(x y2x) dx (y + x2y) dy = 0
x(1 y2) dx = y(1 + x2) dy
$(x"d"x)/(1 + x^2) = (y"d"y)/(1 - y^2)$
Integrating on both sides, we get
$int (x)/(1 + x^2) "d"x = int (y)/(1 - y^2) "d"y$
$int (2x)/(1 + x^2) "d"x = - int (-2y)/(1 - y^2) "d"y$
log |1 + x2| = log |1 y2| + log |c|
log |1 + x2| = $log|"c"/(1 - y^2)|$
1 + x2 = $"c"/(1 - y^2)$
(1 + x2)(1 y2) = c ......(i)
When x = 2, y = 0, we have
(1 + 4)(1 0) = c
c = 5
Substituting c = 5 in (i), we get
(1 + x2)(1 y2) = 5, which is the required particular solution.
