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Singular solutions Separable Eq. and IVP

  1. Feb 19, 2009 #1

    I am a little bit confused about singular solutions and their relationship with IVP's and decided to ask you.

    As far as I understood, the IVP's could be in a form:
    y' = f(x,y)
    y(x0) = y0

    To obtain a general solution, we could use separable eq. method. I have learned that sometimes when we separate the variables, we assume a particular condition to be valid. (that might have an asymptotic behaviour. )

    Also, as far as I know, a singular solution can not be obtained from a general solution.

    My impression was that the general solution shouldn't have worked for y = y0 if we derived it assuming y is not equal to y0. However, that example confuses me;

    y' = 2 (y^1/2)
    y(0) = 0

    By the sep. eq. method and integrating both sides;

    (dy / 2(y^1/2)) = dx

    I have found that y = (x-c)^2. However, while separating, we actually assume that y is not equal to 0.

    What's confusing is, at this general solution, when I plug y = 0 and x = 0; I get a particular solution; which is y = x^2 and that perfectly works well! It does not make sense to get a particular solution which suggests y= 0 when x = 0 using a general solution that is not defined for y = 0. (Because I had assumed y is not equal to 0 while I was separating the variables. )
    Last edited: Feb 19, 2009
  2. jcsd
  3. Feb 19, 2009 #2


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    Science Advisor

    It is sufficient to solve for y with y not 0, then extend the solution to y= 0 since, in order that the derivative exist, y must be continuous.

    But did you notice that y(x)= 0 for all x is also a solution? In fact, if you set y(x)= (x- a)2 for x< a< 0, y(x)= 0 for a< x< b (where b> 0), y(x)= (x-b)2 for 0< b< x is also a solution to the differential equation satisfying y(0)= 0.
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