Answer:

$2n\times J_n(x)= x\times J_{n-1}(x)+x\times J_{n+1}(x)\dots\dots\dots(recurrence \ formulae)$

Putting n=3 in recurrence formulae

$6 \times J_3 (x)={ x\times J_2(x)+ x\times J_4(x)}$

$\dfrac{{6\times J_3(x)-x\times J_2(x)}}{{x}}=J_4(x)$

$\dfrac{6}{x}\times J_3(x) - J_2(x)=J_4(x)\dots\dots(1)$

$Putting \: n=2\: in \:recurrence \:formulae$

$4\times J_2(x)=x \times J_1(x)+x\times J_3(x)$

$\dfrac{4\times J_2(x)-x\times J_1(x)}{x}=J_3(x)$

${\dfrac{4}{x}}\times J_2(x)-J_1(x)=J_3(x)\dots\dots\dots(2)$

Putting equation 2 in equation 1

$\dfrac{6}{x}\times [{\dfrac{4}{x}\times J_2(x)-J_1(x)}]-J_2(x)=J_4(x)$

$\dfrac{24}{x^2}\times J_2(x)-\dfrac{6}{x}\times J_1(x)-J_2(x)=J_4(x)$

$\bigg[\dfrac{24}{x^2}-1\bigg]\times J_2(x)-\dfrac{6}{x}\times J_1(x)=J_4(x)\dots\dots(3)$

Putting n=1 in recurrence formula we get,

$2\times J_1(x)= x\times J_0(x)+x\times J_2(x)$

$\dfrac{2\times J_1(x)-x\times J_0(x)}{x}=J_2(x)$

$\dfrac{2}{x}\times J_1(x)-J_0(x)=J_2(x)\dots\dots(4)$

Substituting eqn (4) in eqn (3) 

$\bigg[\dfrac{24}{x^2}-1\bigg]\times \bigg[\dfrac{2}{x}\times J_1(x)-J_0(x)\bigg]-\dfrac{6}{x}\times J_1(x)=J_4(x)$

$\dfrac{48}{x^3}\times J_1(x)-\dfrac{2}{x}\times J_1(x)-\dfrac{24}{x^2}\times J_0(x)+J_0(x)-\dfrac{6}{x}\times J_1(x)=J_4(x)$

$\bigg[\dfrac{48}{x^3}-\dfrac{8}{x}\bigg]\times J_1(x)-\bigg[\dfrac{24}{x^2}-1\bigg]\times J_0(x)=J_4(x)$

Hence proved