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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 method if we know $f(x)$. And also I know how to find general solution if we know one particular solution ($y_0$)

I am looking for exact analitic solution $y= L({f(x)})$ without knowing a particular solution, if it exists. ///$L$ defines operator such as integral,derivatives, radicals, or any defined function.

Or 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 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 diff equation.

Thank you for answers

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2  
See Ricatti equation. –  Sasha Jan 17 '12 at 13:41
3  
@Sasha: It is Riccati and not Ricatti. –  Jon Jan 17 '12 at 13:59

2 Answers 2

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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I don't think it is possible to show that such an integrating factor doesn't exist. –  Shahab Jan 17 '12 at 15:19
    
@Shahab,Then please prove your assumption... –  pedja Jan 17 '12 at 15:36
    
Oops my bad! I didnt read your comment completely. I thought you said no integrating factors will exist at all, whereas you were only referring to integrating factors in terms of only one variable. –  Shahab Jan 17 '12 at 16:08
    
@Shahab,It always exist only if it is in terms of two variables... –  pedja Jan 17 '12 at 16:09
    
Yeah it was a mistake on my part. –  Shahab Jan 17 '12 at 16:09

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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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

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