in this problem we consider an equation in differential form M dx+N dy=0
(2y-2xy^2)dx+(2x-2x^2y)dy=0
give implicit general solutions to the differential equation.
F(x,y)=
My = 2-4xy
Nx = 2-4xy
F(x,y) = ∫M dx = ∫ 2y-2xy^2 dx = 2xy - x^2y^2 + h(y)
Now, we have
Fy = 2x - 2x^2y + h'(y)
so, h'(y)=0, and h(y) = c
F(x,y) = 2xy - x^2y^2
To find the implicit general solution to the given differential equation, we need to integrate the equation by finding a function F(x, y) such that its partial derivatives with respect to x and y match the given expression.
Step 1: Identify M and N.
From the given equation, we have M = 2y - 2xy^2 and N = 2x - 2x^2y.
Step 2: Check for exactness.
For a differential equation to be exact, the condition ∂M/∂y = ∂N/∂x must hold. Let's check if it is satisfied here:
∂M/∂y = 2 - 4xy
∂N/∂x = 2 - 4xy
Since ∂M/∂y = ∂N/∂x, the equation is exact.
Step 3: Integrate.
To find F(x, y), we integrate M with respect to x, treating y as a constant, and N with respect to y, treating x as a constant:
∫(2y - 2xy^2)dx = 2xy - x^2y^2 + g(y) = F(x, y)
∫(2x - 2x^2y)dy = 2xy - x^2y^2 + h(x) = F(x, y)
Here, g(y) and h(x) are integration functions, which can be determined by taking the partial derivative of F(x, y) with respect to y and x, respectively.
Step 4: Determine the general solution.
Since F(x, y) is the same in the two above equations, we equate them to find the general solution:
2xy - x^2y^2 + g(y) = 2xy - x^2y^2 + h(x)
The integrations functions g(y) and h(x) are arbitrary functions, so let's call them G(y) and H(x):
2xy - x^2y^2 + G(y) = 2xy - x^2y^2 + H(x)
Now we have the implicit general solution:
2xy - x^2y^2 + G(y) = H(x), where G(y) and H(x) can be any arbitrary functions.
Therefore, F(x, y) = 2xy - x^2y^2 + G(y) is the implicit general solution to the given differential equation.