Solving the Differential Equation (4y + yx^2)dy  (2x + xy^2)dx = 0
This article will guide you through solving the differential equation (4y + yx^2)dy  (2x + xy^2)dx = 0. We will use the method of exact differential equations.
1. Identifying the Form
First, let's rearrange the given equation to get it in a standard form:
(2x + xy^2)dx  (4y + yx^2)dy = 0
This equation is in the form M(x, y)dx + N(x, y)dy = 0. We have:
 M(x, y) = 2x + xy^2
 N(x, y) = (4y + yx^2)
2. Checking for Exactness
A differential equation is exact if the following condition holds:
∂M/∂y = ∂N/∂x
Let's calculate these partial derivatives:
 ∂M/∂y = 2xy
 ∂N/∂x = 2xy
Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
3. Finding an Integrating Factor
We can sometimes make a nonexact equation exact by multiplying it by an integrating factor. A common approach is to look for an integrating factor that depends only on x or only on y.
Let's try finding an integrating factor μ(x) that depends only on x. We need:
∂(μM)/∂y = ∂(μN)/∂x
Expanding this, we get:
μ(∂M/∂y) = μ(∂N/∂x) + (dμ/dx)N
Substituting the values of M, N, ∂M/∂y, and ∂N/∂x:
μ(2xy) = μ(2xy) + (dμ/dx)(4y  yx^2)
This simplifies to:
**(dμ/dx) = μ/x
This is a separable differential equation. Solving for μ, we get:
μ(x) = 1/x
4. Making the Equation Exact
Now, multiply the original equation by the integrating factor μ(x) = 1/x:
((2x + xy^2)/x)dx  ((4y + yx^2)/x)dy = 0
This simplifies to:
(2 + y^2)dx  (4y/x + y^2)dy = 0
Now, let's check if the equation is exact. We have:

M(x, y) = 2 + y^2

N(x, y) = (4y/x + y^2)

∂M/∂y = 2y

∂N/∂x = 4y/x^2
Since ∂M/∂y = ∂N/∂x, the equation is now exact.
5. Solving the Exact Equation
Since the equation is exact, there exists a function F(x, y) such that:
 ∂F/∂x = M(x, y)
 ∂F/∂y = N(x, y)
Let's integrate ∂F/∂x = M(x, y) with respect to x:
F(x, y) = ∫(2 + y^2)dx = 2x + xy^2 + g(y)
Here, g(y) is a constant of integration that depends only on y.
Now, differentiate F(x, y) with respect to y:
∂F/∂y = 2xy + g'(y)
We know that ∂F/∂y = N(x, y) = (4y/x + y^2). Equating these, we get:
2xy + g'(y) = (4y/x + y^2)
Solving for g'(y), we find:
g'(y) = 4y/x  2y^2
Integrating g'(y) with respect to y, we get:
g(y) = 2y^2/x  (2/3)y^3 + C
where C is an arbitrary constant.
Therefore, the general solution to the differential equation is:
F(x, y) = 2x + xy^2  2y^2/x  (2/3)y^3 + C = 0
Conclusion
We successfully solved the differential equation (4y + yx^2)dy  (2x + xy^2)dx = 0 using the method of exact differential equations. The general solution is given by:
2x + xy^2  2y^2/x  (2/3)y^3 + C = 0