For a possibly unstable <normal mode>, $c_i\ne0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\widehat w=V^aq$ into the <Taylor–Goldstein equation> and collect terms:
$$
q''+2a\frac{U'}Vq'+\left[-k^2+(a-1)\frac{U''}V+\frac{N^2+a(a-1)(U')^2}{V^2}\right]q=0.
$$
Multiplication by $V^{2a}$ puts its first two terms in <divergence form>:
$$
(V^{2a}q')'-k^2V^{2a}q+\left[(a-1)U''V^{2a-1}+\{N^2+a(a-1)(U')^2\}V^{2a-2}\right]q=0.
$$
Multiply by the <complex conjugate> $\overline q$ and apply <integration by parts>. Since $q(\pm L)=0$, the boundary term vanishes and the <power-transformed Taylor–Goldstein energy identity> is
$$
\boxed{\int_{-L}^LV^{2a}(|q'|^2+k^2|q|^2)dz=\int_{-L}^L\left[\{N^2+a(a-1)(U')^2\}V^{2a-2}+(a-1)U''V^{2a-1}\right]|q|^2dz.}
$$
The weight is $V^{2a}$, not $|V|^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 1 / b / i / Solution

A <differential operator> is in divergence form when its highest derivatives occur inside a <divergence>, for example $Lu=\nabla\cdot(A\nabla u)$ or $(p(x)u')'$. This form makes <integration by parts> express the operator through a lower-order bilinear form and boundary flux.

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