Given:$$y''+\frac{2}{r}y'+(1-\frac{l(l+1)}{r^2})y=0$$$$J(r)=r^{\frac{1}{2}}y(r)$$

Show that$$J''+\frac{1}{r}J'+(1-\frac{l(l+1)+\frac{1}{4}}{r^2})J=0$$

I can get to:$$J''+\frac{1}{r}J'=r^\frac{1}{2}((\frac{l(l+1)}{r^2}-1)y+\frac{1}{4r^2}y)$$

Question

Given:$$y''+\frac{2}{r}y'+(1-\frac{l(l+1)}{r^2})y=0$$$$J(r)=r^{\frac{1}{2}}y(r)$$

Show that$$J''+\frac{1}{r}J'+(1-\frac{l(l+1)+\frac{1}{4}}{r^2})J=0$$

I can get to:$$J''+\frac{1}{r}J'=r^\frac{1}{2}((\frac{l(l+1)}{r^2}-1)y+\frac{1}{4r^2}y)$$

anonymous · Answer

Ok, it looks like your method has overcomplicated things, so I will use a different method.

To begin, write $y=Jr^{-\frac{1}{2}}$ and calculate the derivatives of y in terms of J:
$$\large y=Jr^{-\frac{1}{2}}$$ $$\large y'=J'r^{-\frac{1}{2}}-\frac{1}{2}Jr^{-\frac{3}{2}}$$ $$\large y''=J''r^{-\frac{1}{2}}-J'r^{-\frac{3}{2}}+\frac{3}{4}Jr^{-\frac{5}{2}}$$And then substitute these into your first equation.

anonymous · Answer

Do you understand the differentiation? Are you comfortable with this method; do you not need explanation?

anonymous · Answer

Yes

anonymous · Answer

To further simplify calculations, I would write $1-\frac{l(l+1)}{r^2}=q$.

anonymous · Answer

You should check my calculations yourself.

The rest is pretty simple. Can you figure it out on your own?

anonymous · Answer

Yes, give me a minute, I'll go through everything this way. I'll post results

anonymous · Answer

This worked alot better than what I was doing. I got:$$J''+\frac{1}{r}J'+(1+\frac{\frac{1}{4}+l(l+1)}{r^2})J=0$$So I just made a sign mistake somewhere.