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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.


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)


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


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


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.

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    $\begingroup$ A simple fact about Riccati equations : if $u, v$ solve a linear system of the first order $u'(t)=\alpha(t) u(t) + \beta(t) v(t)$ , $v(t)'=\gamma(t) u(t) + \delta(t) v(t)$, then the ratio $u(t)/v(t)$ solves a Riccati equation easily written. And any Riccati equation can be related this way to such a system (even with some freedom of choice). $\endgroup$ Jul 21, 2014 at 13:33
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    $\begingroup$ There's a current question about the terminology "endless" here: mathoverflow.net/questions/243692/… If you see this comment, perhaps you can enlighten the OP where you saw this terminology, before it disappears through closure + automatic deletion or otherwise. $\endgroup$
    – Todd Trimble
    Jul 5, 2016 at 14:31
  • $\begingroup$ @ToddTrimble : Thank you for information, I gave more information to the OP. Best Regards $\endgroup$
    – Mathlover
    Jul 5, 2016 at 19:19

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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
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  • $\begingroup$ Thank you for great referances.I got new ideas to approach the problem. According to the diffalg1.pdf, it was prooved that $y'+y^{2}=x$ no solutions which can be written using elementary functions, or anti-derivatives of elementary functions, or exponentials of such anti-derivatives, or anti-derivative of those, etc. but it be solved using power series, integrals which depend on a parameter, or Bessel functions of order 1/3. Does it mean if we add a new function to elemantry function group or integrals which depend on a parameter, we can find a closed form solution of the $y'+y^{2}=f(x)$? $\endgroup$
    – Mathlover
    Feb 2, 2012 at 21:56
  • $\begingroup$ @Mathlover, one can always make an equation soluble in closed form in this way by adjoining to the collection of elemementary functions a new function $F(x_0, y_0, x)$ that is the solution to the differential equation passing through $(x_0, y_0)$. $\endgroup$
    – LSpice
    Sep 10, 2015 at 17:49
  • $\begingroup$ @LSpice I especially wonder if we add only one function (such as $J_{\frac{1}{3}}(x)$) into elementary functions would be enough to solve the general equation $y'+y^2=f(x)$ or it has not been enough to solve $y'+y^2=f(x)$ yet. how can define the solubility for $y'+y^2=f(x)$? I have been looking for any such special function or functions (limited number) to add into elementary functions that would be enough to solve $y'+y^2=f(x)$ . Maybe the adding a function into elementary functions would never be enough to solve the general equation $y'+y^2=f(x)$. Thanks for your comments $\endgroup$
    – Mathlover
    Sep 21, 2015 at 13:18
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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

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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)) $

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  • $\begingroup$ Very nice referances. Thank you very much. I found many nice detail about the subject.en.wikipedia.org/wiki/WKB_approximation $\endgroup$
    – Mathlover
    Apr 12, 2012 at 11:39
  • $\begingroup$ I'm guessing $\sqrt{-}$ is a typo --- should it be $\sqrt{-1}$ ? $\endgroup$ Mar 13, 2021 at 17:56

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