I am trying to thoroughly solve the infinite spherical well potential problem that is introduced in Griffith's Introduction to QM, Chapter 4. To solve the Radial part of the equation, one must solve the Bessel equation, which I know is:

$$\frac{d^2Y}{dx^2}+x\frac{dY}{dx}+(x^2-\nu^2)Y=0$$

To arrive at this, I begin manipulating the Radial equation: $$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)-\frac{2mr^2}{\hbar^2}(V(r)-E)R=l(l+1)R$$ Upon applying the derivative and using that inside the spherical well the potential is zero: $$\frac{d^2R}{dr^2}+2r\frac{dR}{dr}+(k^2r^2-l(l+1))R=0$$ If I use a substitution such as $x=kr$, and call $\nu^2={l(l+1)}$ I can arrive at: $$\frac{d^2R}{dx^2}+2x\frac{dR}{dx}+(x^2-\nu^2)R=0$$ but this is not the Bessel Equation! Alternatively, I could have used the substitution $u(r)=rR(r)$, which yields: $$r^2\frac{d^2u}{dr^2}+(k^2r^2-l(l+1))u=0$$ Even after using the same substitution as begore, I only get:

$$x^2\frac{d^2Y}{dx^2}+(x^2-\nu^2)Y=0$$ However, this is not the Bessel equation either. My question is: How can I get to the Bessel equation from either of these? Or is that not possible? In griffith's book, he states that the $u=rR$ form of the radial equation is solved by the spherical Bessel functions, so it should be the Bessel function. And if I can indeed get to the Bessel equation, I would face another problem: The solution for the Bessel equations that agree with Griffith's are those when $\nu=\pm 1/2$, but since $\nu^2=l(l+1)$, I don't see how we could ever get to that result

I am trying to thoroughly solve the infinite spherical well potential problem that is introduced in Griffith's Introduction to QM, Chapter 4. To solve the Radial part of the equation, one must solve the Bessel equation, which I know is:

$$\frac{d^2Y}{dx^2}+x\frac{dY}{dx}+(x^2-\nu^2)Y=0$$

To arrive at this, I begin manipulating the Radial equation: $$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)-\frac{2mr^2}{\hbar^2}(V(r)-E)R=l(l+1)R$$ Upon applying the derivative and using that inside the spherical well the potential is zero: $$\frac{d^2R}{dr^2}+2r\frac{dR}{dr}+(k^2r^2-l(l+1))R=0$$ If I use a substitution such as $x=kr$, and call $\nu^2={l(l+1)}$ I can arrive at: $$\frac{d^2R}{dx^2}+2x\frac{dR}{dx}+(x^2-\nu^2)R=0$$ but this is not the Bessel Equation! Alternatively, I could have used the substitution $u(r)=rR(r)$, which yields: $$r^2\frac{d^2u}{dr^2}+(k^2r^2-l(l+1))u=0$$ Even after using the same substitution as begore, I only get:

$$x^2\frac{d^2Y}{dx^2}+(x^2-\nu^2)Y=0$$ However, this is not the Bessel equation either. My question is: How can I get to the Bessel equation from either of these? Or is that not possible? In griffith's book, he states that the $u=rR$ form of the radial equation is solved by the spherical Bessel functions, so it should be the Bessel function. And if I can indeed get to the Bessel equation, I would face another problem: The solution for the Bessel equations that agree with Griffith's are those when $\nu=\pm 1/2$, but since $\nu^2=l(l+1)$, I don't see how we could ever get to that result