Soi $$ f $$ une solution non nulle de $$ (E): y^{\prime \prime}-\frac{1}{1+t^{2}} y^{\prime}+y=0 $$. On note $$ q: t \in \mathbb{R}^{+} \mapsto \frac{1}{1+t^{2}} $$ 3. Justifier à l'aide du théorème de Cauchy-Lipschitz, que pour tout $$ t \in \mathbb{R},\left(f(t), f^{\prime}(t)\right) \neq(0,0) $$ 4. En déduire qu'il existe deux applications $$ r \in \mathcal{C}^{1}\left(\mathbb{R}^{+}, \mathbb{R}_{+}^{*}\right) $$ et $$ \theta \in \mathcal{C}^{1}\left(\mathbb{R}^{+}, \mathbb{R}\right) $$ telles que $$ \forall t \in \mathbb{R}^{+}, f(t)=r(t) \cos (\theta(t)) \text { et } f^{\prime}(t)=r(t) \sin (\theta(t)) $$

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Schultz Evans · Accepted Answer

For all $$ t $$ in $$ \mathbb{R}^+ $$, $$ f(t) $$ and $$ f'(t) $$ are both non-zero. Therefore, there exist functions $$ r(t) $$ and $$ \theta(t) $$, both continuously differentiable on $$ \mathbb{R}^+ $$, such that: $$ f(t) = r(t) \cos(\theta(t)) \quad \text{and} \quad f'(t) = r(t) \sin(\theta(t)) $$

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