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
(x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0
5 answers
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
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
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.
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.
It was probably a typo in each case, so there was not really a disagreement, especially everyone uses the same method of solution.
Finally answer will be x^3+y^3-6xy(x+y)=c