(x+y)dx+2xdy=0
y/x = u => y = ux; y' = u'x + u
(x+y)dx+2xdy=0
x + y + 2xy' = 0
x + ux + 2x(u'x + u) = 0
x + ux + 2(x^2)u' + 2xu = 0
x + 3ux + 2(x^2)u' = 0
1 + 3u + 2xu' = 0
1 + 3u = -2x(du/dx)
-dx/2x = du/(1 + 3u) | 1 + 3u = 0 => 3y/x = -1; 3y = -x (2)
-1/2 ln|x*C| = 1/3 ln|3u + 1|
-1/2 ln|x*C| = 1/3 ln|3*y/x + 1|
Далле що то не выходет