Question

To calculate the given system of equation.

$\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}{\left(x\right)}^{\mathbf{p}}{\mathbf{J}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{=}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{p}}{\mathbf{J}}_{\mathbf{p}\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Short answer

This equation has been proved.

Step-by-step solution

Concept of Bessel’s Equation:

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

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

Calculation of the equations ddx[(x)pJp(x)] and ddx[(x)-pJp(x)] 

From equation,

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]={\left(\mathrm{x}\right)}^{\mathrm{p}}{\mathrm{J}}_{\mathrm{p}-1}\left(\mathrm{x}\right)$

This equation can be written as follows

$$\begin{gathered} {\left(\mathrm{x}\right)}^{\mathrm{P}}{\mathrm{J}}_{\mathrm{P}}\left(\mathrm{x}\right)+\mathrm{p}{\left(\mathrm{x}\right)}^{\mathrm{p}-1}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{P}}{\mathrm{J}}_{\mathrm{p}-1}\left(\mathrm{x}\right){\mathrm{J}}_{\mathrm{P}}^{\text{'}}\left(\mathrm{x}\right)+\left(\frac{\mathrm{p}}{\mathrm{x}}\right){\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right) \\ ={\mathrm{J}}_{\mathrm{p}-1}\left(\mathrm{x}\right){\mathrm{J}}_{\mathrm{P}}^{\text{'}}\left(\mathrm{x}\right) \\ ={\mathrm{J}}_{\mathrm{p}-1}\left(\mathrm{x}\right)-\left(\frac{\mathrm{p}}{\mathrm{x}}\right){\mathrm{J}}_{\mathrm{P}}^{\text{'}}\left(\mathrm{x}\right) \end{gathered}$$

From equation,

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=-{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)$

This equation can be written as follows:

$$\begin{gathered} -{\mathrm{J}}_{\mathrm{p}}^{\text{'}}\left(\mathrm{x}\right)+\left(\frac{\mathrm{P}}{\mathrm{x}}\right){\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right)\left(\frac{\mathrm{p}}{\mathrm{x}}\right){\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)-{\mathrm{J}}_{\mathrm{p}+1}\left(\mathrm{x}\right) \\ ={\mathrm{J}}_{\mathrm{P}}^{\text{'}}\left(\mathrm{x}\right) \end{gathered}$$