Solving the Differential Equation: A Step-by-Step Guide

Introduction

Differential equations are a complex but fundamental area of mathematics, widely used in various fields such as physics, engineering, and economics. One specific type of differential equation to consider is the equation of the form M(x, y)dx N(x, y)dy 0. This article will guide you through solving a particular differential equation of this form: Mxy dx (y^4 - x^2)dy 0. We will explore the process of transforming this equation into a solvable form using the method of integrating factors.

Understanding the Problem

The given differential equation is:

Mxy dx (y^4 - x^2)dy 0

where: M(x, y) xy N(x, y) y^4 - x^2 My x Nx -2x

The equation is not exact, as Nx - My -3/y is dependent only on y. This suggests the application of an integrating factor.

Finding the Integrating Factor

An integrating factor is a function μ(x, y) that, when multiplied by the differential equation, makes it exact. In this case, we are dealing with a scenario where the integrating factor is a function of y only, μ(y). The equation for the integrating factor is derived from the condition:

Nx - My / M -3/y

Given that M xy, we can find the integrating factor as:

d(μ(y)) / dy -3/y * μ(y)

This is a separable differential equation for μ(y). Solving this, we get:

μ(y) 1 / y3

Therefore, the modified differential equation is:

(1 / y3)Mxy dx (1 / y3)(y^4 - x^2)dy 0

This simplifies to:

x / y2dx (y - x2 / y3)dy 0

Checking for Exactness

To confirm that the transformed equation is exact, we need to check if:

(x / y2)y (y - x2 / y3)x

Calculating the partial derivatives, we get:

(x / y2)y -2x / y3

(y - x2 / y3)x -2x / y3

Since both partial derivatives are equal, the equation is now exact.

Finding the Potential Function

Let F(x, y) be the potential function such that:

Fx(x, y) x / y2

Fy(x, y) y - x2 / y3

Integrating the first equation with respect to x:

F(x, y) x2 / (2y2) g(y)

Now, differentiate this expression with respect to y:

Fy(x, y) -x2 / y3 g'

Setting this equal to the second given equation:

y - x2 / y3

We find:

g'

Integrating for g(y):

g(y) y2 / 2 C

Thus, the potential function is:

F(x, y) x2 / (2y2) y2 / 2 C

Final Solution

The solution to the differential equation is given by setting the potential function equal to a constant C:

x2 / (2y2) y2 / 2 C

This is the final form of the solution to the given differential equation.