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Thread: 2nd order ODE (Help)

  1. #1
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    2nd order ODE (Help)

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

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