Question

\(y=\ln (x),\left(2 e^{y}-x\right) y^{\prime}=1\). Recall: \(\ln (x)\) is \(\log (\mathrm{x})\) in MATLAB.

Short answer

\( y' = \frac{1}{2e^{y} - x} \)

Step-by-step solution

Identify Given Equation

The exercise provides the function \( y = \ln(x) \) and the differential equation \((2e^{y} - x) y' = 1\). Our goal is to find \( y' \).

Isolate \( y' \)

The differential equation is \((2e^{y} - x) y' = 1\). To isolate \( y' \), divide both sides by \(2e^{y} - x\).

Derive Expression for \( y' \)

After isolating \( y' \), we have: \[ y' = \frac{1}{2e^{y} - x} \]

Key concepts

Logarithmic Functions

Logarithmic functions are mathematical functions used to determine the power to which a number, known as the base, must be raised to achieve another number. In this exercise, the base is the natural number, denoted as "e," which is approximately 2.71828. The function is represented as \( y = \ln(x) \), where "ln" is the natural logarithm. In many programming environments like MATLAB, the natural logarithm is denoted as \( \log(x) \) instead.
Understanding logarithmic functions is crucial for solving differential equations, especially when they involve exponential growth or decay. Logarithms allow us to transform multiplicative relationships into additive ones, making it easier to manipulate equations.

• Natural logarithm \( \ln(x) \) is the inverse operation of the exponential function \( e^x \).
• The domain of \( \ln(x) \) is \( x > 0 \).
• Logarithms are useful in simplifying derivatives involving exponential functions.

Isolating Variables

Isolating variables in an equation means arranging the equation so that one variable is on its own on one side of the equation. This is a critical step when solving for unknown variables. In the original problem, we're given the differential equation \( (2e^{y} - x) y' = 1 \). The goal is to solve for \( y' \), which represents the derivative of \( y \) with respect to \( x \).
To isolate \( y' \), divide both sides of the equation by \( 2e^{y} - x \). This manipulation leads us to the expression:

• Step 1: Identify the term to be isolated (\( y' \)).
• Step 2: Use algebraic operations (division in this case) to move other terms to the opposite side.
• Resulting equation: \( y' = \frac{1}{2e^{y} - x} \).

Isolating variables is a fundamental skill in algebra and calculus. It enables us to solve equations and understand relationships between variables better.

MATLAB Programming

MATLAB is a powerful programming environment used for numerical and computational problems. It is highly useful for solving differential equations like the one given in this exercise. It's built-in functions for logarithms and exponentials make it particularly handy.
For logarithmic functions, MATLAB uses \( \log(x) \) to denote the natural logarithm, not to be confused with the common logarithm \( \log_{10}(x) \). This is important to remember when converting mathematical equations for computation in MATLAB.

• To solve the problem using MATLAB, first represent the equation in a format it can understand.
• MATLAB can numerically find solutions using methods like ODE solvers once \( y' \) is computed.
• Be sure to use function handles and control flow structures for efficient programming.

Understanding how to format and input equations is essential for effective MATLAB programming and for handling differential equations.

Exponential Functions

Exponential functions are characterized by the constant base raised to a variable exponent. In this exercise, the function \( e^y \) appears in the differential equation \( (2e^{y} - x) y' = 1 \).
Exponential functions play a key role in modeling situations involving growth or decay, such as population dynamics or radioactive decay. They are also key components when considering the derivatives of logarithmic functions.

• The function \( e^y \) reflects continuous growth, where \( e \) is Euler's number, approximated by 2.71828.
• Inverse of \( e^y \) is \( \ln(y) \), which helps in solving differential equations analytically.
• Because \( e^y \) appears in the denominator of \( y' = \frac{1}{2e^{y} - x} \), it's crucial to understand its influence on the rate of change represented by \( y' \).

Grasping exponential functions and their properties is essential for solving complex problems in calculus and beyond.