Solve differential equation $xyy'=x^4+y^4$ How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=\dfrac{x^4+y^4}{xy}$$
so I expect (maybe) some kind of symetric function for y(x).
Thanks for any help.
 A: Observing that $y y' = \left(\frac{1}{2}y^2\right)'$ we define the new dependent variable $z := \frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing
\begin{equation}
z = - \frac{x}{4} \frac{u'}{u}, \quad z^2 = \frac{x^2}{16} \frac{(u')^2}{u^2}, \quad z' = - \frac{1}{4} \frac{u'}{u} - \frac{x}{4} \frac{u'' u - (u')^2}{u^2},
\end{equation}
which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable
\begin{equation}
\xi := x^2, \quad \frac{d}{dx} = 2 \xi^{1/2} \frac{d}{d\xi}, \quad \frac{d^2}{dx^2} = 2 \frac{d}{d\xi} + 4 \xi \frac{d^2}{d\xi^2},
\end{equation}
we obtain the Bessel differential equation
\begin{equation}
\xi^2 \frac{d^2 u}{d \xi^2} + \xi \frac{d u}{d\xi} + \xi^2 u = 0,
\end{equation}
with fundamental solutions $J_0(\xi)$, $Y_0(\xi)$ (zeroth-order Bessel functions).
