Looking for the solution of first order non-linear differential equation ($y ′+y^{2}=f(x)$) without knowing a particular solution

Question

I have been working on Riccati Equation. I have tried many different methods to find a closed form for the solution of first order non-linear differential equation ($y'+y^{2}=f(x)$) without knowing a particular solution. My aim is to open a topic and to collect all known methods and to progress finding the general solution of Riccati Equation without knowing a particular solution (if possible). May be it can be proved that the solution cannot be expressed in closed form. Actually, I am looking for a similar closed form to linear differential equation ( $y'+y=f(x) $) as known $y=e^{-x}\int{f(x)e^{x}}dx $

Do you know any method to show the closed solution form of ($y'+y^{2}=f(x)$) without knowing a particular solution? If you say, it is not possible to find such closed form or possible to find it, please proof it.

I know how to find a particular solution via endless variable transform or endless integral or endless derivatives or power series. And you can find Wiki link about the subject in link
http://en.wikipedia.org/wiki/Riccati_equation This equation is also related to second order linear differential equation. If we put $y=u'/u$ This equation will turn into $u''(x)-f(x).u(x)=0$. If we find general solution of $y'+y^{2}=f(x)$, it means that $u''(x)-f(x).u(x)=0$ will be solved as well. As we know, many function such as Bessel function or Hermite polinoms and so many special functions are related to Second Order linear differential equation.

I added some solution methods and shew how we can find solution of ($y'+y^{2}=f(x)$).Methods are to find a particular solution and general solution (1-Endless transform, 2-Endless Integral,3-Endless Derivatives,4-Power series) Perhaps, A closed form of general solution can be combination of the methods below or need another kind of approach to the problem.

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1-Endless Transform

$y'+y^{2}=f(x) $

$y=\frac{1}{Z} $

$y'=\frac{-Z'}{Z^{2}} $

$\frac{-Z'}{Z^{2}}+\frac{1}{Z^{2}}=f(x) $

$Z'+Z^{2}f(x)=1 $
$Z=P.Q $

$P'Q+PQ'+P^{2}Q^{2}f(x)=1 $

$P'+P\frac{Q'}{Q}+P^{2}Qf(x)=\frac{1}{Q} $

$Q=\frac{1}{f(x)} $

$P '+P\frac{-f'(x)}{f(x)}+P^{2}=f(x) $

$P=T+\frac{f'(x)}{2f(x)}$

$T '+T^{2}=f(x)+(\frac{-f'(x)}{2f(x)})^{2}+(\frac{-f'(x)}{2f(x)})'$

$y=\frac{1}{Z}=\frac{1}{PQ}=\frac{f(x)}{P}=\frac{f(x)}{\frac{f'(x)}{2f(x)}+T} $

If we define $f_{n+1}(x)=f_n(x)+(\frac{-f_n'(x)}{2f_n(x)})^{2}+(\frac{-f_n'(x)}{2f_n(x)})'$,

$f_0(x)=f(x)$

$y_n(x)=\frac{f_n(x)}{\frac{f_n'(x)}{2f_n(x)}+y_{n+1}} $

$y_0(x)=y_p(x) $ is our particular solution

$y=y_p+\frac{1}{H} $

$y_p'+(\frac{-H'}{H^{2}})+y_p^{2}+\frac{2y_p}{H}+\frac{1}{H^{2}}=f(x) $

$\frac{-H'}{H^{2}}+\frac{2y_p}{H}+\frac{1}{H^{2}}=0 $

$H'-2y_p.H=1 $

$H(x)=e^{2\int{y_p}dx}\int{e^{-2\int{y_p}dx}}dx $

$y(x)=y_p(x)+\frac{e^{-2\int{y_p(x)}dx}}{\int{e^{-2\int{y_p(x)}dx}}dx} $ (This is general solution)

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2-Endless Integral

$y'+y^{2}=f(x) $

$y'=f(x)-y^{2}=$

$y(x)=\int{(f(x)-y^{2})} dx=\int{(f(x)-(\int{[f(x)-y^{2}]}dx)^{2})} dx=..$

The result is endless integral solution. We need iteration to find solution

$y_{n+1}=\int{(f(x)-y_n^{2})} dx$ if we start with $y_0(x)=g(x)$
$y_p(x)=y_{\infty}(x) $
$y_p(x)$ is a particular solution

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3-Endless Derivatives

$y'+y^{2}=f(x) $

$y^{2}=f(x)-y'$

$y=\sqrt{f(x)-y'}$

$y=\sqrt{(f(x)-(\sqrt{f(x)-y'})'} = ..$

$y_{n+1}=\sqrt{f(x)-y_n'}$ if we start with $y_0(x)=g(x)$
$y_p(x)=y_{\infty}(x) $

$y_p(x)$ is a particular solution

The result is endless derivatives solution. We need iteration to find solution

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4-Power series method

$y'+y^{2}=f(x)=f(0)+f'(0)x+\frac{f''(0)x^{2}}{2!}+\frac{f'''(0)x^{3}}{3!}+...$

$y_p(x)$ is a particular solution if $a_0$ is selected any number. if $a_0$ is selected as c constant, the general solution of y(x) can be found depends on x and c.

$y(x)=a_0+a_1x+\frac{a_2x^{2}}{2!}+\frac{a_3x^{3}}{3!}+...$

$y'(x)=a_1+a_2x+\frac{a_3x^{2}}{2!}+\frac{a_4x^{3}}{3!}+...$

$y^{2}(x)=a_0^{2}+(2a_0a_1)x+(2a_0\frac{a_2}{2!}+a_1^{2})x^{2}+...$

$y'+y^{2}=f(x)$
$a_0=c$

$a_0^{2}+a_1=f(0)$

$a_1=f(0)-c^{2}$

$a_2+2a_0a_1=f'(0)$

$a_2=f'(0)-2c(f(0)-c^{2})=f'(0)-2cf(0)+2c^{3}$

(All $a_n$ can be found in that method and depends on c )

$y(x)=c+(f(0)-c^{2})x+\frac{(f'(0)-2cf(0)+2c^{3})x^{2}}{2!}+....$ (This is general solution)

Note:I asked the same question in math.stackexchange.com and I noticed that also theories can be asked here. I decided to open a topic here too you can see the link ( https://math.stackexchange.com/questions/99850/how-can-i-solve-the-differential-equation-yy2-fx )

Thanks for your advices and answers.

Answer 1

You are asking about a very classical problem. The Picard-Vessiot theory was developed to show that, in a certain well-defined sense, there is no `closed form' solution to problems of this kind. You should take a look at the books by Kolchin and Ritt on differential algebra. For a start on the basic ideas, have a look at this paper by Hubbard and Lundell:

http://www.math.cornell.edu/~hubbard/diffalg1.pdf

Answer 2

This is basically searching for a solution to the Schrodinger Equation with an arbitrary potential function fixed in time (using the transformation given taking the Ricatti Differential Equation to a second order linear homogenous differential equation). That alone should tell you something about the solvability… That said, if you are comfortable casting the solution as an infinite series where each term is determined recursively by previous terms then you can write the solution as an infinite sum using the Adomian Decomposition Method.

See 3.3 in this paper:

https://arxiv.org/pdf/2102.10511.pdf#page16

Answer 3

a Riccati equation can be turned into $ u''+f(x)u=0 $ by the transformation $ y= \frac{u'}{u} $ using the WKB ansatz we have the asymptotic solution to 'u' as

$ u(x)\sim C(f(x)^{-1/4})\exp( -\int dx \sqrt{-1}f(x)) $