(x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0
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
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
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.
Finally answer will be x^3+y^3-6xy(x+y)=c
The given equation is a first-order linear homogeneous differential equation. Let's solve it step by step.
First, let's rewrite the given equation as:
(x^2 - 4xy - 2y^2)dx + (y^2 - 4xy - 2x^2)dy = 0
To determine if this equation is exact, we need to check if the partial derivatives of the coefficients with respect to y and x are equal. Let's find these partial derivatives.
Partial derivative of (x^2 - 4xy - 2y^2) with respect to y:
∂/∂y (x^2 - 4xy - 2y^2) = -4x - 4y
Partial derivative of (y^2 - 4xy - 2x^2) with respect to x:
∂/∂x (y^2 - 4xy - 2x^2) = -4y - 4x
Since the partial derivatives are equal, the equation is exact.
Next, we need to find the integrating factor to solve the exact differential equation. The integrating factor, denoted by μ, is given by:
μ = e^(∫[Q(x, y) / P(x, y)]dy)
Let's calculate μ using the equation (y^2 - 4xy - 2x^2) / (x^2 - 4xy - 2y^2):
μ = e^(∫[(y^2 - 4xy - 2x^2) / (x^2 - 4xy - 2y^2)]dy)
To integrate the above expression, we need to factor the numerator and denominator. The numerator can be factored as (y - 2x)(y + 2x), and the denominator can be factored as (x - 2y)(x + 2y).
So, the integrating factor μ is given by:
μ = e^(∫[(y - 2x)(y + 2x) / (x - 2y)(x + 2y)]dy)
Now, by simplifying the expression inside the integral and integrating, we can find the value of μ.
Once we have the value of μ, we multiply both sides of the equation by μ:
μ * [(x^2 - 4xy - 2y^2)dx + (y^2 - 4xy - 2x^2)dy] = 0
This will convert the exact differential equation into an exact differential equation, which is easier to solve.
Now, we can integrate both sides of the equation to obtain the general solution. Once we find the general solution, we can find specific solutions by imposing initial or boundary conditions.