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Can this be solved?

  1. Mar 31, 2008 #1
    This is not a homework problem, I just thought of it when I was looking at problems with it being just y and not y^2. Here's the problem. It's entirely possible that it's not solvable, I'm just curious.
  2. jcsd
  3. Mar 31, 2008 #2


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    That's non-linear so solving it won't be easy. There is, however, a theorem that says every first order d.e. has an "integrating" factor. You can write that as [itex]dy/dx= x- y^2[/itex] so [itex] dy= (x- y^2)dx[/itex] or [itex]dy+ (y^2- x)dx= 0[/itex]. There must exist some function v(x,y) such that [itex]v(x,y)dy+ v(x,y)(y^2-x)dx= 0[/itex] is "exact": that is so that there exist a function f(x,y) so that [itex]df= vdy+ v(y^2-x)dx[/itex]. If that is true then we must have [itex]v_x= (v(y^2-x))_y[/itex].

    But there is no theorem that says it will be easy to find v(x,y)!
  4. Apr 2, 2008 #3
    This equation is of the Riccati type. It can be transformed into a linear one by using the substitution:

    [tex]y(x)=\frac{1}{u(x)}\cdot \frac{du(x)}{dx}[/tex]

    Giving thus as transformed equation:

    [tex]\frac{d^2u}{dx^2}-x\cdot u = 0[/tex]

    Which is the one of Airy, with known solution. After the inverse transformation you get the solution of the original equation.
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