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Thread: the interval for which the solution to DE makes sense

  1. #1
    Junior Member
    May 2010

    the interval for which the solution to DE makes sense

    Here's an exercise from Tenebaum & Pollard's book:

    Differential equation: x \frac{dy}{dx} = 2y
    Solution: y = x^{2}
    Give the interval on which the solution makes sense.
    I know how to show that the bottom one is the solution of the top equation, what I don't understand is why the interval on which the solution works is not:
    -\infty < x < \infty

    but instead:
    x \neq 0 as given the by author.

    Trying to understand this intuitively, I'm thinking that anything multiplied by zero is zero. So when x = 0, anything substituted on the left hand side gives zero. However, if y = x^{2} is the solution, when x = 0, the RHS is always zero.

    I don't see where the restriction x \neq 0 comes from.

    Thanks for your help!
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  2. #2
    Feb 2010
    "The" solution y = x^2 is not entirely correct. Every ODE has an infinite family of solutions, usually parameterized by a constant of integration:

    y = Cx^2

    is a more general solution. Even C=0 works; i.e., the function y \equiv 0 satisfies the differential equation from the book.

    To try to understand the domain where the solution makes sense intuitively, try thinking of solutions as curves (i.e., integral curves) in the xy-plane, and think about the uniqueness of solutions. Given any point (x_0, y_0) where x_0<0, you can can find a constant C_0 such that the curve

    y(x) = C_0x^2

    passes through the point (x_0, y_0). However what happens at x=0? For any (other) constant C_1, the curve z(x)=C_1x^2, x > 0, is also a solution to the DE, and it meets the solution y(x) at the origin. Uniqueness is violated.

    For any point in the domain (-\infty, 0) \times \mathbf{R}, there is a unique solution of the DE passing through that point that is defined for all x \in (-\infty,0). The same is true for the domain (0, \infty)\times \mathbf{R}. This is why the authors exclude x=0.

    Hope that helps!

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