Solve this ordinary differential equation in closed form

Question

I saw this problem is Birkhoff and Rota's ODE.

Express in closed form all solutions of the following differential equation $$y'= \frac{x^2-y^2}{x^2+y^2}.$$

I tried using substitution $y = vx$, which gives $$\int{\frac{1 + v^2}{1 - v -v^2 -v^3}}dv$$ this is not solvable in closed form(at least as far as I know and using Wolfram Alpha). Even wolfram doesn't yield a closed solution to this DE.

Answer 1

Making $y=v x$, you effectively end with $$\int\frac {dx} x=-\int\frac {v^2+1 } {v^3+v^2+v-1 }dv=-\int\frac {v^2+1 } {(v-a)(v^2+b v+c) }dv$$ where $a$ is the only real root of the cubic. Using partial fraction decomposition $$\frac {v^2+1 } {(v-a)(v^2+b v+c) }=\frac{1}{a^2+a b+c}\Bigg[\frac{a^2+1}{v-a} +\frac { (a b+c-1)v+(a c-a-b) }{v^2+b v+c }\Bigg]$$ which does not make any problem.

You have $$a=\frac{1}{3} \left(2 \sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2 \sqrt{2}}\right)\right)-1\right)$$

Answer 2

Hint: The denominator of RHS suggests that one tries polar coordinates $x=r\cos(\theta), y=r\sin(\theta), r\in\mathbb{R}_{>0}$. Note that then

$$\dfrac{dy}{dx} = \dfrac{\partial y / \partial r}{\partial x/\partial r} = \dfrac{\partial y / \partial \theta}{\partial x/\partial \theta}.$$