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Domain of a Differential Equation

  1. Mar 5, 2008 #1
    Find the particular solution y=f(x) to dy/dx=(1+y)/x given that f(-1)=1 and state its domain.

    My answer was: 2|x|-1, x ∈ ℝ/{0}

    Apparently, the domain should be x<0

    Solving the differential equation was not an issue at all, but I have no idea why the domain is restricted to x<0. I understand that zero cannot be included because dy/dx fails to exist at x=0, but why are positive real numbers not included?

    In general, as well, how is the domain of a particular differential equation found that is not obvious (e.g. found by simply looking at dy/dx)
     
  2. jcsd
  3. Mar 5, 2008 #2

    arildno

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    With the domain of a "particular" solution for a diff.eq, we mean those x's for which the given intial conditions yield a UNIQUE solution.

    In your case, an equally good solution is:

    y(x)=2|x|-1, x less than 0, and y(x)=3x-1 for x greater than 0.

    The discontinuity at x=0 wreaks havoc on the uniqueness of the solution that normally is guaranteed by specifying a SINGLE initial condition.
     
  4. Mar 5, 2008 #3

    HallsofIvy

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    Notice that if you were given exactly the same differential equation with the condition f(1)= 1, the domain would be the set of all positive numbers.
     
  5. Mar 5, 2008 #4
    Ah, I see now. For an initial condition problem, the domain of the solution curve is the set of all x that satisfies the differential equation and contains the initial point. So, since there's a singularity at x=0, only the left half of the curve is the solution. Is my reasoning correct?
     
  6. Mar 6, 2008 #5

    arildno

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    Indeed.

    If you were to have a unique solution covering all of R except 0, you'd need TWO initial conditions, one in each of the two disconnected parts of the domain.
     
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