(Bessel Functions) I'm trying to figure out a couple very short steps of algebra in between a webpage on the orthogonality of Bessel Functions, but I'm having a tough time figuring out their result. More information below.

Question

mendicant_bias · Answer

On here: http://www.hit.ac.il/staff/benzionS/Differential.Equations/Orthogonality_of_Bessel_functions.htm From line 94 to 95, what I don't understand is just how they get their result; if I multiplied both equations in 94 by v and u respectively, I wouldn't/can't figure out how i'd be able to get the first terms of either equation to cancel out with one another.

mendicant_bias · Answer

@phi

mendicant_bias · Answer

The way I see it: 
$$(1) \ \ \ x[xu']'+(a^2x^2-m^2)u=0,$$$$(2) \ \ \ x[xv']'+(b^2x^2-m^2)v=0, $$
Multiplying the top and bottom by y and u respectively and subtracting gives:

mendicant_bias · Answer

$$xv[xu']'+(a^2x^2-m^2)uv=0=xv[xu''+u']+(a^2x^2-m^2)uv$$$$xu[xv']'+(b^2x^2-m^2)vu=0=xu[xv''+v']+(b^2x^2-m^2)vu$$

mendicant_bias · Answer

I'm assuming those first two terms are wholly cancelled out due to them not sharing any a's or b's and the final result that we want is; $$(b^2-a^2)xuv=\frac{d}{dx}(vxu'-uxv')$$

mendicant_bias · Answer

I also don't understand how it's appropriate for them to seemingly move stuff inside and outside of the derivative operation. @iambatman