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Three linear first-order ODEs

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Out of a set of differential equations with boundary conditions, there are three (first order) equations I couldn't solve. These are:

    2. Relevant equations
    1. [tex]
    \frac {dy} {dx} = \sqrt{x + y}, y(1) = 0.
    [/tex]
    2. [tex]
    \frac {dy} {dx} = 2y(x \sqrt{y} - 1), y(0) = 1.
    [/tex]
    3. [tex]
    2x^2 \frac {dy} {dx} = x^2 + y^2, y(2) = 4.
    [/tex]

    3. The attempt at a solution
    The first two can probably be solved with a nice substitution. I tried u = x + y for the first one, but this gave me the equation [tex]\frac {du} {dx} = 1 + \sqrt{u}[/tex], which can be solved for u to get, after resubstituting, [tex]2 \sqrt{x+y} - 2 log[1+\sqrt{x+y} - 2 + 2 log[2] = x[/tex], which doesn't seem solvable for y[x].
    For the second one I tried substituting [tex]u = x \sqrt{y} [/tex] and [tex]u = x \sqrt{y} - 1 [/tex], but neither gave an equation that could be written in terms of u only (without x or y).
    For the third one I only noticed that y=x is a general solution, but it doesn't agree with the initial condition y(2) = 4, and neither does any manipulation such as y=2x or y=x+2.
     
  2. jcsd
  3. Apr 13, 2012 #2
    By the way, Mathematica could only solve for (2.) and gave the solution:
    [tex] y[x] = \frac {1}{(1+x+e^{x} C[1])^{2}} [/tex].

    For the first one it couldn't find a solution and for the third one it gave an output of a couple of dozen of lines of amongst other Bessel functions.

    Another method I tried was writing the equation in the form
    [tex] a(x,y) + b(x,y) \frac{dy}{dx} = 0[/tex] and then trying to find a constant E(x,y) such that [tex] \frac{\partial E}{\partial x} = a(x,y), \frac{\partial E}{\partial y} = b(x,y) [/tex]. However, this requires equality of mixed second-order derivatives, and this is not the case for either of the equations. Neither could I easily see a constant [tex] \phi(x,y) [/tex] to multiply the above equation with such that [tex] \phi a(x,y) [/tex] and [tex] \phi b(x,y) [/tex] do satisfy.
    (I'm not sure what this method is called in English; it was mentioned in my Dutch textbook)
     
  4. Apr 13, 2012 #3
    The first one can't be solved for y with elementary functions, so that's about as far as you can go.

    Apparently the second one can be written as a Bernoulli equation according to WolframAlpha, and that's how you would get the solution that Mathematica gave you.

    For the third one, y = ux will turn it into a separable equation.
    I think it's interesting how x works in the differential equation, but you can't get it from the general solution...

    The method you're talking about is exact differential equations I think; it doesn't look like that method will be helpful for this one.
     
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