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Integration by substitution

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Original problem is differential equation dy/dx=(x+2y)/(3y-2x)
    This is part of solving differential equation.

    x(dv/dx) = (1+4v-3v^2)/(3v-2)

    so one way of solving, I take out the negative sign

    x(dv/dx) = -((3v^2-4v-1)/(3v-2)) , separate and bring over
    -∫((3v-2)/(3v^2-4v-1)) dv = ∫1/x dx

    which gets me
    -(1/2)ln(3v^2-4v-1)=ln(x)+C

    however, if i dont take out negative sign,
    x(dv/dx) = (1+4v-3v^2)/(3v-2), separate and bring over
    ∫((3v-2)/(1+4v-3v^2)) = ∫1/x dx

    which gets me
    -(1/2)ln(-3v^2+4v+1)=ln(x)+C result is opposite sign from above.

    checked again and again and i can't seem to find out why even though the working seems ok, the answer is very different...

    i get 3y^2-4yx-x^2=C for one method
    and x^2+4xy-3y^2=C for the other method
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2

    Dick

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    It doesn't make any difference whether you take the sign out or not. Eg. 3y^2-4yx-x^2=1 is the same solution as x^2+4xy-3y^2=(-1).
     
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