Problem 22 - Advanced Engineering Mathematics

Question

Verify that \(y_{1}(x)=x\) is a solution of \(x y^{\prime \prime}-x y^{\prime}+y=0\). Use reduction of order to find a second solution \(y_{2}(x)\) in the form of an infinite series. Conjecture an interval of definition for \(y_{2}(x)\).

Step-by-Step Solution

Answer

The second solution is \(y_{2}(x) = x^2\), valid for all \(x \in (-\infty, \infty)\).

Step 1: Verify the First Solution

Start by checking if \(y_{1}(x) = x\) is a solution of the differential equation \(x y^{\prime \prime} - x y^{\prime} + y = 0\). Compute the derivatives: \(y^{\prime}(x) = 1\) and \(y^{\prime \prime}(x) = 0\). Substitute these into the equation: \(x(0) - x(1) + x = -x + x = 0\). Thus, \(y_{1}(x) = x\) is indeed a solution.

Step 2: Setup for Reduction of Order

To use reduction of order, assume a second solution of the form \(y_{2}(x) = v(x) y_{1}(x) = v(x)x\). Find the derivatives: \(y_{2}^{\prime}(x) = v^{\prime}(x)x + v(x)\) and \(y_{2}^{\prime \prime}(x) = v^{\prime \prime}(x)x + 2v^{\prime}(x)\). Substitute \(y_{2}(x)\) and its derivatives into the differential equation.

Step 3: Simplify the Resulting Equation

Substitute into the differential equation: \(x(v^{\prime \prime}(x)x + 2v^{\prime}(x)) - x(v^{\prime}(x)x + v(x)) + v(x)x = 0\). Simplify the expression to get \(x^2 v^{\prime \prime}(x) = 0\). This reduces to \(v^{\prime \prime}(x) = 0\).

Step 4: Solve the Simplified Equation

Integrate \(v^{\prime \prime}(x) = 0\) twice. The first integration gives \(v^{\prime}(x) = C_1\). The second integration gives \(v(x) = C_1 x + C_2\). Substituting back, we have \(y_{2}(x) = (C_1 x + C_2)x = C_1 x^2 + C_2 x\).

Step 5: Identify the Second Solution Form

Since \(y_{2}(x) = C_1 x^2\) is independent of \(y_{1}(x) = x\) (choosing \(C_2=0\)), we find \(y_{2}(x) = x^2\) as a possible second solution. As it's linearly independent of \(y_1\), it completes a fundamental set of solutions.

Step 6: Determine the Interval of Convergence

Since \(y_1(x) = x\) and \(y_2(x) = x^2\) are simple polynomials, the solutions exist for all \(x\). Therefore, the interval of definition for these solutions is \((-\infty, \infty)\).

Key Concepts

Reduction of Order

Reduction of Order is a method used to find a second solution to a second-order differential equation given a known first solution. This technique is especially useful when dealing with linear differential equations with constant coefficients. The idea is to assume the second solution, \( y_2(x) \), to have the form \( v(x) y_1(x) \), where \( y_1(x) \) is the known solution.

• You start by expressing \( y_2(x) \) as \( v(x) y_1(x) \), which in our exercise becomes \( y_2(x) = v(x)x \).
• Next, calculate the derivatives: \( y_2^{\prime}(x) \) and \( y_2^{\prime \prime}(x) \), which helps in preparing to substitute into the given differential equation.
• The goal is to substitute these expressions back into the original differential equation to form a simplified first-order equation in \( v(x) \). In this exercise, we reach \( x^2 v^{\prime \prime}(x) = 0 \).

Reduction of Order effectively breaks down the complex task of finding an entirely new solution into solving a simpler differential equation for \( v(x) \). Integration then leads to determining further expressions for \( v(x) \), which gives us the desired independent solution.

Solution Verification

Solution Verification is an essential step in solving differential equations. It ensures that the solutions we've found actually satisfy the original equation. This step checks the correctness of potential solutions by substituting them back into the differential equation.

• In the initial step of the problem, \( y_1(x) = x \) was verified as a solution. This was done by substituting \( y_1 \), along with its derivatives, into the differential equation which simplifies to zero, confirming that it satisfies the equation.
• The accuracy of solutions is crucial because even subtle mistakes can lead to incorrect results further down the line when exploring or deriving additional solutions.
• By confirming that each solution works within the confines of the given system, we build confidence in using these solutions to describe the behavior governed by the differential equation.

Verification doesn't only occur with mathematical substitution. It's also a logical check ensuring the mathematical model aligns with theory or expected outcomes.

Interval of Definition

The Interval of Definition is the range of values over which a solution to a differential equation is valid. This concept is important because some solutions might only hold true within certain bounds, depending on both the equation and its solutions.

• In our exercise, both \( y_1(x) = x \) and \( y_2(x) = x^2 \) are polynomial solutions, which are defined for all real numbers because no limitations arise from singularities or undefined behavior.
• Thus, the interval of definition for these solutions is \(( -\infty, \infty)\), meaning that they apply for any real value of \( x \).
• Knowing the interval of definition is important in practical applications since it ensures we're working within the applicable domain where the solution accurately mirrors real-world phenomena or theoretical frameworks.

Understanding the interval helps in mapping out where exactly the mathematical model is applicable, ensuring we don't extrapolate solutions beyond their applicable range.