Question

(x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0

Answer 1

I assume you are looking for a solution?
We want a function F(x,y) such that

Fx(x,y) = x^2 - 4xy - 2y^2
and
Fy(x,y) = y^2 - 4xy - 2x^2

To make sure there is such an F, let
M = x^2 - 4xy - 2y^2
N = y^2 - 4xy - 2x^2

My = -4x - 4y
Nx = -4y - 4x

So, things look good.

Our solution is
F(x,y) = 1/2 x^3 - 2x^2y - 2xy^2 + 1/3 y^3 = C

Answer 2

Your differential equation is of the form
M(x,y) dx + N(x,y) dy = 0
with dN/dx = dM/dy (those being partial derivatives)
The last equation is a necessary and sufficient condition for a solution of the form
f(x,y) = 0
In your case
df/dx (partial) = x^2 -4xy -2y^2
and
df/dy (partial) = y^2 -4xy -2x^2

f(x,y) = x^3/3 -2yx^2 -2y^2x + g(y)
y^2 -4xy -2x^2 = -2x^2 -4yx -dg/dy
-dg/dy = y^2
g(y) = -y^3/3 + C

Thus the solution is
x^3/3 -2yx^2 -2y^2x -y^3/3 = C

Answer 3

Steve and I disagree on a couple of terms. He could be right. My math is rusty. Check the math yourself. There are some good tutorials on the "exact differential" method online.

Answer 4

Steve's first term should have read (1/3)x^3 while the last term should be positive (+(1/3)y^3).

It was probably a typo in each case, so there was not really a disagreement, especially everyone uses the same method of solution.

Answer 5

Finally answer will be x^3+y^3-6xy(x+y)=c