(x^(2)y^(2)+xy+1)ydx+(x^(2)y^(2)-xy+1)xdy=0

Jasonbradley

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Jun 17, 2024

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Solving the Differential Equation: (x²y² + xy + 1)ydx + (x²y² - xy + 1)xdy = 0

This article aims to guide you through the process of solving the given differential equation:

(x²y² + xy + 1)ydx + (x²y² - xy + 1)xdy = 0

This equation is a non-exact differential equation, meaning it cannot be directly integrated. However, we can make it exact by finding an integrating factor. Let's break down the solution step-by-step:

1. Identifying the Equation as Non-Exact

A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if:

∂M/∂y = ∂N/∂x

In our case:

• M(x,y) = (x²y² + xy + 1)y
• N(x,y) = (x²y² - xy + 1)x

Calculating the partial derivatives:

• ∂M/∂y = 2x²y² + 3xy + 1
• ∂N/∂x = 2x²y² - y + 1

Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.

2. Finding an Integrating Factor

We can find an integrating factor μ(x,y) to make the equation exact. There are two common methods:

• Method 1: If (∂M/∂y - ∂N/∂x)/N is a function of x alone, then μ(x) = exp(∫(∂M/∂y - ∂N/∂x)/N dx)
• Method 2: If (∂N/∂x - ∂M/∂y)/M is a function of y alone, then μ(y) = exp(∫(∂N/∂x - ∂M/∂y)/M dy)

In our case:

(∂M/∂y - ∂N/∂x)/N = (4xy + 2)/[(x²y² - xy + 1)x]

This expression is not a function of x alone. Let's check the other method:

(∂N/∂x - ∂M/∂y)/M = (-4xy - 2)/[(x²y² + xy + 1)y]

This is also not a function of y alone. Therefore, we cannot use the standard methods to find an integrating factor directly.

3. Alternative Approach: Grouping Terms

We can manipulate the equation to make it easier to solve. Notice that the terms in the equation can be grouped as follows:

(x²y² + xy + 1)ydx + (x²y² - xy + 1)xdy = 0

(x²y² + 1)(ydx + xdy) + xy(ydx - xdy) = 0

Now, let's introduce new variables:

• u = xy
• v = x²y² + 1

We can rewrite the equation as:

v(du) + u(dv) = 0

This equation is now exact!

4. Solving the Exact Equation

Since the equation is exact, we can integrate both sides:

∫v(du) + ∫u(dv) = C

Where C is the constant of integration.

Integrating both sides:

vu = C

Substituting back the original variables:

(x²y² + 1)(xy) = C

5. Final Solution

Therefore, the general solution to the differential equation is:

(x²y² + 1)(xy) = C

Where C is an arbitrary constant. This represents a family of curves that satisfy the original differential equation.

This approach demonstrates how to solve a non-exact differential equation using an alternative method of grouping terms and introducing new variables to achieve an exact form.