Problem 14 - Advanced Engineering Mathematics

Question

In Problems \(1-16\), the indicated function \(y_{1}(x)\) is a solution of the given equation. Use reduction of order or formula (5), as instructed, to find a second solution \(y_{2}(x)\). $$ \begin{array}{ll} x^{2} y^{\prime \prime}-3 x y^{\prime}+5 y=0 ; & y_{1}=x^{2} \cos (\ln x) \end{array} $$

Step-by-Step Solution

Answer

Use reduction of order with assumed form \( y_2 = v(x) x^2 \cos(\ln x) \).

Step 1: Define Differential Equation

The given differential equation is \( x^2 y'' - 3x y' + 5y = 0 \). We are also given that \( y_1 = x^2 \cos(\ln x) \) is a solution.

Step 2: Use Reduction of Order

For reduction of order, assume a second solution of the form \( y_2 = v(x) y_1 = v(x) x^2 \cos(\ln x) \). We need to find the form of \( v(x) \).

Step 3: Compute Derivatives of Assumed Solution

Calculate the first and second derivatives of \( y_2 = v(x) x^2 \cos(\ln x) \). Let \( u = x^2 \cos(\ln x) \), then \( y_2 = v(x)u \). Find \( y_2' = v'u + vu' \) and \( y_2'' = v''u + 2v'u' + vu'' \).

Step 4: Substitute into Differential Equation

Substitute \( y_2'', y_2', \) and \( y_2 \) back into the original differential equation. Simplify each term and focus on terms involving derivatives of \( v(x) \).

Step 5: Simplify and Solve for \( v(x) \)

Obtain and solve the resulting differential equation for \( v(x) \). Since \( y_1 \) is a solution, terms will cancel out, leading to a simpler equation in \( v \). Integrate the result(s) as needed.

Step 6: Express the Second Solution

Once \( v(x) \) is found, express \( y_2(x) \) in terms of known functions. Often, \( v(x) \) will include simple functions like exponentials or logarithms, leading to \( y_2 \).

Key Concepts

Second Solution

When working with differential equations, finding a second solution is often crucial. This allows us to construct a general solution, which is a combination of all possible specific solutions to the differential equation. In the reduction of order technique, knowing one solution of the equation helps us find another.

To find the second solution, we start with the given first solution, in this problem, it is \( y_1 = x^2 \cos(\ln x) \). We assume the second solution \( y_2(x) \) can be expressed as \( y_2 = v(x)y_1 = v(x)x^2 \cos(\ln x) \), where \( v(x) \) is a function that needs to be determined.

By substituting this expression for \( y_2 \) into the original differential equation, and ensuring all derivatives are appropriately calculated, we can focus on finding \( v(x) \). This strategic method simplifies the complex process of finding all solutions, giving us the needed second solution when the first one is already known.

Differential Equation

A differential equation contains derivatives of a function and helps model various physical phenomena, like motion, heat, or sound. The given equation is \( x^2 y'' - 3x y' + 5y = 0 \).

The goal is to find the function \( y(x) \) that satisfies this equation. When approached correctly, solutions can often be expressed in terms of elementary functions.

• **Homogeneous Equations:** Equations like the one given are called homogeneous linear differential equations. They don't have terms without the dependent variable or its derivatives.
• **Order of the Equation:** This differential equation is second-order because it involves the second derivative, \( y'' \).
• **Constant Coefficients:** In this problem, the coefficients are either constants or involve the independent variable \( x \), making them variable coefficients.

The process typically involves identifying known solutions and using methods like reduction of order or integration to find the function \( y(x) \) that satisfies the full equation.

Derivative Calculation

Calculating derivatives is an essential part of solving differential equations, as they are central to the setup of the equations. Here, \( y_2 = v(x)x^2 \cos(\ln x) \) is differentiated to help find \( v(x) \). Breaking down each part, we differentiate using the product rule.

The product rule gives us:- First derivative: \( y_2' = v'u + vu' \), where \( u = x^2 \cos(\ln x) \) and \( u' \) is its derivative.- Second derivative: \( y_2'' = v''u + 2v'u' + vu'' \)Each part of \( y_2' \) and \( y_2'' \) needs careful computation, especially for the second derivative which includes terms like \( v'' \), \( v' \), and other derivatives.

Once the derivatives are computed, they are substituted back into the differential equation. Substituting these into the equation simplifies matters, focusing on \( v(x) \) terms. Finding derivatives accurately is vital since they dictate the behavior and nature of the differential equation's solutions.