Please Solve this homogenous Equation?

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

it is not exact but make it exact as

[(d/dy ((x^2) - 2(y^2))) + (d/dx 2xy)]/(2xy) as it can only be function of x which great the answer is

-3/x so e^(int -3/x dx) = e^(-3*ln(x)) = 1/x^3

[(1/x) - (2*(y^2)/(x^3))] dx + (2y/x^2) dy = 0

d/dy = -4y/x^3 = d/dx which means procede

int (2y/x^2) dy = (y^2/x^2) + g(x)

d/dx (y^2/x^2) + g(x) = [-2(y^2)/(x^3)] + g'(x)

[-2(y^2)/(x^3)] + g'(x) = [(1/x) - (2*(y^2)/(x^3))

g'(x) = 1/x; g(x) = ln(x)

final answer is

(y^2/x^2) + ln(x) = c

y^2/x^2 = c - ln(x)

y^2 = (x^2)*[c - ln(x)]

y = +/- x*sqrt[c - ln(x)]

x^2 - 2y^2 + 2xyy' = 0
y' = (2y^2 - x^2)/(2xy)
let y = ux
y' = u'x + u = (2u^2 - 1)/(2u) = u - 1/(2u)
u'x = -1/(2u)
uu' = -1/(2x)
udu = -dx/(2x)
u^2 = -ln x + const
y^2/x^2 = ln (k/x)
y^2 = x^2 ln (k/x)

Let f(x,y)= constant
df = partial f/partial x ds+ partial f/partial y dy
=(x^2-y^2) dx + 2xy dy
Comparing the two, you have
partial f/partial = x^2 -y^2
Now integrate wrt x. You will get
x^3/3 - y^2 x=f(x,y)+h(y)
similarly
partial f/ partial y = 2 xy
f(x, y) = x y^2 + g(x)
g(x) and h(y) are added for completeness. Comparing the two solutions you can guess h(y)=0 and g(x) = -x^3/3
The solution looks like
x^3/3- y^2 x= constnat. But your terms are wrong. Can you check the equation.
According ly
(x^2+y^2) dx + 2 y x = constant.