= Solution
Work on an interval with $x\ne0$ and $\cos(y/x)\ne0$, as required by the original equation. Set $v=y/x$, so $y'=v+xv'$, and obtain $xv'=3\tan v$. Where $\sin v\ne0$, <separation of variables> gives $\cot v\,dv=3\,dx/x$ and therefore
$$
\boxed{\sin(y/x)=Cx^3.}
$$
The solutions lost by division by $\tan v$ are $v=k\pi$, and they are included by $C=0$. On a fixed local branch one can write
$$
y=x\left[k\pi+(-1)^k\arcsin(Cx^3)\right],\qquad |Cx^3|<1.
$$
The integer $k$ is fixed along a connected regular branch. Implicit <differentiation> recovers the original equation whenever $x\cos(y/x)\ne0$. The boundary $|Cx^3|=1$ is excluded because the original <tangent function> has a pole there.
Back to article page
