Are there any explicit solutions known for this ODE?

Question

I am considering the following ODE, $$u''+\frac{5}{r}u'+K\left(\frac{2}{1+4r^2}\right)^2u=0$$ where $K>0$ and $u=u(r):\mathbb{R}_{+}\to \mathbb{R}.$ Can we deduce that any solution $u$ to the above ODE decays as follows $$u(r)\leq \frac{C}{r^{\alpha}}?$$

Answer 1

$$u''+\frac{5}{r}u'+K\left(\frac{2}{1+4r^2}\right)^2u=0 \tag 1$$ $x=1+4r^2\quad\implies\begin{cases} \frac{du}{dr}=8r\frac{du}{dx}\\ \frac{d^2u}{dr^2}=16(x-1)\frac{d^2u}{dx^2}+8\frac{du}{dx} \end{cases}$ $$16(x-1)\frac{d^2u}{dx^2}+8\frac{du}{dx}+40\frac{du}{dx}+K\left(\frac{2}{x}\right)^2u=0$$ $$(x-1)\frac{d^2u}{dx^2}+3\frac{du}{dx}+\frac{K}{4x^2}u=0 \tag 2$$ This is an ODE of the hypergeometric kind. In order to reduce it on the standard form the change of function is : $$u(x)=x^\nu y(x)\quad\begin{cases} \frac{du}{dx}=\nu x^{\nu-1}y+x^\nu\frac{dy}{dx}\\ \frac{d^2u}{dx^2}=\nu(\nu-1)x^{\nu-2}y+2\nu x^{\nu-1}\frac{dy}{dx}+x^\nu\frac{d^2y}{dx^2} \end{cases}$$ Putting them into Eq.$(2)$ and after simplification : $$x(x-1)\frac{d^2y}{dx^2}+\big((3+2\nu)x-2\nu\big)\frac{dy}{dx}+\left(\nu(2+\nu)+\frac{K+4\nu-4\nu^2}{4x} \right)y=0 \tag 3$$ With $K+4\nu-4\nu^2=0\quad\implies\quad \nu=\frac12(1\pm\sqrt{1+K})\quad$ : $$x(1-x)y''+(c-(a+b+1)x)y'-aby=0$$ $c=-2\nu\quad;\quad a=2+\nu+\sqrt{\nu(1+\nu)(2+\nu)}\quad;\quad b=2+\nu-\sqrt{\nu(1+\nu)(2+\nu)}$

See https://mathworld.wolfram.com/HypergeometricDifferentialEquation.html for solution in term of hypergeometric function $\:_2F_1(a,b;c;x)$.

Then back to $u(x)$ and $u(r)$.