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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: $\displaystyle x \frac{dy}{dx} = 2y$
    Solution: $\displaystyle y = x^{2}$
    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:
    $\displaystyle -\infty < x < \infty$

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

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

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

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

    $\displaystyle y = Cx^2$

    is a more general solution. Even $\displaystyle C=0$ works; i.e., the function $\displaystyle y \equiv 0$ 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 $\displaystyle xy$-plane, and think about the uniqueness of solutions. Given any point $\displaystyle (x_0, y_0)$ where $\displaystyle x_0<0$, you can can find a constant $\displaystyle C_0$ such that the curve

    $\displaystyle y(x) = C_0x^2$

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

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

    Hope that helps!

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