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Math Help - 2nd order ODE (Help)

  1. #1
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    2nd order ODE (Help)

    Let \phi (x) be a solution of the second order linear differential equation
    y^{''}+(sin x)^{2}y^{'}+y=0
    defined on \mathbb{R} satisfying \phi (0)=\phi (\pi) and \phi^{'} (0)=\phi^{'} (\pi). Prove that \phi (x+\pi)=\phi (x) for all x \in \mathbb{R}.

    How do I go about doing it?
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  2. #2
    Super Member Rebesques's Avatar
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    Let \psi(x)=\phi(x+\pi), \ x\in \mathbb{R}.
    Then \psi(0)=\phi(\pi)=\phi(0), \ \psi'(0)=\phi'(\pi)=\phi'(0).
    By uniqueness, \psi\equiv\phi.
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