Solving Non-Exact Differential Equations of the Form f(x, y)dx g(x, y)dy 0

Non-exact differential equations are a common challenge in mathematics and physics. This article will guide you through the process of determining whether a given differential equation is exact, and if not, how to find an integrating factor to make it exact. The methodology will be applied to a specific differential equation, which will be solved step-by-step.

Introduction

A differential equation of the form (M(x, y)dx N(x, y)dy 0) is considered exact if the partial derivatives (frac{partial M}{partial y}) and (frac{partial N}{partial x}) are equal. If they are not equal, the equation is non-exact, and an integrating factor can sometimes be found to make it exact.

Problem Statement

Consider the following non-exact differential equation:

For the equation (yx^3y2x^2dx 4xy^48y^3dy 0), we can rewrite it as:

[yx^2xy2]dx [x4y^3xy2]dy 0

To simplify the equation, we divide by (xy - 2):

[y/x y2 x^2]dx [x/x y2 4y^3]dy 0

Testing for Exactness

To test for exactness, we calculate the partial derivatives:

(frac{partial}{partial y}(frac{y}{x} - y^2 x^2) 2/x - 2y^2)

(frac{partial}{partial x}(-x/y^2 - 4y^3) 2/x - 2y^2)

Since both partial derivatives are equal, we proceed to find an integrating factor, (mu(x, y)) to make the equation exact.

Integrating Factor and Solution

Let's find an integrating factor, (mu(y)) that depends only on (y):

(frac{partial mu}{partial y} mu(frac{partial M}{partial y} - frac{partial N}{partial x}))

simplifies to:

(frac{partial mu}{partial y} -2ymu)

Integrating both sides, we get:

(ln(mu) -y^2 C)

(mu(y) e^{-y^2})

Applying this integrating factor to the equation, we have:

(e^{-y^2}(frac{y}{x} - y^2 x^2)dx e^{-y^2}(-frac{x}{y^2} - 4y^3)dy 0)

Since the equation is now exact, we can search for the potential function, (psi(x, y)), such that:

(frac{partial psi}{partial x} e^{-y^2}(frac{y}{x} - y^2 x^2))

(frac{partial psi}{partial y} e^{-y^2}(-frac{x}{y^2} - 4y^3))

Integrating the first equation, we get:

(psi(x, y) ln(x) - y^2frac{x^2}{2} frac{x^3}{3} f(y))

By differentiating with respect to (y), we find (f(y)):

(frac{partial psi}{partial y} -2y(x - frac{x^2}{2}) f'(y) -frac{x}{y^2} - 4y^3)

Thus, (f'(y) -4y^3) and (f(y) -y^4).

(psi(x, y) ln(x) - y^2frac{x^2}{2} frac{x^3}{3} - y^4 C)

The solution to the differential equation is:

(ln(x) - y^2frac{x^2}{2} frac{x^3}{3} - y^4 C 0)

(ln(x(y-2)) - frac{x^3}{3} y^4 C)

Alternative Solution Approach

As an alternative, if we rewrite the second term as (x 4xy^4 - 8y^3dy), the solution becomes:

(y^4 frac{x^3}{3} - ln(x(y-2)) C)

This process demonstrates the comprehensive methodology for solving non-exact differential equations, including testing for exactness, finding integrating factors, and integrating the potential function to yield the general solution.

Conclusion

Understanding and solving non-exact differential equations is an essential skill for students and professionals in mathematics, physics, and engineering. The article provided a detailed solution process and two possible solutions to the given problem, demonstrating the importance of methodical and systematic approaches to solving complex equations.