how do i go about integrating

    y
    ----- dy ?
    y+2

    Probably a dumb question, but I took a semester break between this and calc II, so lots of areas are rusty :(

    Simply couldn't get it into du/u form..
     
  6. Sep 15, 2008 #5

    Dick

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    It is dumb. :) Sorry, you are just out of practice. Try substituting u=y+2.
     
  7. Sep 15, 2008 #6
    lol. I appreciate the honesty :rofl:

    u = y + 2

    gives me

    (U-2)
    -------du
    U

    Which i guess can be broken up into

    U
    ---- = 1
    U


    and -2/U

    now all i have to integrate is

    du
    --------
    U

    bringing the -2 outside of the integral, correct?
     
  8. Sep 15, 2008 #7

    Dick

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    Sure. You might just write that as ((u-2)/u)*du=(1-2/u)*du=1*du-2*(1/u)*du. It's a little easier that trying to place the '-----' correctly. Just use parentheses. Or preferably tex if you dare.
     
  9. Sep 15, 2008 #8
    i really appreciate the help.

    so this is where i'm at.
    ln |x+4| = 2*ln |Y+2| + ln C

    which, IIRC gives

    X+4 = (Y+2)^2 + C ?
     
  10. Sep 15, 2008 #9
    and since you are given a certain x y value from the question, you can sub those in to solve for the constant c
     
  11. Sep 15, 2008 #10

    Dick

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    You missed a part. The integral of 1-2/u is u-2ln(|u|). What happened to the u?
     
  12. Sep 15, 2008 #11
    Ok great! Thanks again guys for the help.

    I'll do that and double check my answer with the one in the book.

    plugging in the x and y values. C = 0

    Strangely enough, the answer in the book reads:

    x + 4 = (y + 2)^2*e^(-y+1)

    I understand how they got everything up until the e with exponent.
     
  13. Sep 15, 2008 #12
    hmmm im taking a minute to look at it now. not sure where i messed up.
     
  14. Sep 15, 2008 #13
    ah i think i see it.


    as you stated above, (1/u)*du is seperate from (-2/u)*du correct? So i should integrate those seperately?

    Edit: I see where i forgot u now.
     
    Last edited: Sep 15, 2008
  15. Sep 15, 2008 #14
    Dick, p00nst3r, thanks for not doing the problem for me, and helping me think things through. I hate it when people do all the work for me lol.

    I'll be checking back before I go to bed.
     
  16. Sep 15, 2008 #15
    No problem. Glad to have helped!
    I was actually considering doing the question for you and then I realized that I should guide you to the answer instead.
    Good thing I didn't do it for you!
     
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