Question Number 219591 by SdC355 last updated on 29/Apr/25
$$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\ $$$$\mathrm{Ai}\left({x}\right)\:\mathrm{and}\:\mathrm{Bi}\left({x}\right)\:\mathrm{Airy}\:\mathrm{Function} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({z}\right)\:\mathrm{is}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}{f}\left({z}\right) \\ $$