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

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 Ricatti 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 ( http://math.stackexchange.com/questions/99850/how-can-i-solve-the-differential-equation-yy2-fx )

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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). – Pietro Majer Jul 21 '14 at 13:33

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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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 $y'+y^{2}=f(x)$? – Mathlover Feb 2 '12 at 21:56
@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 $(x_0, y_0)$. – L Spice Sep 10 '15 at 17:49
@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 – Mathlover Sep 21 '15 at 13:18

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

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

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Very nice referances. Thank you very much. I found many nice detail about the subject.en.wikipedia.org/wiki/WKB_approximation – Mathlover Apr 12 '12 at 11:39