Question

Solve the two differential equations in Problem $5.11$of Chapter 13

$r\frac{d}{dr}\left(r\left(\frac{dR}{dr}\right)\right)=n^{2}R$

Short answer

The general solution of the equation is$R=A{r}^n+B{r}^{-n}$

For$n\ne 0,R=A\ln r+B$

For$n=0$$R=A{r}^l+B{r}^{-(l+1)}$

Step-by-step solution

Given information

The given differential equation is$r\frac{d}{dr}\left(r\left(\frac{dR}{dr}\right)\right)=n^{2}R$ .

Differential equation

A differential equation is an equation that relates one or more unknown functions and their derivatives.

Simplify the given equation

Let's start with the first equation

$r\frac{\text{d}}{\text{d}r}\left(r\frac{\text{d}R}{\text{d}r}\right)=n^{2}R$

and rewrite it to

$r^2\frac{{\text{d}}^{2}R}{\text{d}{r}^2}+r\frac{\text{d}R}{\text{d}r}-n^{2}R=0$

Using substitution$r=e^z.$we can rewrite this further to

$$\begin{gathered} \frac{{\text{d}}^{2}R}{\text{d}{z}^2}-\frac{\text{d}R}{\text{d}z}+\frac{\text{d}R}{\text{d}z}-n^{2}R=0 \\ \frac{{\text{d}}^{2}R}{\text{d}{z}^2}-n^{2}R=0. \end{gathered}$$

To solve this we use the auxiliary equation

$D^2-n^2=0$

Which solution is $D=\pm n$. Therefore our solution is of the form

$R=A{r}^n+B{r}^{-n}$

To find the whole solution for, $n=0$we need to go back to the differential equation itself:

$\frac{{\text{d}}^{2}R}{\text{d}{z}^2}=0$

Solution for this equation is trivial and is given by the linear function

role="math" localid="1664347874005" $$\begin{gathered} R=Az+B \\ R=A\ln r+B \end{gathered}$$

The other equation can also be rewritten in a similar way

$$\begin{gathered} \frac{\text{d}R}{\text{d}r}\left(r^2\frac{\text{d}R}{\text{d}r}\right)=l(l+1)R \\ {r}^2\frac{{\text{d}}^{2}R}{U^2}+2r{r}^2\frac{\text{d}R}{1}-l(l+1)R=0. \end{gathered}$$

$R=A{e}^{nz}+B{e}^{-nz}=A{e}^{\ln {\tau }^n}+B{e}^{\ln {\tau }^{-n}}$

Again using the substitution $\left(r=e^z\right).$we get

$\frac{{\text{d}}^{2}R}{\text{d}{z}^2}+\frac{\text{d}R}{\text{d}z}-l(l+1)R=0$

We find the solution using the auxiliary equation

$D^2+D-l(l+1)=0$

Which solutions are

$$\begin{gathered} D_{1,2}=\frac{-1\pm \sqrt{1+4l(l+1)}}{2} \\ =\frac{-1\pm \sqrt{{(2l+1)}^2}}{2} \\ =\frac{-1\pm |2l+1|}{2} \end{gathered}$$

If assume the$I$to be positive we can omit the absolute value and get a solution of the form

$$\begin{gathered} R=A{e}^{lz}+B{e}^{-(l+1)z}=A{e}^{\ln r^l}+B{e}^{\ln r^{-(l+1)}} \\ R=A{r}^l+B{r}^{-(l+1)} \end{gathered}$$