Second Order Differential Equation inhomogeneous $$y''(s) - \frac{s^2}{c^2}y(s) - \frac{g}{s\cdot c^2}= 0$$
I am getting confused with what to do with the 
$$\frac{-g}{s\cdot c^2}$$ 
part
 A: $$y''(s) - \frac{s^2}{c^2}y(s) - \frac{g}{s\cdot c^2}= 0$$
Let $s=\sqrt{\frac{2}{c}}x$
$$y''(x) - \frac{x^2}{4}y(x) - \sqrt{\frac{2}{c}}\frac{g}{x\cdot c}= 0$$
Let $a=\frac{g}{c}\sqrt{\frac{2}{c}}$ 
$$y''(x) - \frac{x^2}{4}y(x) = \frac{a}{x}$$
$Y''(x) - \frac{x^2}{4}Y(x) = 0$ is a parabolic cylinder kind of ODE. The solutions are :
$$Y(x)=c_1 D_{-1/2}(x)+c_2 D_{-1/2}(i\space x)$$
where $D$ is the symbol of the parabolic cylinder function : http://mathworld.wolfram.com/ParabolicCylinderFunction.html
$y(x)=Y(x)+y_p(x)$ where $y_p(x)$ is any particular solution of the ODE.
Let $y_0(x)=f(x)D_{-1/2}(x)=f\space D$
In interest of space, we simplify the notation $D_{-1/2}(x)=D(x)$
$y_0''=f''D+2f'D'+fD''$ together with $D'' - \frac{x^2}{4}D =0$ lead to :
$y_0''(x) - \frac{x^2}{4}y_0(x) =f''D+2f'D'+fD''- \frac{x^2}{4}f\space D =f''D+2f'D' = \frac{a}{x}$
$$f''D+2f'D' = \frac{a}{x}$$
The usual method of solving of this first order linear ODE leads to :
$$f(x)=a\int{\frac{1}{D^2}\left(\int{\frac{D}{x}dx}\right)dx}$$
Since we are looking for only one particular solution, we can forget the integration constants.
$$y(x)=c_1 D_{-1/2}(x)+c_2 D_{-1/2}(i\space x)+a\space D_{-1/2}(x)\int{\frac{1}{D_{-1/2}^2(x)}\left(\int{\frac{D_{-1/2}(x)}{x}dx}\right)dx}$$
Then, comme back to the variable $s$ with $x=\sqrt{\frac{c}{2}}s$ and with $a=\frac{g}{c}\sqrt{\frac{2}{c}}$
Of course, the integrals involving the parabolic cylinder functions are not simple closed forms. Note that the parabolic cylinder functions can be expressed in terms of Bessel functions, but leading to even biger formulas. 
