We subtract \(k^{2}y\) from both sides of equation (1.1) so that the differential equation is converted to

$$ y^{\prime\prime}(x)+ny^{\prime}(x)+\frac{m}{x}y^{\prime }(x)-k^{2}y(x)=f \bigl(x,y(x)\bigr)-k^{2}y(x). $$

(2.1)

Let us now consider the homogeneous type of the above equation, i.e.,

$$ \omega^{\prime\prime}(x)+n\omega^{\prime}(x)+\frac{m}{x} \omega^{\prime }(x)-k^{2}\omega(x)=0,\quad m>0, n\in\mathbb{R}, $$

(2.2)

with two homogeneous conditions

$$ \omega^{\prime}(0)=0, \qquad A\omega(1)+B\omega^{\prime}(1)=0. $$

(2.3)

Suppose that \(u_{m}(x)\) and \(v_{m}(x)\) are given as follows:

$$\begin{aligned} & u_{m}(x)=e^{\frac{1}{2} x (-\sqrt{4 k^{2}+n^{2}}-n )} L_{\alpha }^{m-1} \bigl(x \sqrt{4 k^{2}+n^{2}} \bigr),\qquad \alpha=-\frac{m (\sqrt{4 k^{2}+n^{2}}+n )}{2 \sqrt{4 k^{2}+n^{2}}}, \end{aligned}$$

(2.4)

$$\begin{aligned} & v_{m}(x)=e^{\frac{1}{2} x (-\sqrt{4 k^{2}+n^{2}}-n )} U \bigl(\beta,m,\sqrt{4 k^{2}+n^{2}} x \bigr),\qquad \beta=\frac{m (n+\sqrt{4 k^{2}+n^{2}} )}{2 \sqrt{4 k^{2}+n^{2}}}, \end{aligned}$$

(2.5)

where \(L_{\alpha}^{m-1}\) is the generalized Laguerre polynomial which is related to hydrogen atom wave functions in quantum mechanics. Further, \(U(a,b,z)\) is the hypergeometric function which is a second linearly independent solution to Kummer’s equation and is defined by

$$ U(a,b,z)=\frac{1}{\Gamma(a)} \int_{0}^{\infty}t^{a-1} (t+1)^{b-a-1} e^{-tz}\,\mathrm{d}t. $$

(2.6)

Now, we define

$$\begin{aligned} & u(x)=u_{m}(x), \end{aligned}$$

(2.7)

$$\begin{aligned} & v(x)=v_{m}(x)-Su_{m}(x),\qquad S=\frac{Av_{m}(1)+Bv_{m}^{\prime}(1)}{Au_{m}(1)+Bu_{m}^{\prime}(1)}. \end{aligned}$$

(2.8)

It can be easily seen that \(u(x)\) and \(v(x)\) are the solutions of Eq. (2.2) so that \(u^{\prime}(0)=u_{m}^{\prime}(0)=0\) and \(Av(1)+Bv^{\prime}(1)=0\) hold, respectively. We will give some properties of these functions in the next section.

Assume that \(\varphi(x)\) satisfies

$$\begin{aligned} &\varphi^{\prime\prime}(x)+n\varphi^{\prime}(x)+\frac{m}{x}\varphi ^{\prime}(x)-k^{2}\varphi(x)=0, \end{aligned}$$

(2.9)

$$\begin{aligned} &\varphi^{\prime}(0)=0,\qquad A\varphi(1)+B\varphi^{\prime}(1)=C. \end{aligned}$$

(2.10)

In fact, \(\varphi(x)\) can be expressed in terms of \(u(x)\) as follows:

$$ \varphi(x)=\frac{Cu(x)}{Au(1)+Bu^{\prime}(1)}. $$

(2.11)

Now, consider differential equation (2.1) with the homogeneous boundary conditions

$$ y^{\prime}(0)=0,\qquad Ay(1)+By^{\prime}(1)=0. $$

(2.12)

Then, for \(k^{2}\neq0\), Eq. (2.1) can be converted to an equivalent integral equation by means of the Green’s function appropriate to the operator on the left-hand side as follows:

$$ y(x)=\varphi(x)+ \int_{0}^{1}G(x,t)\bigl[k^{2}y(t)-f \bigl(t,y(t)\bigr)\bigr]\,\mathrm{d}t, $$

(2.13)

where \(G(x,t)\) satisfies

$$\begin{aligned} &G_{xx}+nG_{x}+\frac{m}{x}G_{x}-k^{2}G=- \delta(x-t), \end{aligned}$$

(2.14)

$$\begin{aligned} &G_{x}(0,t)=0,\qquad AG(1,t)+BG_{x}(1,t)=0, \end{aligned}$$

(2.15)

where \(\delta(x)\) is the Dirac delta function. It can be easily discovered from the elementary theory of differential equations that \(G(x,t)\) may be expressed as

$$ G(x,t)=\frac{1}{W(t)} \textstyle\begin{cases} u(x)v(t), & \mbox{$x\leq t$}, \\ v(x)u(t), & \mbox{$x\geq t$,} \end{cases} $$

(2.16)

where \(u(x)\) and \(v(x)\) are given by Eqs. (2.7)-(2.8) and the Wronskian \(W(t)\) is defined by

$$\begin{aligned} W(t)&\equiv v(t)u^{\prime}(t)-u(t)v^{\prime}(t) \\ &=v_{m}(t)u_{m}^{\prime}(t)-u_{m}(t)v_{m}^{\prime}(t) \\ &=\frac{1}{2} e^{-t (\sqrt{4 k^{2}+n^{2}}+n )} \begin{pmatrix} m (\sqrt{4 k^{2}+n^{2}}+n ) U (\frac{n m}{2\sqrt{4 k^{2}+n^{2}}}+\frac{m}{2}+1,m+1,\sqrt{4 k^{2}+n^{2}} t )\\ {}\times L_{\frac {1}{2} m (-\frac{n}{\sqrt{4 k^{2}+n^{2}}}-1 )}^{m-1} (t \sqrt {4 k^{2}+n^{2}} )\\ -2 \sqrt{4 k^{2}+n^{2}} U (\frac{1}{2} m (\frac{n}{\sqrt{4 k^{2}+n^{2}}}+1 ),m,\sqrt{4 k^{2}+n^{2}} t )\\ {}\times L_{-\frac{m n}{2 \sqrt{4 k^{2}+n^{2}}}-\frac{m}{2}-1}^{m} (t \sqrt{4 k^{2}+n^{2}} ) \end{pmatrix} \\ &= W(1)e^{n-nt}t^{-m}, \end{aligned}$$

(2.17)

where, obviously, \(W(1)\) depends on \(n,m\) and *k*.