Solution to ODE Abel Equation I aim to find the exact form solution to the this ODE:
$$\frac{dS}{dw}S = \frac{a}{w}S^2 + \frac{b}{w}S - c$$
where S is a continuous differentiable function of w, real positive and a, b, c are positive, non zero, real values.  
I follow the procedure in:
Panayotounakos, D. E. and Zarmpoutis, T. I. (2011). Construction of Exact Parametric or
Closed Form Solutions of Some Unsolvable Classes of Nonlinear ODEs (Abel's Nonlinear
ODEs of the First Kind and Relative Degenerate Equations). International Journal of
Mathematics and Mathematical Sciences, 2011.
In particular I move from eq. 4.3 in the paper. I obtain a particular solution of the form:
$$S(w) = K w^{-1/2} $$
where K is a combination of the parameters a, b, c. But this form is not a solution to the ODE I started from. 
Could you please help me to find the solution? Where am I wrong?
Thanks.
 A: Approach $1$:
Let $S=w^aT$ ,
Then $\dfrac{dS}{dw}=w^a\dfrac{dT}{dw}+aw^{a-1}T$
$\therefore w^aT\left(w^a\dfrac{dT}{dw}+aw^{a-1}T\right)=\dfrac{aw^{2a}T^2}{w}+\dfrac{bw^aT}{w}-c$
$w^{2a}T\dfrac{dT}{dw}+aw^{2a-1}T^2=aw^{2a-1}T^2+bw^{a-1}T-c$
$w^{2a}T\dfrac{dT}{dw}=bw^{a-1}T-c$
$T\dfrac{dT}{dw}=bw^{-a-1}T-cw^{-2a}$
Let $x=-\dfrac{bw^{-a}}{a}$ ,
Then $w=\dfrac{(-1)^\frac{1}{a}b^\frac{1}{a}}{a^\frac{1}{a}x^\frac{1}{a}}$
$\dfrac{dT}{dw}=\dfrac{dT}{dx}\dfrac{dx}{dw}=bw^{-a-1}\dfrac{dT}{dx}$
$\therefore bw^{-a-1}T\dfrac{dT}{dx}=bw^{-a-1}T-cw^{-2a}$
$T\dfrac{dT}{dx}=T-\dfrac{cw^{1-a}}{b}$
$T\dfrac{dT}{dx}-T=\dfrac{(-1)^\frac{1}{a}b^{\frac{1}{a}-2}c}{a^{\frac{1}{a}-1}x^{\frac{1}{a}-1}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
According to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf, the ODE has simpler form of the general solution when $1-\dfrac{1}{a}=0,-1,-2,-\dfrac{1}{2}$ , i.e. when $a=1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{3}$
Approach $2$:
$S\dfrac{dS}{dw}=\dfrac{a}{w}S^2+\dfrac{b}{w}S-c$
$S\dfrac{dS}{dw}=\dfrac{aS^2+bS-cw}{w}$
$(aS^2+bS-cw)\dfrac{dw}{dS}=Sw$
Let $X=aS^2+bS-cw$ ,
Then $w=\dfrac{aS^2+bS-X}{c}$
$\dfrac{dw}{dS}=\dfrac{2aS+b}{c}-\dfrac{1}{c}\dfrac{dX}{dS}$
$\therefore X\left(\dfrac{2aS+b}{c}-\dfrac{1}{c}\dfrac{dX}{dS}\right)=S\dfrac{aS^2+bS-X}{c}$
$\dfrac{(2aS+b)X}{c}-\dfrac{X}{c}\dfrac{dX}{dS}=\dfrac{aS^3+bS^2-SX}{c}$
$\dfrac{X}{c}\dfrac{dX}{dS}=\dfrac{((2a+1)S+b)X-aS^3-bS^2}{c}$
$X\dfrac{dX}{dS}=((2a+1)S+b)X-aS^3-bS^2$
Assume $a\neq-\dfrac{1}{2}$ , as the case $a=-\dfrac{1}{2}$ had already been solved in Approach $1$ :
Let $R=S+\dfrac{b}{2a+1}$ ,
Then $X\dfrac{dX}{dR}=RX-a\left(R-\dfrac{b}{2a+1}\right)^3-b\left(R-\dfrac{b}{2a+1}\right)^2$
A: This is just a special case solution!!
Using a substitution of the form $S = w^{\alpha}V$, we can transform the equation to a form of
$$
\alpha w^{2\alpha-1}V^{2} +w^{2\alpha}VV' = aw^{2\alpha-1}V^{2} + bw^{\alpha-1}V - C
$$
Clearly setting $\alpha = a$ is the easiest starting point. This reduces the equation to
$$
w^{2\alpha}VV' = bw^{\alpha-1}V - C
$$
Re-arranging leads to
$$
VV' = bw^{-1-\alpha}V - Cw^{-2\alpha}
$$
The next restriction is $\alpha=1$ and then the equation becomes separable and has a solution of the form
$$
\left(b\frac{S}{w}-C\right) +C\ln\left(b\frac{S}{w}-C\right) = -\frac{b^{3}}{w} + \lambda
$$
