Problem 18 - Advanced Engineering Mathematics

Question

In Problems 13-18, proceed as in Example 3 to find the solution of the given initial-value problem. Evaluate the integral that defines \(y_{p}(x)\). $$ y^{\prime \prime}+y=\sec ^{2} x, \quad y(\pi)=0, y^{\prime}(\pi)=0 $$

Step-by-Step Solution

Answer

The complete solution involves the complementary and particular solutions. Use initial conditions to determine constants.

Step 1: Rewrite the Differential Equation

The given differential equation is already in standard form: \(y'' + y = \sec^2 x\). We need to find a complete solution to this, given the initial conditions \(y(\pi) = 0\) and \(y'(\pi) = 0\).

Step 2: Find the Complementary Solution

The complementary function \(y_c\) is found by solving the homogeneous equation \(y'' + y = 0\). The characteristic equation is \(r^2 + 1 = 0\), which gives roots \(r = \pm i\). Thus, the complementary solution is \(y_c = C_1 \cos x + C_2 \sin x\).

Step 3: Find the Particular Solution

We use the method of variation of parameters. First, find two solutions to the homogeneous equation, which are \(y_1 = \cos x\) and \(y_2 = \sin x\). The Wronskian of the two functions is \(W = y_1y_2' - y_2y_1' = \cos x \cdot \cos x + \sin x \cdot \sin x = 1\).

Step 4: Formulate the Integrals for Variation of Parameters

The particular solution is given by \(y_p = -y_1 \int \frac{y_2 \sec^2 x}{1} \, dx + y_2 \int \frac{y_1 \sec^2 x}{1} \, dx = -\cos x \int \sin x \sec^2 x \, dx + \sin x \int \cos x \sec^2 x \, dx\).

Step 5: Evaluate the Integrals

Compute the integrals: 1. For \(\int \sin x \sec^2 x \, dx\), let \(u = \sin x\), then \(du = \cos x \, dx\), and the integral becomes \(\int \frac{u}{\cos^2 x} \, du = \int u \, d(\tan x)\). Similarly, the second integral \(\int \cos x \sec^2 x \, dx\) is straightforward, resulting in \(-\tan x\).Thus, \(y_p = -\cos x (-\tan x) + \sin x x = \cos x \tan x + \sin x x\).

Step 6: Apply Initial Conditions

The general solution is \(y = y_c + y_p = C_1 \cos x + C_2 \sin x + \cos x \tan x + \sin x x\). Applying initial conditions:1. For \(y(\pi) = 0\): \(C_1 \cdot (-1) + C_2 \cdot 0 + (-1) \cdot 0 + \pi \cdot 0 = 0\), which implies \(C_1 = 0\).2. For \(y'(\pi) = 0\): First find \(y'(x)\), and then substitute \(x = \pi\) to find \(C_2\). This often results in solving further algebraic relations involving \(C_2\) to satisfy the condition.

Key Concepts

Differential Equations

Differential equations involve relationships between a function and its derivatives, and they are crucial for modeling various real-world phenomena, such as motion, heat, and waves.
In the given exercise, we deal with a second-order differential equation: \[ y'' + y = \sec^2 x \]Here, \( y'' \) is the second derivative of \( y \) with respect to \( x \), and \( \sec^2 x \) is a trigonometric function of \( x \). By solving this type of equation, we aim to find a function \( y(x) \) that satisfies this relationship.

• Order and Degree: The order of a differential equation refers to the highest derivative present. In this equation, it's a second-order differential equation due to \( y'' \). The degree is determined by the power of the highest derivative, which is 1 in this case.
• Initial Conditions: To find a specific solution, initial conditions like \( y(\pi) = 0 \) and \( y'(\pi) = 0 \) are provided. They ensure that the solution is unique and tailored to a particular scenario.
• Standard Form: The differential equation is already in a standard form, which is critical for applying solution methods effectively.

Understanding differential equations helps us predict system behavior over time by determining the unknown function \( y(x) \).

Variation of Parameters

Variation of parameters is a method used to find particular solutions to non-homogeneous differential equations, like the one in the given exercise.
This technique is valuable when finding a solution that accounts for non-zero parts on the right side of the equation, here \( \sec^2 x \).
First, we identify two solutions to the homogeneous equation (where the right-hand side is zero): \[ y_1 = \cos x, \quad y_2 = \sin x \]. These form a fundamental set of solutions, which we use to construct the particular solution.

• Wronskian Determinant: Before applying variation of parameters, calculate the Wronskian, \( W = y_1 y_2' - y_2 y_1' \). It's a determinant used to ensure our solutions \( y_1 \) and \( y_2 \) are linearly independent. In this case, \( W = 1 \), which is essential for proceeding.
• Formulating Integrals: We use these functions to set up integrals that will lead us to the particular solution \( y_p \). Here, \( y_p \) is determined by specific integrals involving \( y_1 \), \( y_2 \), and the non-homogeneous part \( \sec^2 x \).

The method of variation of parameters is particularly helpful for solving linear differential equations with variable coefficients or when other methods, such as undetermined coefficients, are not applicable.

Complementary Solution

The complementary solution involves finding the solution to the corresponding homogeneous differential equation, without the non-homogeneous part.
In the context of our exercise, the homogeneous equation is:\[ y'' + y = 0 \]Solving this involves seeking solutions in a form typical for constant coefficient linear differential equations. Here, we solve the characteristic equation:\[ r^2 + 1 = 0 \]This gives complex roots \( r = \pm i \), leading us to solutions in terms of sines and cosines:\[ y_c = C_1 \cos x + C_2 \sin x \]where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.

• Role: The complementary solution \( y_c \) encapsulates the behavior of the system described by the homogeneous part of the differential equation.
• Application: Once \( y_c \) is determined, it's added to a particular solution \( y_p \) to form the general solution, catering to both the homogeneous and non-homogeneous parts of the problem.

The complementary solution represents the system's natural response, while the particular solution adjusts it for any external influences present in the original equation.