# 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.)

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$y(x)=\int _0^x\left[ \frac{d}{t}-\frac{t}{y(t)}\right] dt+y(0)$. Is this not what you're looking for? – Jonathan Gleason Sep 7 '11 at 12:53
To those who know, what does the notation $\frac{d}{x}$ mean? Or is this a constant $d$? – mixedmath Sep 7 '11 at 13:13
its a constant@mixedmath – Bhargav Sep 7 '11 at 13:14
it is a very bad choice of a notation but no big deal. – user13838 Sep 7 '11 at 13:22
@Jonathan: I'm looking for an integral that does not contain the variable $y$. – jnm2 Sep 7 '11 at 14:20

First, $\dfrac{dy}{dx}=\dfrac{d}{x}-\dfrac{x}{y}$ itself belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $y=\dfrac{1}{u}$,

Then $\dfrac{dy}{dx}=-\dfrac{1}{u^2}\dfrac{du}{dx}$

$\therefore-\dfrac{1}{u^2}\dfrac{du}{dx}=\dfrac{d}{x}-xu$

$\dfrac{du}{dx}=xu^3-\dfrac{du^2}{x}$

Check whether this ODE satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=5:

$\left(\dfrac{x}{-\dfrac{d}{x}}\right)'=\biggl(-\dfrac{x^2}{d}\biggr)'=-\dfrac{2x}{d}\neq\dfrac{\lambda}{x}$

$\therefore$ not satisfy the special case in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=5

Since the coefficient of $u$ of this ODE is $0$,

$\therefore$ also not satisfy the special case in http://www.hindawi.com/journals/ijde/2010/436860/#EEq2.3