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Factoring in terms of a variable

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    ((1/3)(y^3))+(1/y)=.5*ln((x^2)+1)

    Solve in terms of y

    2. Relevant equations



    3. The attempt at a solution

    I am in college differential equations, and i just can't solve in terms of y. y teacher wants that...i tried wolfram alpha, etc. How on earth do you factor this in terms of y?

    Can you even?
     
  2. jcsd
  3. Mar 13, 2013 #2

    Bacle2

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  4. Mar 14, 2013 #3

    Ray Vickson

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    If I understand correctly what you wrote, you want to solve the 4th degree equation
    [tex] \frac{y^3}{3} + \frac{1}{y} = r,[/tex]
    where ##r = (1/2) \ln( x^2 + 1).##
    There are formulas for the solution of 4th degree polynomials, but they are not pretty. Maple obtains the following four solutions (assuming r > 0)

    Solution 1:
    y = 1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+1/4*2^(1/2)*((-(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)-16*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+12*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

    Solution 2:
    y = 1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)-1/4*2^(1/2)*((-(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)-16*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+12*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

    Solution 3:
    y = -1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+1/4*(-(2*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+32*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+24*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

    Solution 4:
    y = -1/4*2^(1/2)*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)-1/4*(-(2*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+32*(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2)+24*r*2^(1/2)*(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3)/(((36*r^2+4*(-256+81*r^4)^(1/2))^(2/3)+16)/(36*r^2+4*(-256+81*r^4)^(1/2))^(1/3))^(1/2))^(1/2)

    Note: Maple's notation is that a/b*c means (a/b)*c.
     
  5. Mar 14, 2013 #4

    SammyS

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    If you mean solving for x in terms of y, that's not too difficult with this.
     
  6. Mar 14, 2013 #5

    Mark44

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    That's my take, too. "In terms of y" suggests solving for x as a function of y.
     
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