Question

To calculate the given system of equation.$\frac{\mathbf{d}}{\mathbf{dx}}\left[{\left(x\right)}^p{J}_p\left(x\right)\right]\mathbf{=}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{p}}{\mathbf{J}}_{p-1}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Short answer

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

Step-by-step solution

Concept of Bessel’s Equation: 

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

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

Differentiate the equation ddx[x-pJp(x)] with respect to x :

Differentiate the equation with respect to to obtain:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=\frac{\mathrm{d}}{\mathrm{dx}}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+1+\mathrm{p})}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}}}{2^{2\mathrm{n}+\mathrm{p}}}\right] \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+1+\mathrm{p})}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}}}{2^{2\mathrm{n}+\mathrm{p}}}\right] \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}\left(2\mathrm{n}\right)}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+1+\mathrm{p})}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}-1}}{2^{2\mathrm{n}+\mathrm{p}}}\right] \end{gathered}$$

Using the fact that $⌜(\mathrm{n}+1)=\mathrm{n}⌜\left(\mathrm{n}\right)$cancelling the factor 2and n to obtain:

localid="1659271537510" $\frac{d}{dx}\left[{\left(x\right)}^{-p}{J}_p\left(x\right)\right]=\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜\left(n\right)⌜(\mathrm{n}+1+p)}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}-1}}{2^{2\mathrm{n}+\mathrm{p}-1}}\right]$

Multiplication of ddx[x-pJp(x)] on both sides with  :

Multiply both sides of the equation with ${\mathrm{x}}_{\mathrm{p}}$to obtain:

$$\begin{gathered} {\mathrm{x}}_{\mathrm{p}}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]={\mathrm{x}}_{\mathrm{p}}\sum _{\mathrm{n}=1}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜\left(\mathrm{n}\right)⌜(\mathrm{n}+1+\mathrm{p})}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}-1}}{2^{2\mathrm{n}+\mathrm{p}-1}}\right] \\ =\sum _{\mathrm{n}=1}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜\left(\mathrm{n}\right)⌜(\mathrm{n}+1+\mathrm{p})}\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}}{\left(\mathrm{x}\right)}^{2\mathrm{n}-1}}{2^{2\mathrm{n}+\mathrm{p}-1}}\right] \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜\left(\mathrm{n}\right)⌜(\mathrm{n}+1+\mathrm{p})}\left[\frac{{\left(\mathrm{x}\right)}^{2\mathrm{n}-1}}{2^{2\mathrm{n}+\mathrm{p}}}\right] \\ =\sum _{\mathrm{n}=1}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜\left(\mathrm{n}\right)⌜(\mathrm{n}+1+\mathrm{p})}{\left[\frac{\left(\mathrm{x}\right)}{2}\right]}^{2\mathrm{n}+\mathrm{p}+1} \end{gathered}$$

Therefore,

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

Replacement of n by n + 1 :

Replace n by n + 1 to obtain:

$$\begin{gathered} {\mathrm{x}}_{\mathrm{p}}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\left(\mathrm{x}\right)}^{-\mathrm{p}}{\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)\right]=-\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\left(n\right)}}{⌜\left(\mathrm{n}\right)⌜\left(\left(\mathrm{n}+1\right)\left(1+\mathrm{p}\right)\right)}{\left[\frac{x}{2}\right]}^{2\left(n\right)+(p+1)} \\ =J_{\left(p+1\right)}\left(x\right) \end{gathered}$$

Compare with given equation.

Multiply both sides with ${\mathrm{x}}^{-\mathrm{p}}$to obtain:

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

Hence, proved.