Problem 8
Question
Determine whether the given differential equation is exact. If it is exact, solve it. $$ \left(1+\ln x+\frac{y}{x}\right) d x=(1-\ln x) d y $$
Step-by-Step Solution
Verified Answer
The differential equation is not exact.
1Step 1: Understanding Exactness
To determine if the differential equation \((1 + \ln x + \frac{y}{x}) dx = (1 - \ln x) dy\) is exact, we need to verify if the equation fits into the form \(M(x, y)dx + N(x, y)dy = 0\), where \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). First, we must express this equation in standard form.
2Step 2: Rewriting the Equation
Rewrite the equation as: \((1 + \ln x + \frac{y}{x}) dx - (1 - \ln x) dy = 0\).Here, \(M(x, y) = 1 + \ln x + \frac{y}{x}\) and \(N(x, y) = -(1 - \ln x)\).
3Step 3: Checking for Exactness
Calculate \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\).\(\frac{\partial M}{\partial y} = \frac{1}{x}\)\(\frac{\partial N}{\partial x} = \frac{d}{dx}[-(1 - \ln x)]= 0.\)Since \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial x}\), the equation is not exact.
Key Concepts
Partial DerivativesDifferential EquationsExactness Condition
Partial Derivatives
Partial derivatives are an essential concept in calculus, especially when dealing with functions of multiple variables. If you have a function that depends on two or more variables, such as \( f(x, y) \), the partial derivative of \( f \) with respect to one of the variables indicates how the function changes as that particular variable changes, while keeping the others constant.
For instance, consider the function \( M(x, y) = 1 + \ln x + \frac{y}{x} \). To find the partial derivative of \( M \) with respect to \( y \), denoted as \( \frac{\partial M}{\partial y} \), you treat \( x \) as a constant and differentiate \( M \) in terms of \( y \). This results in \( \frac{1}{x} \) because the derivative of \( \frac{y}{x} \) with respect to \( y \) is \( \frac{1}{x} \). The \( 1 + \ln x \) terms vanish since they are independent of \( y \).
Similarly, partial derivatives help us analyze how each variable individually impacts the function, which is crucial for solving complex problems in physics, engineering, and economics.
For instance, consider the function \( M(x, y) = 1 + \ln x + \frac{y}{x} \). To find the partial derivative of \( M \) with respect to \( y \), denoted as \( \frac{\partial M}{\partial y} \), you treat \( x \) as a constant and differentiate \( M \) in terms of \( y \). This results in \( \frac{1}{x} \) because the derivative of \( \frac{y}{x} \) with respect to \( y \) is \( \frac{1}{x} \). The \( 1 + \ln x \) terms vanish since they are independent of \( y \).
Similarly, partial derivatives help us analyze how each variable individually impacts the function, which is crucial for solving complex problems in physics, engineering, and economics.
Differential Equations
Differential equations involve functions and their derivatives. They are central to modeling the behavior of systems over time. Essentially, they describe how a quantity changes with respect to change in another quantity. These equations can be ordinary or partial, depending on the type of derivatives involved.
For the problem at hand, we need to analyze whether the given differential equation is exact. This is done by checking if the equation fits the specific form \( M(x, y)dx + N(x, y)dy = 0 \) and whether it satisfies the exactness condition: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
The given equation \((1 + \ln x + \frac{y}{x}) dx = (1 - \ln x) dy\) represents a type of differential equation we can potentially solve once we confirm exactness. However, if the condition doesn't hold, the usual methods for solving exact differential equations won't apply.
For the problem at hand, we need to analyze whether the given differential equation is exact. This is done by checking if the equation fits the specific form \( M(x, y)dx + N(x, y)dy = 0 \) and whether it satisfies the exactness condition: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
The given equation \((1 + \ln x + \frac{y}{x}) dx = (1 - \ln x) dy\) represents a type of differential equation we can potentially solve once we confirm exactness. However, if the condition doesn't hold, the usual methods for solving exact differential equations won't apply.
Exactness Condition
The exactness condition is a crucial concept when solving certain types of differential equations. For an equation \(M(x, y)dx + N(x, y)dy = 0\) to be considered exact, the following must be true: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). This condition ensures that there exists a well-defined function \( \psi(x, y) \) such that \( \frac{\partial \psi}{\partial x} = M \) and \( \frac{\partial \psi}{\partial y} = N \).
In our example, where \(M(x, y) = 1 + \ln x + \frac{y}{x}\) and \(N(x, y) = -(1 - \ln x)\), we calculated the partial derivatives: \( \frac{\partial M}{\partial y} = \frac{1}{x} \) and \( \frac{\partial N}{\partial x} = 0 \). Clearly, these derivatives are not equal, showing that the equation is not exact.
When this condition fails, alternative methods, like finding an integrating factor, may be necessary to solve the equation. The exactness condition thus serves as a quick test for determining the applicable methods for solving differential equations.
In our example, where \(M(x, y) = 1 + \ln x + \frac{y}{x}\) and \(N(x, y) = -(1 - \ln x)\), we calculated the partial derivatives: \( \frac{\partial M}{\partial y} = \frac{1}{x} \) and \( \frac{\partial N}{\partial x} = 0 \). Clearly, these derivatives are not equal, showing that the equation is not exact.
When this condition fails, alternative methods, like finding an integrating factor, may be necessary to solve the equation. The exactness condition thus serves as a quick test for determining the applicable methods for solving differential equations.
Other exercises in this chapter
Problem 8
Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are
View solution Problem 8
Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=\frac{x+3 y}{3 x+y} $$
View solution Problem 8
\(e^{x} y \frac{d y}{d x}=e^{-y}+e^{-2 x-y}\)
View solution Problem 8
Use computer software to obtain a direction field for the given differential equation. By hand, sketch an approximate solution curve passing through each of the
View solution