y’ = (x^2+3x^2)/(2xy)
sol:
y’ = (x^2+3x^2)/(2xy) = x/2y + 3y/2x = 1/2(y/x) ^ (-1) +3/2(y/x)
this equation is homogenous
u = y/x y = ux
y’ = du/dx*x +u
du/dx*x +u = 1/2u^ (-1) +3/2u
separable
(2u/(1+u^2)) du = 1/x dx
Integral both sides
ln|1+u^2| = ln|x|
1+u^2 = cx
y^2/x^2 +1 = cx
y^2+x^2-cx^3=0
