Since $\mathcal{G}_{k}(z)$ and $\mathcal{D}_{k}(z)$ solve the homogeneous equation

(5.84)

$$\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=0$$

which misses the first order term. Immediately, for all $z\geq 0$,

(5.85)

$$\frac{d}{dz}\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=0,$$

which implies that the Wronskian $\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))$ is a constant. So it suffices to calculate it at $z=0$. By the definition of the Wronskian,

	$\displaystyle\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=$	$\displaystyle e^{j_{k}z^{n}}\cdot\Big(\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$
(5.86)			$\displaystyle-\frac{d}{dz}\Ku(\beta,\alpha,-j_{k}z^{n})\cdot\Tri(\beta,\alpha,-j_{k}z^{n})\Big).$

To calculate $\frac{d}{dz}\Tri(\beta,\alpha,-j_{k}z^{n})$, we will apply Kummer’s transformation law to relate $\Tri$ and $\Ku$, that is,

(5.87)				$\displaystyle\Tri(\beta,\alpha,-j_{k}z^{n})$
			$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\mathcal{U}(\alpha-\beta,\alpha,{j_{k}}z^{n})$
			$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot({j_{k}}z^{n})^{1-\alpha}\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
			$\displaystyle=$	$\displaystyle e^{-{j_{k}}z^{n}}\cdot\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\alpha-\beta,\alpha,{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\beta,2-\alpha,{j_{k}}z^{n})\Big)$
			$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}z\cdot\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n}).$

So it follows that

(5.88)				$\displaystyle\frac{d}{dz}\Tri(\beta,\alpha,-{j_{k}}z^{n})$
			$\displaystyle=$	$\displaystyle\frac{\Gamma(1-\alpha)}{\Gamma(1-\beta)}\cdot\frac{d}{dz}\Ku(\beta,\alpha,-{j_{k}}z^{n})+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Big(\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})$
			$\displaystyle+$	$\displaystyle z\cdot\frac{d}{dz}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})\Big).$

Since $n\geq 2$, it directly follows from the definition of $\Ku$ that

(5.89)		$\displaystyle\frac{d}{dz}\Big|_{z=0}\Ku(\beta,\alpha,-{j_{k}}z^{n})=0,$
(5.90)		$\displaystyle\frac{d}{dz}\Big|_{z=0}\Ku(1-\alpha+\beta,2-\alpha,-{j_{k}}z^{n})=0.$

Therefore,

(5.91)		$\displaystyle\frac{d}{dz}\Big|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})$	$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta)}\cdot{j_{k}}^{\frac{1}{n}}\cdot\Ku(1-\alpha+\beta,2-\alpha,0)$
			$\displaystyle=$	$\displaystyle\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.$

Now evaluate (5.86) at $z=0$, we have

(5.92)

$$\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(z)=\mathcal{W}(\mathcal{G}_{k},\mathcal{D}_{k})(0)=\frac{d}{dz}\Big|_{z=0}\Tri(\beta,\alpha,-{j_{k}}z^{n})=\frac{\Gamma(\alpha-1)\cdot{j_{k}}^{\frac{1}{n}}}{\Gamma(\alpha-\beta)}.$$

∎