# 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?

-
It's nonlinear because of the square root, it's not even an ODE because of that $y(e+fx)$. I think it's what's called a "differential-delay equation". By the way, I admire what you did on that bus in 1955. Welcome to m.se – Gerry Myerson Feb 21 '12 at 11:33
@Gerry Merson: why Differential delay equation? I understand $f$ is a constant. – Riccardo.Alestra Feb 21 '12 at 12:03
Riccardo Alestra is right, $f$ is not a function, but a constant. I ran out of unbiased letters there. I'm sorry for that! With $f$ being a constant, this is a nonlinear ODE again, isn't it? – user25425 Feb 21 '12 at 20:00
I have edited the equation so as to make it less likely that I will misread it in the future. – Gerry Myerson Feb 21 '12 at 22:18
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? – user25506 Feb 22 '12 at 18:37

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 $y=\dfrac{u^2+b}{a}$

$y'=\dfrac{2uu'}{a}$

$\therefore fx(1-x)\dfrac{2uu'}{a}+(e+fx)\dfrac{u^2+b}{a}+u=dx+c$

$\dfrac{2fx(1-x)uu'}{a}+\dfrac{(e+fx)u^2}{a}+\dfrac{be+bfx}{a}+u=dx+c$

$\dfrac{2fx(1-x)uu'}{a}=-\dfrac{(e+fx)u^2}{a}-u-\dfrac{(bf-ad)x+be-ac}{a}$

$uu'=-\dfrac{(e+fx)u^2}{2fx(1-x)}-\dfrac{au}{2fx(1-x)}-\dfrac{(bf-ad)x+be-ac}{2fx(1-x)}$

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.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $u'=-\dfrac{v'}{v^2}$

$\therefore-\dfrac{v'}{v^3}=-\dfrac{e+fx}{2fx(1-x)v^2}-\dfrac{a}{2fx(1-x)v}-\dfrac{(bf-ad)x+be-ac}{2fx(1-x)}$

$v'=\dfrac{((bf-ad)x+be-ac)v^3}{2fx(1-x)}+\dfrac{av^2}{2fx(1-x)}+\dfrac{(e+fx)v}{2fx(1-x)}$

Approach $2$:

$fx(1-x)y'+(e+fx)y+\sqrt{ay-b}=dx+c$

$(fy-d)x+ey-c+\sqrt{ay-b}=fx(x-1)y'$

$((fy-d)x+ey-c+\sqrt{ay-b})x'=fx^2-fx$

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

Let $u=x+\dfrac{ey-c+\sqrt{ay-b}}{fy-d}$ ,

Then $x=u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}$

$x'=u'+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}$

$\therefore(fy-d)u\biggl(u'+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}\biggr)=f\biggl(u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}\biggr)^2-f\biggl(u-\dfrac{ey-c+\sqrt{ay-b}}{fy-d}\biggr)$

$(fy-d)uu'+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}u=fu^2-\dfrac{f((2e+f)y-2c-d)+2f\sqrt{ay-b}}{fy-d}u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^2}+\dfrac{f(ey-c)+f\sqrt{ay-b}}{fy-d}$

$(fy-d)uu'=fu^2-\biggl(\dfrac{f((2e+f)y-2c-d)+2f\sqrt{ay-b}}{fy-d}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)\sqrt{ay-b}}\biggr)u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^2}+\dfrac{f(ey-c)+f\sqrt{ay-b}}{fy-d}$

$uu'=\dfrac{fu^2}{fy-d}-\biggl(\dfrac{f((2e+f)y-2c-d)+2f\sqrt{ay-b}}{(fy-d)^2}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}\biggr)u+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{f(ey-c)+f\sqrt{ay-b}}{(fy-d)^2}$

Let $u=\dfrac{1}{v}$ ,

Then $u'=-\dfrac{v'}{v^2}$

$\therefore-\dfrac{v'}{v^3}=\dfrac{f}{(fy-d)v^2}-\biggl(\dfrac{f((2e+f)y-2c-d)+2f\sqrt{ay-b}}{(fy-d)^2}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}\biggr)\dfrac{1}{v}+\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{f(ey-c)+f\sqrt{ay-b}}{(fy-d)^2}$

$v'=-\biggl(\dfrac{f(ey-c+\sqrt{ay-b})^2}{(fy-d)^3}+\dfrac{f(ey-c)+f\sqrt{ay-b}}{(fy-d)^2}\biggr)v^3+\biggl(\dfrac{f((2e+f)y-2c-d)+2f\sqrt{ay-b}}{(fy-d)^2}+\dfrac{a(fy+d)-2bf-2(cf-de)\sqrt{ay-b}}{2(fy-d)^2\sqrt{ay-b}}\biggr)v^2-\dfrac{fv}{fy-d}$

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 $C$.