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Math Help - Initial Value Problem

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    Super Member Aryth's Avatar
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    Initial Value Problem

    Let c > 0. Show that the function \phi(x) = (c^2 - x^2)^{-1} is a solution to the initial value problem \frac{dy}{dx} = 2xy^2, y(0) = \frac{1}{c^2}, on the interval -c < x < c. Note that this solution becomes unbounded as x approaches \pm c. Thus, the solution exists on the interval (-\delta, \delta) with \delta = c, but not for larger \delta.
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    Quote Originally Posted by Aryth View Post
    Let c > 0. Show that the function \phi(x) = (c^2 - x^2)^{-1} is a solution to the initial value problem \frac{dy}{dx} = 2xy^2, y(0) = \frac{1}{c^2}, on the interval -c < x < c. Note that this solution becomes unbounded as x approaches \pm c. Thus, the solution exists on the interval (-\delta, \delta) with \delta = c, but not for larger \delta.
    If \phi (x) = (c^2 - x^2)^{-1} for |x| < c then \phi '(x) = 2x(c^2 - x^2)^{-2} by the chain rule, therefore, \phi '(x) = 2x \phi(x)^2. Therefore, \phi solves the equation y' = 2xy^2 with y(0) = \tfrac{1}{c^2}. This solution becomes unbounded because \lim _{x\to c^-}\tfrac{1}{c^2-x^2} = \infty and \lim_{x\to -c^+} \tfrac{1}{c^2 - x^2} = \infty. Now say we want to solve y' = 2xy^2, y(0) = \tfrac{1}{c^2} with |x| < \delta (where \delta > c). Then by the existence and uniqueness theorem it means y will coincide with the solution on (-c,c) i.e. y(x) = (c^2 - x^2)^{-1} where |x| < c. But this is a problem because (c^2 - x^2)^{-1} becomes unbounded and so y cannot be differenciable on (-\delta,\delta) if \delta > c.
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    Super Member Aryth's Avatar
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    Thanks for the help. I appreciate it.
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