Solve $(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0$ I was trying to solve 

$$(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0.$$ 

I ended up to the point of obtaining $$x^4\, \text{d}x+xy^4\frac{\text{d}(e^{1/2y^2}x)}{e^{1/2y^2}x}=0.$$ Couldn't proceed further. Can someone help me?.

$$x^4+y^4=x^2\frac{2ydy}{2xdx}$$
Let $\quad\begin{cases}x^2=X\\y^2=Y\end{cases}$
$$X^2+Y^2=X\frac{dY}{dX}$$
$$\frac{dY}{dX}=X+\frac{Y^2}{X}$$
Riccati ODE. Let : $\quad Y(X)= -X\frac{F'(X)}{F(X)} \quad\implies\quad Y'=-\frac{F'}{F} -X\frac{F''}{F}+X\frac{(F')^2}{F^2}  $
$\frac{dY}{dX}=X+\frac{Y^2}{X}=-\frac{F'}{F} -X\frac{F''}{F}+X\frac{(F')^2}{F^2}=X+\frac{\left(-X\frac{F'}{F} \right)^2}{X}$
$-\frac{F'}{F} -X\frac{F''}{F}=X$
$$F''(X)+\frac{1}{X}F'(X)+F(X)=0$$
Bessel equation :
$$F(X)=c_1J_0(X)+c_2Y_0(X)$$
$J_0(X)$ and $Y_0(X)$ are the Bessel functions of first and second kind, of order $0$.
$\frac{dJ_0(X)}{dX}=-J_1(X)$ and $\frac{dJ_0(X)}{dX}=-Y_1(X)$
$J_1(X)$ and $Y_1(X)$ are the Bessel functions of first and second kind, of order $1$.
$\frac{F'(X)}{F(X)}=-\frac{c_1J_1(X)+c_2Y_1(X)}{c_1J_0(X)+c_2Y_0(X)} = -\frac{C\,J_1(X)+Y_1(X)}{C\,J_0(X)+Y_0(X)} $ where $C=\frac{c_1}{c_2}$
$Y(X)=-X\frac{F'(X)}{F(X)}=X\frac{C\,J_1(X)+Y_1(X)}{C\,J_0(X)+Y_0(X)}$
$$y(x)=\pm \,x\,\sqrt{\frac{C\,J_1(x^2)+Y_1(x^2)}{C\,J_0(x^2)+Y_0(x^2)}}$$
$C$ is an arbitrary constant.
