To show ${\mathbf{J}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}{\mathbf{J}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

The solution of Bessel's equation is,${\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{"}\mathbf{}\mathbf{+}\mathbf{}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{n}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{0}$.

${\mathbf{J}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}}{\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{┌}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{n}}$

It is known that:

${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}+\mathrm{p}}}{⌜(\mathrm{n}-\mathrm{p}+1)\left(\mathrm{n}!\right)}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\mathrm{p}}$

Therefore, simplify as follows:

localid="1659265689411" $$\begin{gathered} {\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{┌(\mathrm{n}+1)┌\left(\mathrm{n}+{\displaystyle \frac{3}{2}}\right)}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\left(\frac{1}{2}\right)} \\ =\sqrt{\frac{2}{\mathrm{x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{2}^{\mathrm{n}+1}(\mathrm{n}!)}{2^{2\mathrm{n}+1}(\mathrm{n}!)(2\mathrm{n}+1)!\mathrm{\tau \tau }}{\left(\mathrm{x}\right)}^{2\mathrm{n}+1} \\ =\sqrt{\frac{2}{\mathrm{x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{2}^{\mathrm{n}+1}(\mathrm{n}!)}{2^{2\mathrm{n}+1}(\mathrm{n}!)(2\mathrm{n}+1)!\mathrm{\tau \tau }}{\left(\mathrm{x}\right)}^{2\mathrm{n}+1} \\ =\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{(2\mathrm{n}+1)}{\left(\mathrm{x}\right)}^{2\mathrm{n}+1} \end{gathered}$$

It is known that:

${\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{┌(\mathrm{n}+1)┌\left(\mathrm{n}+{\displaystyle 1-p}\right)}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}-\mathrm{p}}$

Therefore, simplify as follows:

$$\begin{gathered} {\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{2^{2n-1}(\mathrm{n}!)┌\left(\mathrm{n}+{\displaystyle \frac{1}{2}}\right)}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\left(\frac{1}{2}\right)} \\ =\sqrt{\frac{2}{\mathrm{x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^n}{2^{2\mathrm{n}+1}(\mathrm{n}!)┌\left(\mathrm{n}+\frac{1}{2}\right)}{\left(\mathrm{x}\right)}^{2\mathrm{n}-1} \\ =\sqrt{\frac{2}{\mathrm{x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{2}^{\mathrm{n}+1}(\mathrm{n}!)}{2^{2\mathrm{n}-1}(\mathrm{n}!)\left(2\mathrm{n}\right)!\mathrm{\tau \tau }}{\left(\mathrm{x}\right)}^{2\mathrm{n}-1} \\ =\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)1}{\left(\mathrm{x}\right)}^{2\mathrm{n}-1} \end{gathered}$$

Therefore,

${\mathrm{J}}_{-\frac{1}{2}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\cos \mathrm{x}$

Now, substitute values to obtain:

$$\begin{gathered} {\mathrm{J}}_{\frac{3}{2}}\left(\mathrm{x}\right)=\frac{1}{2\mathrm{x}}\left(\mathrm{x}\right)-\mathrm{J}{\text{'}}_{\frac{1}{2}}\left(\mathrm{x}\right) \\ =-\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\cos \mathrm{x}+\frac{1}{2\mathrm{x}}\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\sin \mathrm{x}+\frac{1}{2\mathrm{x}}\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\sin \mathrm{x} \\ =-\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\cos \mathrm{x}+\frac{1}{\mathrm{x}}\sqrt{\frac{2}{\mathrm{\tau \tau x}}}\sin \mathrm{x} \\ =\frac{1}{\mathrm{x}}{\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)-{\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right) \end{gathered}$$

Hence, its proved.