Osborne Valdez
03/30/2024 · Middle School
On considère l'équation différentielle \[ (E): y^{\prime \prime}-\frac{1}{1+t^{2}} y^{\prime}+y=0 \] On admet le théorème de Cauchy-Lipschitz Pour tout \( \left(t_{0}, y_{0}, y_{0}^{\prime}\right) \in \mathbb{R}^{3} \), il existe une unique solution de \( (E) \) vérifiant : \( y\left(t_{0}\right)=y_{0} \) et \( y^{\prime}\left(t_{0}\right)=y_{0}^{\prime} \) Question : Justifier que si \( y \) est une solution de \( (E) \) vérifiant \( y\left(t_{0}\right)=y^{\prime}\left(t_{0}\right)=0 \), alors \( y \) est nulle Partie I : Théorème de Relèvement anominam
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Si une solution \( y \) de l'équation différentielle \( y'' - \frac{1}{1 + t^{2}} y' + y = 0 \) vérifie \( y(t_{0}) = 0 \) et \( y'(t_{0}) = 0 \), alors \( y \) est identiquement nulle.
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