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I really need some specific processes on how to solve the equation.
How to relate the equation $\frac{dy}{dx}=x^2+y^2$ to the Bessel function?

I have already had some clues that a Riccati equation $y^{'}=y^2+g(x)y+h(x)$ can be converted into a new equation :$u^{''}-g(x)u^{'}+h(x)u=0$ by doing the transformation: $u(x)=e^{\int y(x) dx}$.

Wentao Zhang
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  • Might be easier to solve in polar co-ordinates. Did you try those? – whoisit Nov 06 '22 at 09:46
  • Thank you for your advice. I will try it. – Wentao Zhang Nov 06 '22 at 10:00
  • The usual Ricatti transformation $y=-u'/u$ transforms it to $u''+x^2u=0$. What is the type of this equation? – Bob Dobbs Nov 06 '22 at 10:31
  • See https://math.stackexchange.com/questions/2348022/riccati-d-e-vertical-asymptotes, https://math.stackexchange.com/questions/2835521/limit-of-function-as-x-to-infty-when-fx-is-given, https://math.stackexchange.com/questions/3455568/solving-an-ode-using-picards-iteration-technique, not all about this questions, but on different facets of this classical example of a Riccati equation. – Lutz Lehmann Nov 06 '22 at 10:33
  • Thank you very much! – Wentao Zhang Nov 14 '22 at 10:35

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