We know that, $2n J_{n}(x) = xJ_{n + 1}(x) + x J_{n - 1}(x)\\ \therefore J_{n + 1}(x) = \frac{2n}{x}J_{n}(x) - J_{n - 1}(x)......(A)$

Put n = $\frac{3}{2}$ in equation (A)

$J_{\frac{3}{2}}(x) = \frac{3}{x} J_{\frac{3}{2}}(x) - J_{\frac{1}{2}}(x).......(B)$

Now put n = $\frac{1}{2}$ in equation (A)

$J_{\frac{3}{2}}(x) = \frac{1}{x} J_{\frac{1}{2}}(x) - J_{\frac{1}{2}}(x).......(C)$

Substitute (C) in equation (B)

$\therefore J_{\frac{5}{2}}(x) = \frac{3}{x}\bigg[\frac{1}{x}J_{\frac{1}{2}}(x) - J_{\frac{-1}{2}}(x)\bigg] - J_{\frac{1}{2}}(x) \\ J_{\frac{5}{2}}(x) = \frac{3}{x^2} J_{\frac{1}{2}}(x) - \frac{3}{x} J_{\frac{-1}{2}}(x) - J_{\frac{1}{2}}(x) \\ J_{\frac{5}{2}}(x) = \bigg(\frac{3}{x^2} - 1 \bigg) J_{\frac{1}{2}}(x) - \frac{3}{x}J_{\frac{-1}{2}}(x).....(D)$

But we know

$J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}sinx$ and

$J_{\frac{-1}{2}}(x) = \sqrt{\frac{2}{\pi x}}cosx$

$\therefore J_{\frac{5}{2}}(x) = \bigg(\frac{3 - x^2}{x^2}\bigg)\sqrt{\frac{2}{\pi x}}sinx - \frac{3}{x}\sqrt{\frac{2}{\pi x}}cosx \\ \therefore J_{\frac{5}{2}}(x) = \sqrt{\frac{2}{\pi x}}\bigg[\bigg(\frac{3 - x^2}{x^2}\bigg) sinx - \frac{3}{x} cosx\bigg]$

Hence proved.