Integral form of $\frac{dy}{dx} = \frac{d}{x} - \frac{x}{y}$ How can I write $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{d}{x} -\dfrac{x}{y}$ in integral form not containing $y$?
(Its solution represents the family of curves orthogonal to the family of curves $y^2 \cos(2a) - 2dy + x^2 = 0$ in $a$.  I'm fairly sure there isn't a closed form.)
 A: Case $1$: $d=0$
Then $\dfrac{dy}{dx}=-\dfrac{x}{y}$
$y~dy=-x~dx$
$\int y~dy=-\int x~dx$
$\dfrac{y^2}{2}=-\dfrac{x^2}{2}+c$
$y^2=C-x^2$
Case $2$: $d\neq0$
Then $\dfrac{dy}{dx}=\dfrac{d}{x}-\dfrac{x}{y}$
$y\dfrac{dy}{dx}=\dfrac{dy}{x}-x$
This belongs to an Abel equation of the second kind.
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\therefore2xy\dfrac{dy}{du}=\dfrac{dy}{x}-x$
$(dy-x^2)\dfrac{du}{dy}=2x^2y$
$(dy-u)\dfrac{du}{dy}=2yu$
Let $v=dy-u$ ,
Then $u=dy-v$
$\dfrac{du}{dy}=d-\dfrac{dv}{dy}$
$\therefore v\left(d-\dfrac{dv}{dy}\right)=2y(dy-v)$
$dv-v\dfrac{dv}{dy}=2dy^2-2yv$
$v\dfrac{dv}{dy}=(2y+d)v-2dy^2$
Let $s=y+\dfrac{d}{2}$ ,
Then $\dfrac{dv}{dy}=\dfrac{dv}{ds}\dfrac{ds}{dy}=\dfrac{dv}{ds}$
$\therefore v\dfrac{dv}{ds}=2sv-2d\left(s-\dfrac{d}{2}\right)^2$
$v\dfrac{dv}{ds}=2sv-2ds^2+2d^2s-\dfrac{d^3}{2}$
Let $t=s^2$ ,
Then $\dfrac{dv}{ds}=\dfrac{dv}{dt}\dfrac{dt}{ds}=2s\dfrac{dv}{dt}$
$\therefore2sv\dfrac{dv}{dt}=2sv-2ds^2+2d^2s-\dfrac{d^3}{2}$
$v\dfrac{dv}{dt}=v-ds+d^2-\dfrac{d^3}{4s}$
$v\dfrac{dv}{dt}=v\pm d\sqrt{t}+d^2\pm\dfrac{d^3}{4\sqrt{t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
