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Find the general solution of Bernoulli Eq

  1. Jul 11, 2009 #1
    Problem:y' + y(1/x) = 3 x^2 y^2
    Solution:p(x)= 1/x q(x) = 3x^2 <--- These are countinuous functions on the interval (0,+inf)

    y^-2[y' + y(1/x) = 3 x^2 y^2] A

    => y'y^-2 + y^-1(1/x) = 3x^2

    v = y^(-n+1) = y^-1 v' = -y^-2 y'

    Plugin to problem A
    v' + v/x = 3x^2 B

    p(x) = 1/x q(x) =3x^2 <---- theres are continuous on the interval (0,+inf)

    Find integral factor

    h(x) = Int:[ 1/x dx] = ln(x) e^h(x) = e^(ln(x)) = x

    multiply B through by factor

    xv' + v = 3x^3
    [xv]' = 3x^3
    xv = Int:[3x^3 dx] = x^4 + C
    solve for v

    v = x^3 + C/x
    replace v = y^-1

    y^-1 = x^3 + C/x

    y = (x^3 + C/x)^-1


    Book listed solution = 2/(Cx - 3x^3)

    Not sure what im doing wrong there.
     
  2. jcsd
  3. Jul 11, 2009 #2
    On a side question: does dy/dx = y' ?
     
  4. Jul 11, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That is standard notation. Surely you knew that?

    A question back to you: "y(1/x)" means "y times 1/x" and not "y of 1/x"?
     
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