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Homework Help: DE general solution

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    [itex]y' = \frac{y+y^2}{x+x^2}[/itex]

    2. Relevant equations

    separation of variables

    3. The attempt at a solution

    I start with
    [itex]y' = \frac{y+y^2}{x+x^2}[/itex]
    which is
    [itex]\frac{dy}{dx} = \frac{y+y^2}{x+x^2}[/itex]
    next step is

    [itex]dy = \frac{y+y^2}{x+x^2}dx[/itex]

    than I divide both sides by [itex]y+y^2[/itex]

    so gives

    [itex]\frac{dy}{y+y^2} = \frac{1}{x+x^2}dx[/itex]

    so then I integrate both sides.

    [itex]\int\frac{dy}{y+y^2} = \int\frac{1}{x+x^2}dx[/itex]

    which gives

    [itex]ln\right[\frac{\mid y\mid}{\mid y+1\mid}\left][/itex]=[itex]ln\right[\frac{\mid x\mid}{\mid x+1\mid}\left][/itex]

    Is this right so far?
     
  2. jcsd
  3. Aug 4, 2010 #2

    vela

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    Don't forget the arbitrary constant.
     
  4. Aug 5, 2010 #3
    Sorry,

    [itex] ln\left[\frac{\mid y\mid}{\mid y+1\mid}+ C \right][/itex] = [itex]ln\left[\frac{\mid x\mid}{\mid x+1\mid}+C \right][/itex]


    I now multiply both sides by [itex]\mid y+1\mid[/itex]

    [itex]
    ln\left[\frac{\mid y+1\mid \mid y\mid}{\mid y+1\mid} \right]
    [/itex] = [itex]
    ln\left[\frac{\mid y+1\mid \mid x\mid}{\mid x+1\mid}+ C \mid y+1\mid \right]
    [/itex]
     
  5. Aug 5, 2010 #4
    Divide by |y+1| Equals

    [itex]
    ln\left y \right]
    [/itex] = [itex]
    ln\left[\frac{\mid x\mid}{\mid x+1\mid}\right]
    [/itex]
     
  6. Aug 5, 2010 #5
    Do I then take the inverse log to make

    [itex]y = \frac{\mid x\mid}{\mid x+1\mid} + C [/itex]
     
  7. Aug 5, 2010 #6
    How did you knew straight away that the integral of 1/(x+x^2) is ln(|x/(x+1)|) ?
    :)
     
  8. Aug 5, 2010 #7

    vela

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    Not quite. You can combine the two arbitrary constants into one to get

    [tex]\log \left|\frac{y}{y+1}\right| = \log \left|\frac{x}{x+1}\right| + c[/tex]

    Now take the inverse log.
     
  9. Aug 5, 2010 #8

    Char. Limit

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    No, not yet...

    You need a log(A) = log(B), but you have a log(A)=log(B)+C. So, say C=log(D), then say that log(B)+log(D)=log(BD), then your equation can be inverse logged. But not before.
     
  10. Aug 5, 2010 #9

    vela

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    Sure it can.

    log A = log B + c
    → elog A = elog B + c = elog Bec
    → A = Bec
     
  11. Aug 5, 2010 #10

    Char. Limit

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    Good point. I just feel it's easier for me if I do it my way.
     
  12. Aug 5, 2010 #11
    log A = log B + c
    → elog A = elog B + c = elog Bec
    → A = Bec




    So i've got

    [tex]\log \left|\frac{y}{y+1}\right| = \log \left|\frac{x}{x+1}\right| + c[/tex]

    So i take the inverse log of both sides

    [tex]e^{\log \left|\frac{y}{y+1}\right|} = e^\log \left|\frac{x}{x+1}\right| + c}[/tex]
    [tex]e^{\log \left|\frac{y}{y+1}\right|} = e^\log \left|\frac{x}{x+1}\right|} e^c[/tex]

    in turn gives


    [tex]\left|\frac{y}{y+1}\right|} = \left|\frac{x}{x+1}\right|}e^c[/tex]



    How do I get rid of the [tex]y+1[/tex] in the dinominator of LHS?
     
  13. Aug 5, 2010 #12

    Char. Limit

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    First, make some restrictions on x so you can lose the absolute values, then multiply both sides by y+1. You can solve for y from there.
     
  14. Aug 5, 2010 #13
    muitiply both sides by [tex]y+1[/tex] for [tex]x \neq 0[/tex]

    gives

    [tex]y= \frac{x(y+1)e^c}{x+1}[/tex]

    I keep rearranging but I can't seem to get the (y+1) out of the RHS it just seems to be changing sides
     
  15. Aug 5, 2010 #14

    vela

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    First, to get rid of the absolute values, recall that |a|=|b| means a=±b.

    Second, assume x≠0 and y≠0 for the moment. If you take the reciprocal of both sides, you get

    [tex]\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|[/tex]

    Can you see where to go from there?
     
  16. Aug 5, 2010 #15
    Using that |a|=|b| means a=±b.

    [tex]\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|[/tex]


    becomes



    [tex]\frac{-(y+1)e^{-c}}{y}) = \frac{-(y-1)e^{-c}}{y}) [/tex]

    so


    [tex](-e^{-c}+\frac{-1}{y}) = (-e^{-c}-\frac{-1}{y})[/tex]

    is that right so far
     
    Last edited: Aug 5, 2010
  17. Aug 5, 2010 #16

    vela

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    Nope. Check your algebra.
     
  18. Aug 5, 2010 #17
    Using that |a|=|b| means a=±b.

    [tex]\left|1+\frac{1}{y}\right| = e^{-c}\left|1+\frac{1}{x}\right|[/tex]


    becomes



    [tex]\frac{-(y+1)e^{-c}}{y} = \frac{-(y-1)e^{-c}}{y} [/tex]

    so


    [tex]\frac{-(y+1)e^{-C)}{y}) = \frac{-(y-1)e^{-C)}{y}) [/tex]

    is that right so far
     
    Last edited: Aug 5, 2010
  19. Aug 5, 2010 #18
    I mean,

    [tex]\frac{-(y+1)e^{-C)}{y} = \frac{-(y-1)e^{-C}{y}[/tex]
     
  20. Aug 5, 2010 #19
    [tex]\frac{-(y+1)e^{-C}}{y} = \frac{-(y-1)e^{-C}}{y}[/tex]
     
  21. Aug 5, 2010 #20
    i s that better

    [itex]\frac{-(y+1)e^{-C}}{y} = \frac{-(y-1)e^{-C}}{y}[/itex]
     
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