$$y'=x^2+y^2$$ I know, that this is a kind of Riccati equation, but is it possible to solve it with only simple methods? Thank you
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Are exact equations simple enough? That's what I'd try. – Git Gud Jul 18 '13 at 22:27
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Looking at the solution this doesn't look promising. – Nathaniel Bubis Jul 18 '13 at 22:29
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@nbubis Implicitly it could be easy. – Git Gud Jul 18 '13 at 22:29
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@GitGud - always happy to be proven wrong :) – Nathaniel Bubis Jul 18 '13 at 22:30
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@nbubis I can't be arsed $\ddot \smile$, not until the OP let's me know wether it is worth trying to transform it in an exact equation. – Git Gud Jul 18 '13 at 22:32
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It can be solved (or, rather, transformed into a recognizable form) using simple methods, but the result can only be expressed in terms of special functions.
Namely, let us write $\displaystyle y=-\frac{v'}{v}$, then $\displaystyle y'=-\frac{v''}{v}+\frac{v'^2}{v^2}$ so that the equation becomes linear: $$v''+x^2v=0.\tag{1}$$ If we further introduce $$v(x)=\sqrt{x}\cdot f\left(\text{$\frac{x^2}{2}$}\right),\tag{2}$$ then (1) transforms into a particular case of the Bessel equation for $f(t)$: $$t^2f''+tf'+\left(t^2-\frac{1}{16}\right)f=0.$$ It has the general solution $$f(t)=C_1 J_{1/4}(t)+C_2J_{-1/4}(t).\tag{3}$$ Substituting this into (2), we find $v(x)$, and then $y(x)$ is given by its logarithmic derivative. Clearly, $y(x)$ will depend only on the ratio $C_1/C_2$ (instead of $C_1$, $C_2$ separately). This ratio plays the role of integration constant for the initial first order equation $y'=x^2+y^2$.
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2This book is very helpful ---> Handbook of Differential Equations. Daniel Zwillinger. Academic Press 1997. – Felix Marin Aug 20 '13 at 20:05