" 5."(i)(y^(2)-2xy)dx=(x^(2)-2xy)dyquad " (ii) "2xydx+

Solve : (y^2-2xy)dx=(x^2-2xy)dy

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

(x^(2)+xy)dy=(x^(2)+y^(2))dx

(1-xy+x^(2)y^(2))dx=x^(2)dy

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

Find the general solution the following differential equations (i) (dy)/(dx) = (y^(2) - 2xy)/(x^(2) - xy) (ii) (y^(2) - 2xy)dx + (2xy - x^(2))dy = 0 (iii) (3x^(2) + y^(2))dy + (x^(2) + 3y^(2))dx = 0

(dy)/(dx)=(x^(2)-y^(2))/(xy)

x^(2)(dy)/(dx)+y^(2)=xy