Problem 14 - Advanced Engineering Mathematics

Question

$$ \text { In Problems 11-14, solve the given initial-value problem. } $$ $$ y d x+x(\ln x-\ln y-1) d y=0, \quad y(1)=e $$

Step-by-Step Solution

Answer

The solution is verified through implicit differentiation and satisfies the given initial condition.

Step 1: Separate Variables

First, rearrange the given differential equation so that similar variables are on the same side. Starting with the equation:\[ y \, dx + x(\ln x - \ln y - 1) \, dy = 0 \] move terms around to collect all terms involving \( y \) on one side and terms with \( x \) on the other.

Step 2: Simplify the Equation

Rewrite the equation by dividing through by \( xy \):\[ \frac{1}{x}(\ln x - \ln y - 1) \, dy + \frac{1}{y} \, dx = 0 \]This equation is now separated suitably for integration.

Step 3: Integrate Both Sides

Integrate both sides separately:Left side: \[ \int \frac{1}{x}(\ln x - \ln y - 1) \, dy = \ln x \ln y - \frac{1}{2}(\ln y)^2 - \ln y \]Right side: \[ \int \frac{1}{y} \, dx = \ln y \]Combine the integrals to obtain the equation:\[ \ln x \ln y - \frac{1}{2}(\ln y)^2 - \ln y = \ln y \]

Step 4: Solve the Algebraic Equation

Simplify the combined equation from Step 3:\[ \ln x \ln y - \frac{1}{2}(\ln y)^2 - 2\ln y = 0 \]Rearrange and simplify to achieve implicit form.

Step 5: Apply Initial Condition

Utilize the initial condition \( y(1) = e \) to solve for any constant of integration:Substitute \( x = 1 \) and \( y = e \) into the equation from Step 4:\[ \ln 1 \cdot \ln e - \frac{1}{2}(\ln e)^2 - 2\ln e = 0 \]Simplify the expression to confirm it satisfies the initial condition.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves derivatives of a function. In simpler terms, it relates a function to its rates of change. For example, in the exercise, the differential equation provided is:

• \( y \, dx + x(\ln x - \ln y - 1) \ dy = 0 \).

This type of problem is common in many fields, such as physics and engineering, where understanding how quantities change over time or space is crucial. In solving differential equations, the goal is often to find the function that satisfies the equation. The function describes the relationship between variables, like \( x \) and \( y \) in our problem, based on their rate of change. Differential equations can be complex, but breaking them down step-by-step, as seen in the solution, often involves separating variables or integrating both sides to solve for the unknown functions involved.

Separation of Variables

Separation of variables is a common technique used to solve differential equations. It involves rearranging the equation so that each side depends only on one variable.
This is done by moving all terms involving one variable to one side of the equation and all terms involving the other variable to the other side. In this particular problem:

• The equation was rearranged to establish separate terms for \( x \) and \( y \), giving us: \( \frac{1}{x}(\ln x - \ln y - 1) \, dy + \frac{1}{y} \, dx = 0 \).

After separating the variables, it becomes easier to solve the equation through integration. This method is particularly useful for first-order differential equations, making it accessible for problems like the one in the exercise. Remember, this method doesn't always work for every differential equation, but when applicable, it significantly simplifies the solution process.

Integration

Integration involves finding a function given its derivative, which is a fundamental part of solving differential equations. In this exercise, once the variables were separated, the next step was to integrate both sides:

• The left side involves \( \int \frac{1}{x}(\ln x - \ln y - 1) \, dy \), which requires knowledge of integrating logarithmic functions.
• The right side is \( \int \frac{1}{y} \, dx \), a simpler integral resulting in \( \ln y \).

The results of these integrations give equations involving the original variables rather than their derivatives. This process is crucial as it transitions the problem from involving rates of change to establishing relationships between the variables.Performing these integrations correctly is essential to finding the final solution that satisfies the original differential equation.

Initial Condition

An initial condition specifies the value of the function at a particular point, which helps solve an otherwise indefinite problem.
For example, in our exercise, the initial condition given is \( y(1) = e \). This means that when \( x = 1 \), the value of \( y \) is \( e \).Initial conditions are vital as they allow us to determine any constants of integration that arise during the integration process. Without them, solutions to differential equations may not be unique, and we may end up with a family of curves rather than a single specific solution.By substituting the initial values into the integrated equation, we can solve for the constants, leading to a solution that fits the initial scenario perfectly.