differential equation nondevelopable I try to solve this differential equation whose solution seems not to be constructable in power series
$y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers.
If one can help me please?
 A: The differential equation
$$y''+ \left(x+ \frac{a}{x^2}+b \right)y=0$$
has a regular singular point at $x=0$. In such a case, it is not always possible to construct a power series solution. However, it is always possible to find a solution of the form
$$y = x^\alpha p(x)$$
with $\alpha \in \mathbb{C}$ and $p(x) = \sum_{n=0} p_n x^n$. 
A: Hint:
$y''+\left(x+\dfrac{a}{x^2}+b\right)y=0$
$x^2y''+(x^3+bx^2+a)y=0$
Let $y=x^ku$ ,
Then $y'=x^ku'+kx^{k-1}u$
$y''=x^ku''+kx^{k-1}u'+kx^{k-1}u'+k(k-1)x^{k-2}u=x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u$
$\therefore x^2(x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u)+(x^3+bx^2+a)x^ku=0$
$x^{k+2}u''+2kx^{k+1}u'+(x^3+bx^2+k(k-1)+a)x^ku=0$
$x^2u''+2kxu'+(x^3+bx^2+k(k-1)+a)u=0$
Choose $k(k-1)+a=0$ , i.e. $k=\dfrac{1\pm\sqrt{1-4a}}{2}$ , the ODE becomes
$x^2u''+(1\pm\sqrt{1-4a})xu'+(x^3+bx^2)u=0$
$xu''+(1\pm\sqrt{1-4a})u'+(x^2+bx)u=0$
$4x\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}+(2\pm4\sqrt{1-4a})\dfrac{du}{dx}+(4x^2+4bx)u=0$
Let $x=r^2$ ,
Then $\dfrac{du}{dr}=\dfrac{du}{dx}\dfrac{dx}{dr}=2r\dfrac{du}{dx}$
$\dfrac{d^2u}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{du}{dx}\right)=2r\dfrac{d}{dr}\left(\dfrac{du}{dx}\right)+2\dfrac{du}{dx}=2r\dfrac{d}{dx}\left(\dfrac{du}{dx}\right)\dfrac{dx}{dr}+2\dfrac{du}{dx}=2r\dfrac{d^2u}{dx^2}2r+2\dfrac{du}{dx}=4r^2\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}=4x\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}$
$\therefore\dfrac{d^2u}{dr^2}+\dfrac{1\pm2\sqrt{1-4a}}{r}\dfrac{du}{dr}+(4r^4+4br^2)u=0$
Let $u=e^{mr^3}v$ ,
Then $\dfrac{du}{dr}=e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}v$
$\dfrac{d^2u}{dr^2}=e^{mr^3}\dfrac{d^2v}{dr^2}+3mr^2e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v=e^{mr^3}\dfrac{d^2v}{dr^2}+6mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v$
$\therefore e^{mr^3}\dfrac{d^2v}{dr^2}+6mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v+\dfrac{1\pm2\sqrt{1-4a}}{r}\left(e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}v\right)+(4r^4+4br^2)e^{mr^3}v=0$
$\dfrac{d^2v}{dr^2}+\left(6mr^2+\dfrac{1\pm2\sqrt{1-4a}}{r}\right)\dfrac{dv}{dr}+((9m^2+4)r^4+4br^2+3mr(3\pm2\sqrt{1-4a}))v=0$
Choose $9m^2+4=0$ , i.e. $m=\pm\dfrac{2i}{3}$ , the ODE becomes
$\dfrac{d^2v}{dr^2}+\left(\pm4ir^2+\dfrac{1\pm2\sqrt{1-4a}}{r}\right)\dfrac{dv}{dr}+(4br^2\pm2ir(3\pm2\sqrt{1-4a}))v=0$
A: Another way of solving this equation will be a series expansion in the parameter $b$. We know that for $b=0$ the two independent solutions can be easily found by means of the series expansion method.Those solutions are related to Bessel functions. Let us therefore assume that the whole solution reads:
\begin{equation}
y(x) = \sum\limits_{j=0}^\infty b^j y^{(j)}(x)
\end{equation}
We call the $j=0$ function the basic solution and the $j>0$ functions the corrections of order $j$.
Inserting the ansatz into the ODE we get:
\begin{equation}
y^{(0)}_\pm(x) = \sqrt{x} J_{\pm \frac{1}{3} \sqrt{1-4 a}} \left(\frac{2}{3} x^{\frac{3}{2}}\right) =
\frac{x^\alpha}{{\mathcal A}_\pm} F_{0,1}[1\pm \frac{1}{3} \sqrt{1-4 a};-\frac{x^3}{9}]
\end{equation}
where $\alpha=1/2(1\pm \sqrt{1-4 a})$ and ${\mathcal A}_\pm = 3^{\frac{\pm}{3} \sqrt{1-4 a} } \left(\frac{\pm}{3} \sqrt{1-4 a} \right)!$.
