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First order ODE, The Homogeneous Method.

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{1}{xy}[/tex] [tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{(x^2 + 3y^2)}[/tex]


    2. Relevant equations

    used the substitutions:

    v = [tex]\frac{x}{y}[/tex] ,and
    [tex]\frac{dy}{dx}[/tex] = [tex]v + x[/tex] [tex]\frac{dv}{dx}[/tex]

    3. The attempt at a solution

    took out a factor of xy on the denominator of the term on the right hand side and multiplied through by xy, made the substitutions and at the moment I have something that looks like this:

    [tex]\frac{1}{v + 3/v}[/tex] = v + x [tex]\frac{dv}{dx}[/tex]

    as it stands I cant see a way to do this by separation of variables, is it solvable by integrating factor? anyone got any ideas?
     
  2. jcsd
  3. Nov 24, 2009 #2

    Mark44

    Staff: Mentor

    You used the substitution v = x/y, but your work in calculating dy/dx looks like you started with y = vx, which is not equivalent to v = x/y.
     
  4. Nov 24, 2009 #3
    ah thanks, i'll give it another go.
     
  5. Nov 24, 2009 #4

    Mark44

    Staff: Mentor

    I was able to get the equation separated with the substitution v = y/x.
     
  6. Nov 24, 2009 #5
    I have the substitutions:

    v = [tex]\frac{y}{x}[/tex]

    [tex]\frac{dy}{dx}[/tex] = v + x [tex]\frac{dv}{dx}[/tex]

    so now my equation looks a bit like this:

    [tex]\frac{1}{1/v + 3v}[/tex] = v + x [tex]\frac{dv}{dx}[/tex]

    I played with it a bit and still cant see how to seperate it, can anyone point me in the right direction?

    thanks
     
  7. Nov 24, 2009 #6

    Mark44

    Staff: Mentor

    What you have is fine, but needs cleaning up using ordinary algebra techniques.

    1. Rewrite 1/(1/v + 3v)) as a rational expression - one numerator and one denominator, with no fractions in the top or bottom.
    2. Subtract v from both sides, so that you have x*dv/dx on one side, and a rational expression on the other.
    3. Divide both sides by x. You should now be able to move dv or dx so that everything in v or dv is on one side, and everything in x or dx is on the other. The equation is now separated, and you can integrate to find the solution.
     
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