Find the Particular solution to the Cauchy problem (non-linear $1$st-order ODE variable coefficients) let $y(x)$ be a continuous function $0\le y(x) \le1$ over the support $x_\min\le x \le x_\max$, for $x_\min < x_\max$ strictly positive values. 
let $y'(x)$ be the first order derivative of $y(x)$.
let $y(x)^k$ be the $k$-th power of the function $y(x)$.
Solve the Cauchy problem:
$$a(x)y'(x) = b(x)y(x)^{-3} - cy(x)^{-2} - y(x)^{-1}$$
$$y(x_0) = y_0$$
where $a(x)=a_1 + a_2x$ with $a_1>0, a_2<0$ and $b(x)=b_1 + b_2x$ with $b_1<0, b_2>0$ 
and $x_0$, $y_0$, $c$ are positive constant real values. 
I am interested in a solution such that $y'(x)<0$. I proved existence and uniqueness, but I am looking for the analytic expression of $y(x)$. 
 A: My academic interest are PDEs and ODEs. Personally, I'd like to say that there would be explicit solutions for such nonlinear ODE in the first place.
A: $(a_1+a_2x)y'(x)=(b_1+b_2x)y(x)^{-3}-cy(x)^{-2}-y(x)^{-1}$
$(a_1+a_2x)\dfrac{dy}{dx}=\dfrac{b_1+b_2x}{y^3}-\dfrac{c}{y^2}-\dfrac{1}{y}$
$\left(\dfrac{b_2x}{y^3}-\dfrac{1}{y}-\dfrac{c}{y^2}+\dfrac{b_1}{y^3}\right)\dfrac{dx}{dy}=a_2x+a_1$
$\left(x-\dfrac{y^2}{b_2}-\dfrac{cy}{b_2}+\dfrac{b_1}{b_2}\right)\dfrac{dx}{dy}=\dfrac{a_2y^3x}{b_2}+\dfrac{a_1y^3}{b_2}$
This belongs to an Abel equation of the second kind.
Let $u=x-\dfrac{y^2}{b_2}-\dfrac{cy}{b_2}+\dfrac{b_1}{b_2}$ ,
Then $x=u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{2y}{b_2}+\dfrac{c}{b_2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{2y}{b_2}+\dfrac{c}{b_2}\right)=\dfrac{a_2y^3}{b_2}\left(u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}\right)+\dfrac{a_1y^3}{b_2}$
$u\dfrac{du}{dy}+\dfrac{(2y+c)u}{b_2}=\dfrac{a_2y^3u}{b_2}+\dfrac{a_2y^5}{b_2^2}+\dfrac{a_2cy^4}{b_2^2}-\dfrac{a_2b_1y^3}{b_2^2}+\dfrac{a_1y^3}{b_2}$
$u\dfrac{du}{dy}=\dfrac{(a_2y^3-2y-c)u}{b_2}+\dfrac{a_2y^5}{b_2^2}+\dfrac{a_2cy^4}{b_2^2}+\dfrac{(a_1b_2-a_2b_1)y^3}{b_2^2}$
