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ODE with substitution

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]y' = \frac{2x+3y-5}{x+4y}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    First of all, I switched it to another coordinates, a and b, giving:

    [tex]b' = \frac{2a+3b}{a+4b}[/tex]
    where [itex]a = x-4[/itex] and [itex]b = y+1[/itex].
    Then using the substitution [itex]z = \frac{b}{a}[/itex] and some algebra I get:
    [tex]z' = \frac{2z+2-4z^2}{a(1+4z)}[/tex]
    the integral is:
    [tex]\frac{1}{2} \int \frac{(1+4z)dz}{2+1-2z^2} = \int \frac{da}{a}[/tex]
    solving the integral (using partial fractions) gives me:
    [tex]5\ln{|z-1|} + \ln{|z+\frac{1}{2}|} = 6\ln{|a|}+C[/tex]
    getting rid of the logarithm, and putting back a and b inside:
    [tex]\left(\frac{b-a}{a}\right)^5 \left(\frac{2b+a}{2a}\right)=C(a)^6[/tex]
    making it back in x and y:
    [tex]\left(\frac{y-x+5}{x-4}\right)^5 \left(\frac{2y+x-2}{2x-8}\right)=C(x-4)^6[/tex]

    Which should be the final answer, but when I look at the answers, it's different, being:
    [tex](y-x+5)^5(2y+x-2)=C[/tex]

    And my question is - where did I go wrong?
    I had a similar problem where I got similar results, me having denominators which didn't appear in the given answer. Am I doing something wrong in my technique, or both of the answers are equal (and I'm missing some algebra here)?

    Thanks!
     
  2. jcsd
  3. Nov 21, 2009 #2

    tiny-tim

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    Hi manenbu! :smile:

    No, it's the same (if you keep cancelling) …

    except that their C is twice your C (which of course doesn't matter). :wink:
     
  4. Nov 22, 2009 #3
    Ok thanks, but how is it the same?
    I'm ok with solving the ODE but then I guess that the algebra part is a bit hard for me.
    I tried getting from one form to another but no success.
     
  5. Nov 22, 2009 #4

    tiny-tim

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    Hi manenbu! :smile:

    Because on the bottom left, you have (x - 4)5(2x - 8), which is 2(x- 4)6, and on the right …

    ohhh … you need a minus there (i didn't notice that before) …

    you missed out a "minus" on the LHS when you changed to partial fractions. :wink:
     
  6. Nov 22, 2009 #5
    Oh so assuming I get the minus right that's -6 and -6 powers on both sides. Ok got it!
    Thanks. :)
     
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