Problem 3 - 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 \(y_{2}(x)\). $$ y^{\prime \prime}+16 y=0 ; \quad y_{1}=\cos 4 x $$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = \frac{1}{4} \sin(4x) \).

Step 1: Identify the Given Differential Equation and Known Solution

The given differential equation is \( y'' + 16y = 0 \) and one known solution is \( y_1(x) = \cos(4x) \). Our task is to find a second solution \( y_2(x) \) that is linearly independent of the first solution \( y_1(x) \).

Step 2: Recognize the Homogeneous Linear Differential Equation

This is a second-order linear homogeneous differential equation with constant coefficients. We see that the characteristic equation is used to determine the solutions to this equation.

Step 3: Use the Reduction of Order Formula

According to reduction of order, if one solution \( y_1(x) \) is known, a second linearly independent solution \( y_2(x) \), can be found using the formula: \[ y_2(x) = y_1(x) \int \frac{e^{-\int P(x) \, dx}}{(y_1(x))^2} \, dx \]Here, the equation is of the form \( y'' + P(x)y' + Q(x)y = 0 \), simplified to our case, \( P(x) = 0 \) since the equation is \( y'' + 16y = 0 \).

Step 4: Substitute and Simplify

Substitute into the formula:\[ y_2(x) = \cos(4x) \int \frac{1}{\cos^2(4x)}\, dx \]This simplifies to:\[ y_2(x) = \cos(4x) \int \sec^2(4x)\, dx \]

Step 5: Integrate the Expression

The integral of \( \sec^2(4x) \) is \( \frac{1}{4}\tan(4x) \). Therefore:\[ y_2(x) = \cos(4x) \cdot \frac{1}{4} \tan(4x) \]Upon simplification, this becomes:\[ y_2(x) = \frac{1}{4} \sin(4x) \]

Step 6: Confirm the Linear Independence

The Wronskian of \( y_1(x) = \cos(4x) \) and \( y_2(x) = \frac{1}{4} \sin(4x) \) should be nonzero:\[ W(y_1, y_2) = \begin{vmatrix} \cos(4x) & \frac{1}{4} \sin(4x) \ -4\sin(4x) & \frac{1}{4} \cdot 4\cos(4x) \end{vmatrix} \]This determinant evaluates to \( 1 \), confirming that \( y_2(x) \) is a valid second solution.

Key Concepts

Differential Equation

A differential equation is an equation that involves the derivatives of a function. Simply put, it relates a function to its rate of change. In this exercise, we deal with a type of differential equation known as a second-order linear homogeneous differential equation:

• **Second-order:** This refers to the highest derivative in the equation, which in this case is the second derivative, denoted as \( y'' \).
• **Linear:** The equation can be written in the form \( a_2(x)y'' + a_1(x)y' + a_0(x)y = 0 \), where \( a_2(x) \), \( a_1(x) \), and \( a_0(x) \) are coefficients that can be functions of \( x \). Here, our coefficients are constants, making it easier to solve.
• **Homogeneous:** This tells us that every term in the equation involves the function \( y \) or its derivatives. There is no additional function on the right-hand side. This generally implies that the solution will pass through the origin if \( y = 0 \) is a solution.

These characteristics help determine the methods used for finding solutions, like the characteristic equation and reduction of order.

Linearly Independent Solution

In differential equations, finding linearly independent solutions is crucial. Linearly independent solutions cannot be expressed as a simple multiple of each other. They allow the complete general solution of a differential equation to be constructed.

Here's why linear independence matters:

• For second-order differential equations like ours, we always look for two linearly independent solutions, \( y_1(x) \) and \( y_2(x) \).
• These solutions form the basis for the solution space, meaning any other solution to the differential equation can be expressed as a linear combination of these two.

The process we use here, called reduction of order, helps find the second linearly independent solution when one is already known.
Knowing \( y_1(x) = \cos(4x) \), we aim to find \( y_2(x) \) that does not overlap it in the solution space, allowing us to span the entire set of solutions.

Characteristic Equation

The characteristic equation is a tool that simplifies finding solutions to differential equations with constant coefficients. It originates from assuming a solution of the form \( y = e^{rx} \), where \( r \) is a constant we need to solve for.

Here's how it works in our exercise:

• Given the differential equation \( y'' + 16y = 0 \), we substitute \( y = e^{rx} \), leading to \( r^2e^{rx} + 16e^{rx} = 0 \).
• This simplifies to the characteristic equation \( r^2 + 16 = 0 \). Solving for \( r \), we find \( r = \pm 4i \), indicating a complex pair.

With complex roots, the general solution is expressed using sine and cosine functions. The initial real solution \( y_1(x) = \cos(4x) \) comes from this. Another complementary solution, typically involving the sine function, can be derived, supporting the problem's use of reduction of order.

Wronskian

The Wronskian is an important determinant used to confirm the linear independence of solutions in differential equations. Here's how it applies:

• The Wronskian of two functions, \( y_1(x) \) and \( y_2(x) \), is calculated as a determinant: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]
• If the Wronskian is non-zero at any point in the interval of interest, the solutions \( y_1 \) and \( y_2 \) are linearly independent, meaning they form a valid basis for the solution space.
• For our solutions, \( y_1(x) = \cos(4x) \) and \( y_2(x) = \frac{1}{4} \sin(4x) \), the Wronskian evaluates to 1. This confirms that these two solutions are indeed linearly independent.

Thus, employing the Wronskian ensures that we've obtained a valid set of solutions for our differential equation, helping conclude the solution process reliably.