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Solve a given initial-value problem, bernoullis equation

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    x2*(dy/dx)-2xy=3y4
    y(1)=1/2


    3. The attempt at a solution
    the most I have it reduced to du/dx+2u/3x=-1/(u^8*x^2)
     
  2. jcsd
  3. May 27, 2012 #2
    This is a Bernoulli DE. Did you use the substitution [itex]u = y^{1 - 4} = y^{-3}[/itex]? Find [itex]dy/dx[/itex] in terms of [itex]du/dx[/itex] and [itex]y[/itex] and substitute it in the differential equation. Eventually, you'll have a new DE of u and x and you can solve it using whatever way it should be.
     
  4. May 27, 2012 #3
    thats what i did, and when i reduced it down i got this
    du/dx+2u/3x=-1/(u^8*x^2)
    but ive rechecked and rechecked and i cant see if i did something wrong or what I am supposed to do next?
     
  5. May 27, 2012 #4
    What I got is different:

    [itex]dy/dx = -(1/3) y^4 (du/dx)[/itex] ; we substitute this into the differential equation, and we get: [itex]x^2 (-1/3) y^4 du/dx - 2xy = 3y^4[/itex]. Now divide by [itex]x^2[/itex] then multiply by [itex]-3y^{-4}[/itex] and we get: [itex]du/dx + (6u/x) = -9/x^2[/itex]. From here, find the integrating factor and solve the DE. Your mistake is [itex]u^8[/itex]; you multiplied by [itex]-3y^{-4}[/itex] so [itex]y^4[/itex] and [itex]y^{-4}[/itex] will cancel.

    I might have some calculation mistakes, so I'd wait for someone else to confirm this.
     
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