Problem 13 - Advanced Engineering Mathematics

Question

Determine whether the given differential equation is exact. If it is exact, solve it. $$ x \frac{d y}{d x}=2 x e^{x}-y+6 x^{2} $$

Step-by-Step Solution

Answer

The differential equation is not exact.

Step 1: Rewrite the Differential Equation

First, rewrite the given differential equation: \[ x\frac{dy}{dx} = 2x e^x - y + 6x^2 \] in the standard form of an exact equation. Move all terms to one side:\[ x\frac{dy}{dx} + y = 2x e^x + 6x^2 \] This takes the form \[ M(x,y) \, dx + N(x,y) \, dy = 0 \] where \[ M = -(2x e^x + 6x^2) \] and \[ N = x \].

Step 2: Check for Exactness

To verify that the equation is exact, check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).Compute the derivatives:- \( \frac{\partial M}{\partial y} = 0 \)- \( \frac{\partial N}{\partial x} = 1 \)Since \( \frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x} \), the differential equation is not exact.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. They are used to describe various physical and natural phenomena such as heat, sound, electricity, and more.

There are different types of differential equations, including ordinary, partial, linear, and non-linear. In this exercise, we are dealing with an **ordinary differential equation** (ODE), which consists of a function of one independent variable and its derivatives.

The given equation is: 1. \( x \frac{dy}{dx} = 2x e^x - y + 6x^2 \)

To solve an ODE, it is often necessary to perform certain operations to either separate variables or transform the equation into an easier-to-solve form.

A specific type of ODE is the **exact differential equation**. It holds a specific property that allows it to be integrated directly, which makes the solution process straightforward. Understanding whether or not an equation is exact is crucial for determining the method required for solving the problem.

Partial Derivatives

Partial derivatives are used when dealing with functions of more than one variable. They measure the rate at which a function changes as one of its variables changes, keeping the other variables constant.

In the context of exact differential equations, we specifically compute partial derivatives to verify a condition called the **exactness condition**.

Consider equations in the form \( M(x,y) \, dx + N(x,y) \, dy = 0 \). We need to calculate the following partial derivatives: - \( \frac{\partial M}{\partial y} \) - \( \frac{\partial N}{\partial x} \)

These derivatives help us check the criteria required for exact differential equations. If these are equal, then the differential involves a potential function, whose derivatives are \( M(x,y) \) and \( N(x,y) \) respectively.

Exactness Condition

The exactness condition for differential equations is used to identify whether an equation can be integrated directly across its entire domain. This condition allows us to know if there exists a function where its differential is the given equation.

To determine exactness, for an equation \( M(x,y) \, dx + N(x,y) \, dy = 0 \), the condition is: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \]

In the exercise, we computed these two derivatives:
- \( \frac{\partial M}{\partial y} = 0 \)
- \( \frac{\partial N}{\partial x} = 1 \)

Since they are not equal, the equation is not exact. If this condition was fulfilled, we could solve the equation by finding a potential function \( \psi(x,y) \), such that: \( \frac{\partial \psi}{\partial x} = M(x,y) \) and \( \frac{\partial \psi}{\partial y} = N(x,y) \).

Recognizing whether an equation is exact or not, helps select appropriate methods for solving—either through direct integration or seeking alternative transformations.