(4y+yx^2)dy-(2x+xy^2)dx=0

5 min read Jun 16, 2024

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 non-exact 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