We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$. For $x \in \mathbb{R}$, we introduce the vectors $$n(x) = \frac{T^{\prime}(x)}{\lambda}, \quad b(x) = T(x) \wedge n(x)$$ so that $(T(x), n(x), b(x))$ forms a direct orthonormal basis.
(a) Using question 15, show that for all $x \in \mathbb{R}$, we have $2b^{\prime}(x) = -x\, n(x)$.
(b) Deduce that $n^{\prime}(x) = -\lambda T(x) + \frac{x}{2} b(x)$.
(c) Show that $G$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$