Solving a first order nonlinear nonhomogeneous ODE Edit: from the answers, I have learnt that the differential equation can be solved by expressing it as being a hypergeometric differential equation. My question now is that, how many a function in the form of
$x(1-x)^2y''(x)+(1-x)^2y'(x)+ay(x)=0$
be transformed into the form of
$\eta(1-\eta)f''(\eta)+(b-c\eta)f'(\eta)+df(\eta)=0$?
Original question: How may one solve a differential equation in the form of:
$\frac{dy}{dx}=P(x) -ky^2$
I have attempted at reducing it into a second order homogeneous equation in the form of
$\frac{d^2u}{dx^2}=-kP(t)u$
by making the substitution of
$ky= \frac{\frac{du}{dx}}{u}$
However, I am still unable to solve this.
Are there any methods for solving either equation?
If it helps, P(x) is the derivative of:
$f(x)=\frac{a-be^{cx+d}}{1-e^{cx+d}}$ where a,b,c,d are constants
Additionally,
y=0 when x=0, and y=0 as x$\rightarrow$ infinity
A numerical approach to solving the equation with randomly chosen values for constants substituted in gives the following graph:
Link
which looks like (maybe) a chi-square distribution....?
 A: The equation you are working with is the Ricatti equation:
$$y'+ky^{2}=P(x)$$
Where
$$P(x)=\frac{d}{dx}\frac{a-be^{cx+d}}{1-e^{cx+d}}=\frac{c(a-b)e^{cx+d}}{(1-e^{cx+d})^{2}}$$
By doing your substitution 
$$y(x)=\frac{1}{k}\frac{\frac{du(x)}{dx}}{u(x)}$$
you indeed have the second order linear ode
$$\frac{d^{2}u}{dx^{2}}=kP(x)u(x)=k\frac{c(a-b)e^{cx+d}}{(1-e^{cx+d})^{2}}u(x)$$
First it is nice to clean the equation a little bit by letting $z=c{x}+d$ and $k(a-b)/c=\alpha$, then
$$\frac{d^{2}u(z)}{dz^{2}}=\alpha\frac{e^{z}}{(1-e^{z})^{2}}u(z)$$
Then you may also want a change of variables
$$\xi=e^{z}$$
So that
$$\xi(1-\xi)^{2}\frac{d^{2}u(\xi)}{d\xi^{2}}+(1-\xi)^{2}\frac{du(\xi)}{d\xi}-\alpha{u}(\xi)=0$$
Then you let $1-\sqrt{4\alpha+1}=\gamma$ and do the following substitutions
$$\sigma=(x-1)$$
$$u(\xi)=\sigma^{\frac{1-\gamma}{2}}f(\sigma)$$
To give the Gauss hypergeometric ode
$$\sigma(1-\sigma)\frac{d^{2}}{d\sigma^{2}}f(\sigma)+(\gamma+(1-\gamma)\sigma)\frac{d}{d\sigma}f(\sigma)-\frac{1}{4}\gamma^{2}f(\sigma)=0$$
