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Separable equations

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data

    step 1. 2((2/3)y^(3/2) = 2x^(1/2) + C1

    step 2. (2/3)y^(3/2) - x^(1/2) = C, where C = 1/2C1



    3. The attempt at a solution

    I don't understand the where C = 1/2C1 - what is that? I understand everything else, except that.
     
  2. jcsd
  3. May 15, 2012 #2
    Re: seperable equations

    step 1: 2((2/3)y^(3/2) = 2x^(1/2) + C1

    step 1.33: 2((2/3)y^(3/2) - 2x^(1/2) = C1
    (bring 2x^(1/2) term over)

    step 1.67: ((2/3)y^(3/2) - x^(1/2) = C1 /2
    (divide both sides by 2)

    step 2. (2/3)y^(3/2) - x^(1/2) = C, where C = (1/2)C1
    (replace C1 /2 with C)
     
  4. May 15, 2012 #3
    Re: seperable equations

    I see. I originally thought C was some sort of magic number that stays as is, but your solution is more logical.
     
  5. May 15, 2012 #4
    The C's are just constants, so we are free to define new constants in terms of the old ones to make it convenient.
     
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