Verify $J'_0(x)=-J_1(x)$

Question

I need some help with my math coursework. I am given a Bessel function, $x^2y'' + xy' + (x^2 − n^2)y = 0$, and I am told that it has two independent solutions, $J_n(x)$ and $Y_n(x)$. I am told I need to verify the title equation, $J'_0(x)=-J_1(x)$, by showing both $J'_0(x)$ and $J_1(x)$ satisfy the same differential equation. However, I don't know what this equation is. Could somebody please help me?

Answer 1

$$ x^2y''+xy'+(x^2)y=0. $$ Differentiating both sides, $$ x^2 y'''+3xy''+(x^2+1)y'+2xy=0,\\ \Rightarrow x^2 y'''+xy''+(x^2-1)y'+2xy''+2y'+2xy=0,\\ \Rightarrow x^2 y'''+xy''+(x^2-1)y'=0. $$ So $Cy'$ is the solution of the last DE above. $C$ is an arbitrary constant. Letting $y=J_0$ you get the result you want. $J_0=CJ_1$. The Bessel function is actually defined in such a way so that $J_0=-J_1$. It is just a convention. So you should look back on you definition your examiers use, and find $C$.