# Bessel identities

$$a(J_1(a)Y_0(a)-J_0(a)Y_1(a))=\frac{2}{\pi}$$ for any $a$.
$J$ and $Y$ are bessel functions of the first and second kind respectively.
Thank you.

-
what is "*" in your notation? Too much MATLAB recently ? :-) –  Dima Pasechnik Mar 12 at 1:32
This is equivalent to computing the Wronskian of $J_0$ and $Y_0$, since $J'_0 = -J_1$ and $Y'_0 = -Y_1$. $$W(x) = \begin{vmatrix} J_0 & Y_0 \\\ J'_0 & Y'_0 \end{vmatrix} = \begin{vmatrix} J_0 & Y_0 \\\ -J_1 & -Y_1 \end{vmatrix} = J_1 Y_0 - J_0 Y_1.$$ But the zeroth-order Bessel equation is: $$\frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} + y = 0.$$ By Abel's formula, $W(a) = \frac{W(1)}{a}$. You now have to compute any value of $W$.
$$W(a)=W(1)e^{-\int_1^a\frac{1}{x}dx}=\frac{W(1)}{a}$$ –  Vadim Winebrand Mar 12 at 9:09
Thank you.<br> The proof is almost complete now it is required to prove that: $$(J_1(1)Y_0(1)-J_0(1)Y_1(1))=\frac{2}{\pi}$$ –  Vadim Winebrand Mar 12 at 11:18