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Homework Help: Implicit function / Separable ODE

  1. Sep 13, 2012 #1

    The final step of solving a separable ODE is to find a function, f, defined implicitly by a relation

    G(y) = H(x).

    Say G(y) isn't defined at y = a and H(x) isn't defined at x = b, it appears to me that when rearranging such a relation to put y in terms of x, the point at which G(y) isn't defined isn't 'important' because it doesn't correspond to a point where y = f(x) is defined anyway.

    Is this the case and if so, why?


    H(y) = G(x) = 1 / (y-1) = x
    H(y) defined for all real values except y = 0
    G(x) defined for all real values.

    Rearrangement gives,

    y = f(x) = 1 + 1 / x

    and no value of x will solve f(x) = 1

  2. jcsd
  3. Sep 13, 2012 #2
    The point where G(y) is not defined is usually important. This is because you get G(y) by integrating some g(y), which is typically made by dividing some f(y), originally on the left, by h(y), originally on the right. And h(y) = 0 may well correspond to the singular points of G(y). What needs to be done is research whether h(y) results in any special solutions of the original equation (which is significantly simplified in this case). For example:

    y' = y^2 => y'/y^2 = 1 => -1/y = t + C => y = -1/(t + C) (*)

    However, y^2 = 0 => y = 0, which is a singular solution not given by (*).
  4. Sep 14, 2012 #3
    Thanks very much for your response but, to use your example, suppose the original differential equation is

    y' / y^2 = 1 instead of y' = y^2

    what I'm wanting to find out is why the rearranged solution of y' / y^2 doesn't seem to go through the singular point of 1 / y^2

  5. Sep 14, 2012 #4
    Because the function on the right would have to be infinite. In our example, C has to be infinite, or the independent variable has to be infinite. Among other things, that means that the solution is not unique (it is satisfied for any initial value of x). The loss of uniqueness is the more general property of singular solutions. For example, y'^2 = 4y => y = (x + c)^2. There are no infinities involved, however at x = -c all the solutions go through y = 0, which is a singular solution as well. In this case, by the way, no constant of integration can be manipulated to obtain the singular solution.
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