Question

Verify that the following functions are solutions to the given differential equation.\(y=4+\ln x\) solves \(x y^{\prime}=1\)

Short answer

The function \( y = 4 + \ln x \) solves the differential equation \( x y' = 1 \).

Step-by-step solution

Find the Derivative of the Given Function

First, we need to find the derivative of the function given by \( y = 4 + \ln x \). The derivative of \( y \) with respect to \( x \) is \( y' = \frac{d}{dx}(4 + \ln x) = 0 + \frac{1}{x} = \frac{1}{x} \).

Substitute the Derivative into the Differential Equation

Next, substitute \( y' = \frac{1}{x} \) into the left side of the differential equation \( x y' = 1 \). This becomes: \[ x \times \frac{1}{x} = 1. \]

Simplify the Equation

Simplify the expression from Step 2: \( x \times \frac{1}{x} = 1 \) simplifies to \( 1 = 1 \).

Verify the Solution

Since the simplified equation \( 1 = 1 \) holds true, this verifies that the function \( y = 4 + \ln x \) is indeed a solution to the differential equation \( x y' = 1 \).

Key concepts

Understanding the Natural Logarithm

The natural logarithm is denoted by \( \ln \) and is an important concept in calculus and mathematical analysis. It's the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm has unique properties and is the inverse of the exponential function.

• When we see an expression like \( \ln x \), it represents the power to which \( e \) must be raised to achieve \( x \).
• One important property is \( \ln(e) = 1 \) because \( e^1 = e \).
• Similarly, \( \ln(1) = 0 \) since \( e^0 = 1 \).

The function \( y = 4 + \ln x \) shows the natural logarithm combined with a constant, meaning that any logarithmic behavior of \( y \) depends on the behavior of \( \ln x \). This combination can shift the graph of the logarithmic function upward by 4 units.

Calculating and Understanding Derivatives

Derivatives measure how a function changes as its input changes. They are foundational in calculus for understanding rates of change and finding tangent lines to curves.To find the derivative of a function like \( y = 4 + \ln x \):

• First, remember that the derivative of a constant (like 4) is 0. It doesn't change as \( x \) changes.
• The derivative of \( \ln x \), a basic function, is \( \frac{1}{x} \). This rule comes from the limit definition of a derivative and is fundamental in calculus.

So, the derivative of \( y = 4 + \ln x \) is \( y' = \frac{1}{x} \). This derivative tells us how sharply or gently our function \( y \) changes as \( x \) increases, with steeper slopes as \( x \) decreases.

Verifying Solutions to Differential Equations

Verifying a solution involves proving that a given function satisfies a specific differential equation. This process ensures the function correctly models the relationship presented by the differential equation.For \( x y' = 1 \):

• Start with finding the derivative of the proposed solution, \( y = 4 + \ln x \), as we did earlier: \( y' = \frac{1}{x} \).
• Next, substitute this derivative back into the given differential equation. Replace \( y' \) to get \( x \cdot \frac{1}{x} \).
• Simplifying \( x \cdot \frac{1}{x} \) yields 1, which matches the right side of the equation \( 1 = 1 \).

This verification shows that every step holds true and confirms that the original function is indeed a solution. This method is crucial for validating mathematical models in science and engineering.