I'm currently trying to find the solutions to this second order ODE:
R'' + R'/r + c*R = 0, where R(r) and c is an arbitary constant.
I found this page https://physics.stackexchange.com/questions/9231/heat-equation-on-ball-one-dimensional-description where they had already solved this equation. I'm pretty sure he had his dr should've been d/dr but whatever. I am currently at a complete loss as to how he got to that equation.
$$\frac{d^2R}{dr^2}+\frac{1}{r}\frac{dR}{dr}+cR=0 \tag{1}$$ Case $c>0$ :
Change of variable $x=\sqrt{c}\:r \quad\to\quad \frac{d^2R}{dx^2}+\frac{1}{x}\frac{dR}{dx}+R=0$
This is the standard Bessel differential equation which general solution is : $$R(x)=c_1J_0(x)+c_2Y_0(x)$$ Thus the solution of equation (1) is : $$R(r)=c_1J_0(\sqrt{c}\:r)+c_2Y_0(\sqrt{c}\:r) $$ Bessel functions of first and second kind.
Case $c=0$ : $$R(r)=c_1\ln|r|+c_2$$
Case $c<0$ :
Do the same with change of variable $x=\sqrt{-c}\:r$ which leads to $$R(r)=c_1I_0(\sqrt{-c}\:r)+c_2K_0(\sqrt{-c}\:r) $$ Modified Bessel functions of first and second kind.