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

I know how to find endless series solution via endless integral or endless derivatives and power series solution if we know $f(x)$. I also know how to find general solution if we know one particular solution ($y_0$).

I am looking for an exact analitic solution $y= L({f(x)})$ without knowing a particular solution, if it exists. (Here $L$ defines an operator such as integral, derivative, radical, or any defined function.) If it does not exist, could you please prove why we cannot find it?

Note: This equation is related to second order differantial linear equation. If we put $y=u'/u$, this equation will turn into $u''(x)-f(x).u(x)=0$. If we the 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 functions such as Bessel function or Hermite polinoms and so many special functions are related to Second order linear differential equations.

Thank you for answers.

EDIT:

I asked the question in mathoverflow too. You can also find the link below for details. (1-Endless transform, 2-Endless Integral,3-Endless Derivatives,4-Power series) and answers about the subject.

http://mathoverflow.net/questions/87041/looking-for-the-solution-of-first-order-non-linear-differential-equation-y-y

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5  
See Ricatti equation. – Sasha Jan 17 '12 at 13:41
6  
@Sasha: It is Riccati and not Ricatti. – Jon Jan 17 '12 at 13:59

As was pointed out by @Sasha, this is a Riccati equation which linearizes after the change of variable $y=\psi'/\psi$. This actually gives you Schroedinger equation $$\psi''=f(x)\psi,$$ with general potential $f(x)$. This is considered unsolvable - if you manage to find general solution, your name will appear in every course of quantum mechanics and ODEs.

On the other hand every known solvable quantum mechanical potential $f(x)$ leads to general solution of your equation. Several examples are:

  • Pöschl-Teller potential $f(x)=A+\frac{B}{\cosh^2 C x}$ (hypergeometric functions)

  • harmonic oscillator $f(x)=Ax^2+Bx+C$, $A\ne 0$ (parabolic cylinder functions)

  • constant electric field $f(x)=Ax+B$, $A\ne 0$ (Airy functions)

  • free case $f(x)=A\ne 0$ (plane waves)

A few more examples can be found here.

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Interesting. In Maple I tried $y'+y^2 = \sin(x)$, and the solution involves Mathieu functions $S, C, S', C'$.

I tried $y'+y^2=x$, and the solution involves Airy functions Ai, Bi.

I tried $y'+y^2=1/x$, and the solution involves Bessel functions $I_0, I_1, K_0, K_1$.

This is a Riccati equation. For more info, in particular how its solutions are related to solutions of a second-order linear equation, look that up.

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1  
any second order linear diff equation ($y''(x)+p(x)y'(x)+r(x)y(x)=0$) can be transform into ($u''(x)-f(x)u(x)=0$). Thus all second order equations such as Bessel, Airy etc is directly related to my question. the equation is minimum term of all second order diff equations.I believe it is like Abel impossibilty theorm that he shew for quintic (such as $x^5+x+a=0$)cannot be solved by radicals. But I dont know how to prove that it is impossible – Mathlover Jan 17 '12 at 19:42

In general case differential equations of the form : $\frac{dy}{dx}=f(x,y)$ can be rewritten as :

$M(x,y)dx+N(x,y)dy=0$

So , we have that :

$(y^2-f(x))dx +1\cdot dy=0$

Since this equation isn't exact it must be non-exact and therefore we have to find out if there exist integrating factor in terms of just one variable ($x$ or $y$) .

But , one can show that such integrating factor doesn't exist , therefore you cannot apply integrating factor method on this differential equation . So , you have to use "Riccati" method which means that particular solution has to be known .

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