Problem 8 - Calculus and its applications

Question

Show that \(y=x \ln x-5 x+7\) is a solution of \(y^{\prime \prime}-\frac{1}{x}=0\)

Step-by-Step Solution

Answer

Yes, \( y = x \ln x - 5x + 7 \) is a solution because it satisfies the differential equation.

Step 1: Differentiate y to find y'

First, differentiate the function \( y = x \ln x - 5x + 7 \) with respect to \( x \). To differentiate \( x \ln x \), use the product rule: \( \frac{d}{dx} [x \ln x] = \ln x + 1 \). The derivative of \(-5x\) is \(-5\), and the derivative of a constant (7) is 0. Thus, \( y' = \ln x + 1 - 5 \). So, \( y' = \ln x - 4 \).

Step 2: Differentiate y' to find y''

Now, differentiate \( y' = \ln x - 4 \) with respect to \( x \). The derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of a constant (-4) is 0. Therefore, \( y'' = \frac{1}{x} \).

Step 3: Verify y'' satisfies the given differential equation

Substitute \( y'' = \frac{1}{x} \) into the differential equation \( y'' - \frac{1}{x} = 0 \). You get \( \frac{1}{x} - \frac{1}{x} = 0 \). This simplifies to \( 0 = 0 \), showing that the equation is satisfied.

Key Concepts

Calculus

Calculus is a branch of mathematics that explores how things change. It's founded on two major concepts: differentiation and integration. For our exercise, we delve into differentiation, which involves finding the rate at which a function changes at any given point.

In simple terms, differentiation provides us with a derivative that can tell us how steep a curve is at any point or how the slope of a line is changing. This is particularly useful in real-world applications like finding acceleration in physics, where you start from a velocity equation and want to find how speed changes over time.

• In calculus, differentiation is graphically represented as the slope of the tangent line to the curve of a function.
• It is calculated using various rules or formulas, depending on the function's structure.

Mastering calculus basics is key to unraveling more advanced topics. Once comfortable with how functions change and grow, you open the door to solving complex differential equations, much like the one in our exercise.

Differentiation

Differentiation is a fundamental technique in calculus used to compute the derivative of a function. A derivative is the measure of how a function's output changes with respect to a change in its input. It's like finding out how fast something is changing right at a particular instant.

When you differentiate a function, you apply rules that simplify the process. In the given problem, forming derivatives helps to demonstrate that \( y = x \ln x - 5x + 7 \) is a solution to the differential equation \( y'' - \frac{1}{x} = 0 \). Here are a few key differentiation rules used:

• The power rule for simple algebraic expressions.
• The product rule when differentiating functions like \( x \ln x \).
• Constant rule where the derivative of any constant is zero.

Understanding these rules aids in quickly obtaining derivatives of even complex functions, making differentiation an essential skill for problem-solving.

Product Rule

When you encounter a function that is a product of two different functions, the product rule is an essential tool in differentiation. It's essentially a formula that allows you to find the derivative of a product of two functions.

The product rule states: If you have a function \( f(x) \cdot g(x) \), its derivative is \( f'(x)g(x) + f(x)g'(x) \). In our exercise, the term \( x \ln x \) needed differentiation, which required the product rule. Here's how it's applied:

• Identify: Recognize \( x \ln x \) is a product of \( x \) and \( \ln x \).
• Apply Rule: Derivative \( \frac{d}{dx} [x \ln x] = 1 \cdot \ln x + x \cdot \left(\frac{1}{x}\right) \).
• Result: This simplifies to \( \ln x + 1 \).

Utilizing the product rule correctly is vital for tackling more intricate functions composed of multiple parts. It efficiently breaks down each part to simplify finding the overall derivative.