I am solving some differential equations "solvable for $p$" and I came up with the following equation:
$$p^2 + p \left( \frac{y}{x} - \frac{x}{y} \right) - 1 = 0$$
I never came across such an equation before. Is this a quadratic equation and why? because as far as I know the form should be the following:
$$ax^2 + bx + c = 0$$
Where $a, b, c$ are constants but in this case it seems different.
The equation is a quadratic in variable $p$. To solve it, you can either complete the square or use the quadratic formula
Quadratic formula:
$$ p = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
where $a = 1$, $b = \frac{y}{x} -\frac{x}{y} = \frac{y^2-x^2}{xy}$, $c=-1$
You can then make the necessary substitution to obtain a solution for $p$
$$ p = \frac{1}{2} \left(\frac{x^2-y^2}{xy}\pm \sqrt{\left(\frac{y^2-x^2}{xy}\right)^2 + 4}\right)$$
$$ p = \frac{1}{2} \left(\frac{x^2-y^2}{xy} \pm \sqrt{\frac{y^4+2x^2y^2+x^4}{x^2y^2}}\right) $$
$$p = \frac{1}{2} \left( \frac{x^2-y^2}{xy} \pm \frac{x^2+y^2}{xy}\right) $$
$$ p = \frac{x}{y} \ \text{or} \ -\frac{y}{x}$$
where $y \ne 0$, $x \ne 0$