We have:
\begin{equation}
\left[\frac{d^2}{d x^2} + (x + \frac{a}{x^2})\right] y^{(j)}(x) = -y^{(j-1)}(x)
\end{equation}
for $j=1,2,\dots$. The above equation can be solved by means of Greens functions. The solution reads:
\begin{equation}
y^{(j)}(x) = \int\limits_0^x \left(\frac{y^{(0)}_-(x) y^{(0)}_+(\xi) - y^{(0)}_+(x) y^{(0)}_-(\xi)}{{\mathcal W}\left[y^{(0)}_+,y^{(0)}_-\right](\xi)}\right) \cdot (-) y^{(j-1)}(\xi) d \xi 
\end{equation}
Here ${\mathcal W}[y^{(0)}_+,y^{(0)}_-]$is the Wronskian. It is a constant as a function of $\xi$ and it reads:
\begin{equation}
{\mathcal W}[y^{(0)}_+,y^{(0)}_-](\xi) = - \frac{\sin\left[\frac{\pi}{3} \sqrt{1- 4 a}\right]}{\frac{\pi}{3}}
\end{equation}
Writing the above equation in a compact way we get:
\begin{equation}
y^{(J)}(x) = \frac{1}{{\mathcal W}^J} \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot y^{(0)}(\xi) d \xi
\end{equation}
where 
\begin{equation}
{\mathcal K}^{(J)}(x,\xi) := \int\limits_\xi^x {\mathcal K}^{(J-1)}(x,\eta) \cdot {\mathcal K}(\eta,\xi) d \eta
\end{equation}
for $J=2,3,\dots$ and
\begin{equation}
{\mathcal K}^{(1)}(x,\xi) := \left| \begin{array}{rr} y_+^{(0)}(x) & y_-^{(0)}(x) \\  y_+^{(0)}(\xi) & y_-^{(0)}(\xi) \end{array} \right|
\end{equation}
Now, in order to construct the higher order corrections we compute a following quantity(to be termed moments):
\begin{equation}
{\mathcal M}_{J,l}(x) := \int\limits_0^x {\mathcal K}^{(J)}(x,\xi) \cdot \xi^{l+\alpha} d \xi
\end{equation}
where $J=1,2,\dots$ and $l=0,1,2,\dots$. Integrating by parts twice we obtain a following  recursion relation for the moments:
\begin{equation}
(l+2)(l+2\pm \sqrt{1-4 a}){\mathcal M}_{J,l} + {\mathcal M}_{J,l+3} = -{\mathcal W}\cdot\left({\mathcal M}_{J-1,l+2} 1_{J>1} + \delta_{J,1} x^{l+2+\alpha}\right)
\end{equation}
for $J=1,2,\dots$. The solutions to the recursion relations are pretty straightforward and read:
\begin{eqnarray}
{\mathcal M}_{1,l} &=& {\mathcal W} \cdot x^{l+\alpha-1} \cdot \left\{ F_{1,2}\left[\begin{array}{rr} 1 \\
\frac{l+2}{3} & \frac{l+2\pm\sqrt{1-4 a}}{3}\end{array};-\frac{x^3}{3^2}\right]-1\right\} \\
{\mathcal M}_{J,l} &=& {\mathcal W} \sum\limits_{j=0}^\infty  \left(\frac{-1}{3^2}\right)^{j+1} \cdot \frac{{\mathcal M}_{J-1,l+3 j+2}}{\left(\frac{l+2}{3}\right)^{(j+1)} \left(\frac{l+2\pm \sqrt{1-4 a}}{3}\right)^{(j+1)}}
\end{eqnarray}
for $J>1$. Using the expressions for the moments along with the expression for the correction of order $j$ we obtain the first order correction:
\begin{eqnarray}
&& y_\pm^{(1)}(x) = 
\frac{x^{\alpha+2}}{(-9)^1 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^2 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)! 
\sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{2}{3}+j)!(\frac{2\pm\sqrt{1-4 a}}{3}+j)!}\\
&&\sum\limits_{j_1=0}^j \binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\end{eqnarray}
Likewise the second order correction reads:
\begin{eqnarray}
 &&y_\pm^{(2)}(x) = \frac{x^{\alpha+4}}{(-9)^2 {\mathcal A}_\pm} \left[(-\frac{1}{3})!\right]^4 \left(\pm \frac{\sqrt{1-4 a}}{3}\right)! 
\sum\limits_{j=0}^\infty \frac{(-\frac{x^3}{9})^j)}{(\frac{4}{3}+j)!(\frac{4\pm\sqrt{1-4 a}}{3}+j)!}\\
&&\sum\limits_{0\le j_1 \le j_2 \le j}^j 
\binom{j_1-\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_1-\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\binom{j_2+\frac{1}{3}}{-\frac{1}{3}} \cdot \binom{j_2+\frac{1}{3} \pm \frac{\sqrt{1-4 a}}{3}}{-\frac{1}{3}}
\end{eqnarray}
It is now easy to see what is the pattern for all higher order corrections. As such the problem is in principle solved. It would be nice to reduce the multiple sums to single sums though.
