Vtyjkyuk

$fx(1-x)y'+(e+fx)y+sqrtay-b=dx+c$, 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?

For $fx(1-x)y'+(e+fx)y+sqrtay-b=dx+c$ , where $a,fneq0$ :

Approach $1$:

Let $u=sqrtay-b$ ,

Then $y=dfracu^2+ba$

$dfracdydx=dfrac2uadfracdudx$

$therefore fx(1-x)dfrac2uadfracdudx+(e+fx)dfracu^2+ba+u=dx+c$

$dfrac2fx(1-x)uadfracdudx+dfrac(e+fx)u^2a+dfracbe+bfxa+u=dx+c$

$dfrac2fx(x-1)uadfracdudx=dfrac(e+fx)u^2a+u+dfrac(bf-ad)x+be-aca$

$udfracdudx=dfrac(e+fx)u^22fx(x-1)+dfracau2fx(x-1)+dfrac(bf-ad)x+be-ac2fx(x-1)$

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 $2$:

$fx(1-x)y'+(e+fx)y+sqrtay-b=dx+c$

$(fy-d)x+ey-c+sqrtay-b=fx(x-1)dfracdydx$

$((fy-d)x+ey-c+sqrtay-b)dfracdxdy=fx^2-fx$

This also belongs to an Abel equation of the second kind.

Let $u=x+dfracey-c+sqrtay-bfy-d$ ,

Then $x=u-dfracey-c+sqrtay-bfy-d$

$dfracdxdy=dfracdudy+dfraca(fy+d)-2bf-2(cf-de)sqrtay-b2(fy-d)^2sqrtay-b$

$therefore(fy-d)ubiggl(dfracdudy+dfraca(fy+d)-2bf-2(cf-de)sqrtay-b2(fy-d)^2sqrtay-bbiggr)=fbiggl(u-dfracey-c+sqrtay-bfy-dbiggr)^2-fbiggl(u-dfracey-c+sqrtay-bfy-dbiggr)$

$(fy-d)udfracdudy+dfraca(fy+d)-2bf-2(cf-de)sqrtay-b2(fy-d)sqrtay-bu=fu^2-dfracf(f+2e)y-(2c+d)f+2fsqrtay-bfy-du+dfracf(ey-c+sqrtay-b)^2(fy-d)^2+dfracefy-cf+sqrtay-bfy-d$

$(fy-d)udfracdudy=fu^2-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)sqrtay-bu+dfracf(ey-c+sqrtay-b)^2(fy-d)^2+dfracefy-cf+sqrtay-bfy-d$

$udfracdudy=dfracfu^2fy-d-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)^2sqrtay-bu+dfracf(ey-c+sqrtay-b)^2(fy-d)^3+dfracefy-cf+sqrtay-b(fy-d)^2$

Let $u=(fy-d)v$ ,

Then $dfracdudy=(fy-d)dfracdvdy+fv$

$therefore(fy-d)vleft((fy-d)dfracdvdy+fvright)=dfracf(fy-d)^2v^2fy-d-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)^2sqrtay-b(fy-d)v+dfracf(ey-c+sqrtay-b)^2(fy-d)^3+dfracefy-cf+sqrtay-b(fy-d)^2$

$(fy-d)^2vdfracdvdy+f(fy-d)v^2=f(fy-d)v^2-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)sqrtay-bv+dfracf(ey-c+sqrtay-b)^2(fy-d)^3+dfracefy-cf+sqrtay-b(fy-d)^2$

$(fy-d)^2vdfracdvdy=-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)sqrtay-bv+dfracf(ey-c+sqrtay-b)^2(fy-d)^3+dfracefy-cf+sqrtay-b(fy-d)^2$

$vdfracdvdy=-dfrac5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)sqrtay-b2(fy-d)^3sqrtay-bv+dfracf(ey-c+sqrtay-b)^2(fy-d)^5+dfracefy-cf+sqrtay-b(fy-d)^4$

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 1sqrt -1+x+frac
y left( x right) fsqrt -1+xsqrt ay left( x right) +C=0
$
for arbitrary constant $C$.