convert/Kelvin converts, when possible, special functions admitting a 1F1 or 0F1 hypergeometric representation into Kelvin functions. The Kelvin functions are

FunctionAdvisor( Kelvin );

The 6 functions in the "Kelvin" class are:

$\left[\mathrm{KelvinBei}\,\mathrm{KelvinBer}\,\mathrm{KelvinHei}\,\mathrm{KelvinHer}\,\mathrm{KelvinKei}\,\mathrm{KelvinKer}\right]$

$\mathrm{AiryAi}\left(z\right)$

$\mathrm{AiryAi}\left(z\right)$

$\mathrm{convert}\left(\&comma;\mathrm{Kelvin}\right)\mathbf{assuming}0<\mathrm{\Re }\left(z\right)$

$\frac{\sqrt{3}\sqrt{z}\left(\mathrm{KelvinKer}\left(\frac{1}{3}\&comma;\left(\frac{1}{3}-\frac{\mathrm{I}}{3}\right)z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{2}\right)+\mathrm{I}\mathrm{KelvinKei}\left(\frac{1}{3}\&comma;\left(\frac{1}{3}-\frac{\mathrm{I}}{3}\right)z^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{2}\right)\right)\left(\sqrt{3}+\mathrm{I}\right)}{6\mathrm{\pi }}$

$\mathrm{sol}≔y\left(z\right)=\mathrm{\_C1}\mathrm{KummerM}\left(\frac{1}{2}+a\&comma;1+2a\&comma;z\right)+\mathrm{\_C2}\mathrm{KummerU}\left(\frac{1}{2}+a\&comma;1+2a\&comma;z\right)$

$\mathrm{sol}≔y\left(z\right)=\mathrm{\_C1}\mathrm{KummerM}\left(\frac{1}{2}+a\&comma;1+2a\&comma;z\right)+\mathrm{\_C2}\mathrm{KummerU}\left(\frac{1}{2}+a\&comma;1+2a\&comma;z\right)$

$\mathrm{PDEtools}\left[\mathrm{dpolyform}\right]\left(\mathrm{sol}\&comma;\mathrm{no\_Fn}\right)$

$\left[\frac{{\&DifferentialD;}^2}{\&DifferentialD;z^2}y\left(z\right)=\frac{\left(-2a+z-1\right)\left(\frac{\&DifferentialD;}{\&DifferentialD;z}y\left(z\right)\right)}{z}+\frac{\left(1+2a\right)y\left(z\right)}{2z}\right]\&where\left[\right]$

$\mathrm{convert}\left(\mathrm{sol}\&comma;\mathrm{Kelvin}\right)$

$y\left(z\right)=\frac{\mathrm{\_C1}{\&ExponentialE;}^{\frac{z}{2}}\mathrm{\Gamma }\left(1+a\right)\left(\mathrm{KelvinBer}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)+\mathrm{I}\mathrm{KelvinBei}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)\right)2^a}{{\left(\frac{\mathrm{I}}{2}z\right)}^a}+\frac{\mathrm{\_C2}\left(\frac{2^a\mathrm{KelvinBer}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{4{\left(\frac{\mathrm{I}}{2}z\right)}^a{2}^{-1+2a}}-\frac{{\left(\frac{\mathrm{I}}{2}z\right)}^a{2}^{1+2a}\mathrm{KelvinBer}\left(-a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{4z^{2a}{2}^a}-\frac{\mathrm{I}\mathrm{KelvinBei}\left(-a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right){\left(\frac{\mathrm{I}}{2}z\right)}^a{2}^{1+2a}}{4z^{2a}{2}^a}+\frac{\mathrm{I}\mathrm{KelvinBei}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)2^a}{4{\left(\frac{\mathrm{I}}{2}z\right)}^a{2}^{-1+2a}}\right)\sqrt{\mathrm{\pi }}{\&ExponentialE;}^{\frac{z}{2}}}{\sin \left(\mathrm{\pi }\left(1+a\right)\right)}$

$\mathrm{collect}\left(\&comma;\left[\mathrm{KelvinBer}\&comma;\mathrm{KelvinBei}\right]\&comma;\mathrm{simplify}\right)$

$y\left(z\right)=\frac{{\&ExponentialE;}^{\frac{z}{2}}{\left(\mathrm{I}z\right)}^{-a}\left(2\mathrm{\_C1}\mathrm{\Gamma }\left(1+a\right)\sin \left(\mathrm{\pi }a\right)4^a-\mathrm{\_C2}\sqrt{\mathrm{\pi }}\right)\mathrm{KelvinBer}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{2\sin \left(\mathrm{\pi }a\right)}-\frac{\mathrm{\_C2}{z}^{-2a}{\left(\mathrm{I}z\right)}^a\sqrt{\mathrm{\pi }}{\&ExponentialE;}^{\frac{z}{2}}\mathrm{KelvinBer}\left(-a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{2\sin \left(\mathrm{\pi }\left(1+a\right)\right)}+\frac{\mathrm{I}{\&ExponentialE;}^{\frac{z}{2}}{\left(\mathrm{I}z\right)}^{-a}\left(2^{2+2a}\mathrm{\_C1}\mathrm{\Gamma }\left(1+a\right)\sin \left(\mathrm{\pi }a\right)-2\mathrm{\_C2}\sqrt{\mathrm{\pi }}\right)\mathrm{KelvinBei}\left(a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{4\sin \left(\mathrm{\pi }a\right)}-\frac{\mathrm{I}\mathrm{\_C2}{z}^{-2a}{\left(\mathrm{I}z\right)}^a\sqrt{\mathrm{\pi }}{\&ExponentialE;}^{\frac{z}{2}}\mathrm{KelvinBei}\left(-a\&comma;\left(\frac{1}{4}-\frac{\mathrm{I}}{4}\right)z\sqrt{2}\right)}{2\sin \left(\mathrm{\pi }\left(1+a\right)\right)}$