Problem 04 | Elimination of Arbitrary Constants

$cy^2 = x^2 + y$       → equation (1)

$2cy~dy = 2x~dx + dy$

$c = \dfrac{2x~dx + dy}{2y~dy}$
 

Substitute c to equation (1)
$\left( \dfrac{2x~dx + dy}{2y~dy} \right) y^2 = x^2 + y$

$\left( \dfrac{2x~dx + dy}{2~dy} \right) y = x^2 + y$

$(2x~dx + dy)y = 2(x^2 + y)~dy$

$(2x~dx + dy)y = 2(x^2 + y)~dy$

$2xy~dx + y~dy = 2x^2~dy + 2y~dy$

$2xy~dx - y~dy - 2x^2~dy = 0$

$2xy~dx - (y + 2x^2)~dy = 0$           answer

$cy^2 = x^2 + y$
 

Divide by y²
$c = \dfrac{x^2}{y^2} + \dfrac{1}{y}$

$0 = \dfrac{y^2(2x~dx) - x^2(2y~dy)}{(y^2)^2} + \dfrac{-dy}{y^2}$

$0 = \dfrac{2xy^2~dx - 2x^2y~dy}{y^4} - \dfrac{dy}{y^2}$

$0 = \dfrac{2(xy~dx - x^2~dy)y}{y^4} - \dfrac{dy}{y^2}$

$0 = \dfrac{2(xy~dx - x^2~dy)}{y^3} - \dfrac{dy}{y^2}$
 

Multiply by y³
$0 = 2(xy~dx - x^2~dy) - y~dy$

$0 = 2xy~dx - 2x^2~dy - y~dy$

$2xy~dx - (2x^2~dy + y~dy) = 0$

$2xy~dx - (2x^2 + y)~dy = 0$           okay