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 )

Thanks for your advices and answers.