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ODE with 2 parameterized families

  1. May 30, 2013 #1
    1. The problem statement, all variables and given/known data
    problem:
    Find a 1-parameter family of solutions of each of the following equa-
    tions. Assume in each case that the coefficient of [itex]dy \neq 0[/itex].

    [itex](x + \sqrt{ y^2 - xy}) \mathop{dy} - y \mathop{dx} = 0[/itex]

    answer:
    [itex]y = ce^{-2\sqrt{1 - x/y}}, \;\;\; y >0, \, x< y; \;\;\; y = ce^{2\sqrt{1 - x/y}}, \;\;\; y<0, \,x >y[/itex]

    Edit: This is Exercise 7 - 3 from Tennenbaum & Pollard's Ordinary Differential Equations.

    2. Relevant equations



    3. The attempt at a solution
    my work:
    [itex]y\left(\frac{x}{y} + \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; \frac{x}{y} < 1, \, y \neq 0[/itex]
    [itex]x = udy \;\;\; dx =y\mathop{du}+ u\mathop{dy}[/itex]
    [itex]y\left(u + \sqrt{ 1 - u}\right)\mathop{dy} - y \left(y\mathop{du}+ u\mathop{dy}\right) = 0[/itex]
    [itex]\sqrt{ 1 - u}\mathop{dy} - ydu = 0 [/itex]
    [itex]\int\frac{dy}{y} = \int\frac{du}{\sqrt{ 1 - u}} = -\int\frac{dv}{\sqrt{v}}[/itex]
    [itex] ln(y) = -2\sqrt{v} + c = -2\sqrt{1-u} + c = -2\sqrt{1 - x/y} + c[/itex]
    [itex]y = c_1e^{-2\sqrt{1 - x/y}}[/itex]

    I understand how [itex]\frac{x}{y} < 1, \, y \neq 0 [/itex] is the same as [itex]y<0,\, x > y[/itex] or [itex]y >0,\, x< y[/itex]. However I am not sure how to discover the 2nd 1-parameter solution with the positive exponent. I think I'm failing to consider something crucial. Also I may not feel as confident as I'd like about discovering multiple solutions to separable first order ODEs that are not part of the same family (generated by changing a parameter). So general advice or techniques might be appreciated.
     
    Last edited: May 31, 2013
  2. jcsd
  3. May 31, 2013 #2
    I think the two solutions split off before the first step when I factor out a y from within the square root.

    [itex]y\frac{x}{y}\mathop{dy} + \sqrt{ y^2\left(1 - \frac{x}{y}\right)} \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; \frac{x}{y} < 1, \, y \neq 0[/itex]

    Because square roots of positive numbers are always positive, if we factor out a y which happens to be negative, we'd need to have a minus sign.

    [itex]y\left(\frac{x}{y} + \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; y > 0, \, x<y[/itex]
    OR
    [itex]y\left(\frac{x}{y} - \sqrt{ 1 - \frac{x}{y}}\right) \mathop{dy} - y \mathop{dx} = 0, \;\;\;\; y < 0, \, x>y[/itex]


    The lesson I learned is to be mindful that [itex]\sqrt{y^2(\dots)} = \left|y\right|\sqrt{\dots}[/itex]
     
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