Problem 03 | Elimination of Arbitrary Constants

$x^2y = 1 + cx$       → equation (1)

$x^2~dy + 2xy~dx = c~dx$
 

Divide by dx
$c = x^2~y' + 2xy$
 

Substitute c to equation (1)
$x^2y = 1 + (x^2~y' + 2xy)x$

$x^2y = 1 + x^3~y' + 2x^2y$

$1 + x^3~y' + x^2y = 0$
 

Multiply by dx
$dx + x^3~dy + x^2y~dx = 0$

$(1 + x^2y)~dx + x^3~dy = 0$           answer

$x^2y = 1 + cx$

$xy = x^{-1} + c$

$x~dy + y~dx = -x^{-2}~dx$

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

$x^3~dy + x^2y~dx = -dx$

$x^3~dy + dx + x^2y~dx = 0$

$(dx + x^2y~dx) + x^3~dy = 0$

$(1 + x^2y)~dx + x^3~dy = 0$           okay