Answer to Question #102654 in Calculus for BIVEK SAH

Question

Show that
J0(x)+J2(x)=2d/dx[J1(x)]

Answer 1

Show that

J₀(x)-J₂ (x)=2d/dx[J₁ (x)]

Solution:-

2$J'_1(x)=J$_1-1(x)-$J$₁₊₁(x)

$2J_1'(x)=J_0(x)-J_2(x)$

We Know that :-

$xJ'_1(x)=J_1(x)-xJ$ ₂(x) ...........(1)

$xJ_1'(x)=-J_1(x)+xJ$ ₀(x) ............(2)

Adding Equation 1 and 2 we have

2x$J'_1(x)=J_1(x)-J_1(x)-xJ$ ₂(x)$+xJ$₀(x)

$2xJ'_1(x)=x(J$ ₀(x)-$J$ ₂(x))

$2J'_1(x)=J$₀(x)-$J$₂ (x)