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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?

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    I appreciate your help and thank you a lot, guys! Figuring out whether there exists a closed form solution at all seemed like an impossible task to me... now I know that a general closed form solution is out of reach. The general form of the nonlinear differential equation above stems from the following special case, where there is more structure imposed on the constants... $(\lambda_1-\lambda_0)x(1-x)y'+(\lambda_0+ (\lambda_1-\lambda_0)x)y+\sqrt{2ay-2b_0} =a^{-1}(\lambda_1b_1-\lambda_0b_0)x+a^{-1}\lambda_0b_0$ Is there a solution to that special case of the differential equation above?2012-02-22

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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 $\dfrac{du}{dy}=(fy-d)\dfrac{dv}{dy}+fv$

$\therefore(fy-d)v\left((fy-d)\dfrac{dv}{dy}+fv\right)=\dfrac{f(fy-d)^2v^2}{fy-d}-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}(fy-d)v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$

$(fy-d)^2v\dfrac{dv}{dy}+f(fy-d)v^2=f(fy-d)v^2-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$

$(fy-d)^2v\dfrac{dv}{dy}=-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^2}$

$v\dfrac{dv}{dy}=-\dfrac{5afy+ad-6bf+2(f(f+2e)y+de-(3c+d)f)\sqrt{ay-b}}{2(fy-d)^3\sqrt{ay-b}}v+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^5}+\dfrac{efy-cf+\sqrt{ay-b}}{(fy-d)^4}$

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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 $C$.