In the range $0 \leq r < \infty$, find the solution of the equation $$\frac{d^{2}u}{dr^{2}} + \frac{2}{r} \frac{du}{dr} - \frac{n(n+1)}{r^{2}} u = a \delta(r-R),$$ where $n$ is an integer and $0<R<\infty$ is a constant, which is regular at $r=0$ and at infinity and vanishes for $a=0$.
I started by finding the homogeneous solution $u_{h} = r^{n}, r^{-n-1}$ (by plugging in the ansatz $u_{h} = r^{\alpha}$). How to proceed from here?
$$\frac{d^{2}u}{dr^{2}} + \frac{2}{r} \frac{du}{dr} - \frac{n(n+1)}{r^{2}} u = a \delta(r-R)$$ Let $u(r)=r^n f(r)$ then $u'=n r^{n-1}f +r^n f'$ and $u''=n(n-1)r^{n-2}f+2n r^{n-1}f'+r^n f''$ $$ n(n-1)r^{n-2}f+2n r^{n-1}f'+r^n f''+ 2(n r^{n-2}f +r^{n-1} f') - n(n+1)r^{n-2} f = a \delta(r-R)$$ $$r^n f''+2(n+1) r^{n-1}f' = a \delta(r-R)$$ $$\frac{d}{dr}(r^{2(n+1)}f')=r^{2(n+1)}f''+2(n+1) r^{2n+1}f'= a \delta(r-R)r^{n+2} $$ Supposing that $\delta$ is the Dirac delta function : $$r^{2(n+1)}f'=a\int \delta(r-R)r^{n+2}dr = a R^{n+2}H(r-R)+constant$$ H is the Heaviside step function. We are looking for only one particular solution. Then, no need for any constant. $$f'=a R^{n+2}r^{-2(n+1)}H(r-R)$$ $$f=a R^{n+2}\int r^{-2(n+1)}H(r-R)dr=a R^{n+2}\frac{r^{-2n-1}-R^{-2n-1}}{-2n-1}H(r-R)$$ A particular solution is $r^nf(r)$ $$u(r)=c_1 r^n+c_2 r^{-n-1}+a R^{n+2}r^n\frac{r^{-2n-1}-R^{-2n-1}}{-2n-1}H(r-R)$$ $$u(r)=c_1 r^n+c_2 r^{-n-1}+a R\frac{(\frac{r}{R})^n-(\frac{r}{R})^{-n-1}}{2n+1}H(r-R)$$