Does a closed form solution to this nonlinear ODE exist? $fx(1-x)y'+(e+fx)y+\sqrt{ay-b}=dx+c$, where $y=y(x)$
EDIT, Will Jagy: $a,b,c,d,e,f$ are  real constants.
I have never solved a nonlinear ODE before, although I'm familiar with many of the techniques applied for solving linear ODEs. There is one special case of the equation above that I managed to solve, but that is the easy case where $f$ is equal to zero. For $f$ not being equal to zero I have the first derivative inside, which then makes it a linear ODE. I read that there does not have to be a closed form solution. Is there any way how I can check whether or not such a solution exists for the equation above?
 A: For $fx(1-x)y'+(e+fx)y+\sqrt{ay-b}=dx+c$ , where $a,f\neq0$ :
Approach $1$:
Let $u=\sqrt{ay-b}$ ,
Then $y=\dfrac{u^2+b}{a}$
$\dfrac{dy}{dx}=\dfrac{2u}{a}\dfrac{du}{dx}$
$\therefore fx(1-x)\dfrac{2u}{a}\dfrac{du}{dx}+(e+fx)\dfrac{u^2+b}{a}+u=dx+c$
$\dfrac{2fx(1-x)u}{a}\dfrac{du}{dx}+\dfrac{(e+fx)u^2}{a}+\dfrac{be+bfx}{a}+u=dx+c$
$\dfrac{2fx(x-1)u}{a}\dfrac{du}{dx}=\dfrac{(e+fx)u^2}{a}+u+\dfrac{(bf-ad)x+be-ac}{a}$
$u\dfrac{du}{dx}=\dfrac{(e+fx)u^2}{2fx(x-1)}+\dfrac{au}{2fx(x-1)}+\dfrac{(bf-ad)x+be-ac}{2fx(x-1)}$
This mostly belongs to an Abel equation of the second kind, unless when $bf=ad$ and $be=ac$ this ODE can reduce to a bernoulli equation.
Approach $2$:
$fx(1-x)y'+(e+fx)y+\sqrt{ay-b}=dx+c$
$(fy-d)x+ey-c+\sqrt{ay-b}=fx(x-1)\dfrac{dy}{dx}$
$((fy-d)x+ey-c+\sqrt{ay-b})\dfrac{dx}{dy}=fx^2-fx$
This also belongs to an Abel equation of the second kind.
Let $u=x+\dfrac{ey-c+\sqrt{ay-b}}{fy-d}$ ,
Then $x=u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}$
$\therefore(fy-d)u\biggl(\dfrac{du}{dy}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}\biggr)=f\biggl(u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}\biggr)^2-f\biggl(u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}\biggr)$
$(fy-d)u\dfrac{du}{dy}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}u=fu^2-\dfrac{f(f+2e)y-(2c+d)f+2f\sqrt{ay-b}}{fy-d}u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^2}+\dfrac{efy-cf+\sqrt{ay-b}}{fy-d}$
$(fy-d)u\dfrac{du}{dy}=fu^2-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^2}+\dfrac{efy-cf+\sqrt{ay-b}}{fy-d}$
$u\dfrac{du}{dy}=\dfrac{fu^2}{fy-d}-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$
Let $u=(fy-d)v$ ,
Then $\dfrac{du}{dy}=(fy-d)\dfrac{dv}{dy}+fv$
$\therefore(fy-d)v\left((fy-d)\dfrac{dv}{dy}+fv\right)=\dfrac{f(fy-d)^2v^2}{fy-d}-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}(fy-d)v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$
$(fy-d)^2v\dfrac{dv}{dy}+f(fy-d)v^2=f(fy-d)v^2-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$
$(fy-d)^2v\dfrac{dv}{dy}=-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$
$v\dfrac{dv}{dy}=-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)^3\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^5}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^4}$
A: I doubt that there are closed-form solutions in general.  Maple doesn't come up with one.  Of course if $a=0$ you have a linear equation, and that one does have a (rather complicated) closed-form solution.  If $b=c=d=e=0$ there is a closed-form solution
$\arctan \left( \sqrt {-1+x} \right) +{\frac {1}{\sqrt {-1+x}}}+{\frac 
{y \left( x \right) f}{\sqrt {-1+x}\sqrt {ay \left( x \right) }}}+C=0
$
for arbitrary constant $C$.
