Problem 48 - Calculus - AP Edition

Question

Show that if \(x \neq 0,\) then \(y=1 / x\) satisfies the equation \(x^{3} y^{\prime \prime}+x^{2} y^{\prime}-x y=0\)

Step-by-Step Solution

Answer

Yes, \( y = \frac{1}{x} \) satisfies the equation for \( x \neq 0 \).

Step 1: Understand the Function

Identify the function \( y = \frac{1}{x} \) and its role in the differential equation \(x^3 y'' + x^2 y' - xy = 0\). The goal is to demonstrate that this function satisfies the equation when \(x eq 0\).

Step 2: Compute the First Derivative

Calculate the first derivative of \( y \). Since \( y = \frac{1}{x} \), we apply the power rule: \( y = x^{-1} \). The derivative is given by \( y' = -x^{-2} \).

Step 3: Compute the Second Derivative

Calculate the second derivative of \( y \). Differentiating \( y' = -x^{-2} \), we find \( y'' = 2x^{-3} \).

Step 4: Substitute the Derivatives into the Equation

Substitute \( y = \frac{1}{x} \), \( y' = -x^{-2} \), and \( y'' = 2x^{-3} \) into the differential equation \( x^3 y'' + x^2 y' - xy = 0 \).

Step 5: Simplify the Equation

Substitute the values into the equation: \[x^3(2x^{-3}) + x^2(-x^{-2}) - x(\frac{1}{x}) = 2 + (-1) - 1 = 0.\] The result simplifies to zero, confirming that the equation is satisfied.

Key Concepts

First Derivative

The first derivative of a function gives us the rate at which the function's value is changing at any point with respect to its input variable. For the function given in the exercise, which is expressed as \( y = \frac{1}{x} \), finding the derivative involves using derivative rules.
To compute the first derivative, we apply the power rule. Transform the function into a power form: \( y = x^{-1} \). The power rule tells us that the derivative of \( x^n \) is \( n \cdot x^{n-1} \). Therefore, applying this rule here, we find that:

• \( y' = -1 \cdot x^{-2} \)
• Simplifies to \( y' = -\frac{1}{x^2} \)

This result indicates that the slope of the function \( y = \frac{1}{x} \) decreases as \( x \) increases due to the negative sign and it decreases at a rate that is inversely proportional to the square of \( x \). Understanding the first derivative is crucial to tackling differential equations as it serves as a foundational step before pursuing higher-order derivatives.

Second Derivative

The second derivative provides insights into the curvature or the concavity of a function. It dictates how the rate of change (given by the first derivative) itself is changing. In our problem, after identifying the first derivative as \( y' = -x^{-2} \), we move on to compute the second derivative.
To find this, we differentiate the first derivative:

• The second derivative of \( y' = -x^{-2} \) is calculated using the power rule.
• Differentiate \( -x^{-2} \) to get \( y'' = 2x^{-3} \).

Thus, \[ y'' = 2 \cdot x^{-3} \]This indicates the concavity of the curve described by \( y = \frac{1}{x} \) is upward for all \( x eq 0 \). This information, after calculating both first and second derivatives, is fundamental. They form parts of the differential equation that should be satisfied by the function \( y \).

Satisfying Differential Equations

A differential equation involves derivatives and seeks to find a function that satisfies the given relationship among these derivatives and the functions. Here, our task was to verify that \( y=1/x \) is a solution to the differential equation \( x^3 y'' + x^2 y' - xy = 0 \).
Begin by substituting the function and its derivatives into the equation:

• The second derivative \( y'' = 2x^{-3} \).
• The first derivative \( y' = -x^{-2} \).
• The function itself \( y = \frac{1}{x} \).

Plug these into the original equation:\[ x^3 (2x^{-3}) + x^2 (-x^{-2}) - x (\frac{1}{x}) = 0 \]This simplifies to:\[2 - 1 - 1 = 0 \]Through substitution, the equation simplifies correctly to zero. This confirms that the function satisfies the given differential equation. Understanding this concept is vital when working with differential equations as it ensures that the given function properly fulfills the equation based on its derivatives.