Problem 15 - Advanced Engineering Mathematics

Question

In Problems 1-26, solve the given differential equation by undetermined coefficients. $$ y^{\prime \prime}+y=2 x \sin x $$

Step-by-Step Solution

Answer

Solve using method of undetermined coefficients to find a particular solution and combine with the homogeneous solution.

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y'' + y = 2x \sin x \). This is a non-homogeneous linear second-order differential equation with constant coefficients.

Step 2: Solve the Homogeneous Equation

First, solve the associated homogeneous equation: \( y'' + y = 0 \). The characteristic equation is \( r^2 + 1 = 0 \), giving complex roots \( r = \pm i \). The general solution to the homogeneous equation is \( y_h = c_1 \cos x + c_2 \sin x \), where \( c_1 \) and \( c_2 \) are constants.

Step 3: Determine the Form of Particular Solution

The inhomogeneous term is \( 2x \sin x \). We use the method of undetermined coefficients to propose a particular solution of the form \( y_p = (Ax + B)x \cos x + (Cx + D)x \sin x \). This choice accounts for the non-polynomial and trigonometric nature of the inhomogeneous term.

Step 4: Compute Derivatives of the Particular Solution

Calculate the first derivative: \( y_p' = [(Ax + B) - (Cx + D)]x \sin x + [(Cx + D) + (Ax + B)]x \cos x + (A + C)x^2 \cos x - (A + C)x^2 \sin x \). Then, compute the second derivative and simplify accordingly.

Step 5: Substitute into the Original Equation

Substitute \( y_p \) and its derivatives into the original equation \( y'' + y = 2x \sin x \). This will produce a new equation where we compare coefficients for each corresponding component (\( \sin x \) and \( x \sin x \), etc.)

Key Concepts

Non-Homogeneous Linear Differential Equations

Non-homogeneous linear differential equations are a class of equations that play an important role in mathematical modeling. They are characterized by an equation that is not equal to zero, differing from homogeneous equations. Their form is typically given by \( y'' + p(x)y' + q(x)y = g(x) \), where \( g(x) \) is the non-zero part known as the non-homogeneous term. The presence of \( g(x) \) implies that the solutions are not simply multiples of a single function.
To solve these equations, we typically employ two primary steps:

• First, find the general solution to the associated homogeneous equation \( y_h \).
• Then, identify a particular solution \( y_p \) to the non-homogeneous equation, which effectively caters to the non-homogeneous component \( g(x) \).

Once these solutions are found, they are summed to give the general solution of the non-homogeneous equation, expressed as \( y = y_h + y_p \).
This methodology allows us to account for external forces or "input" functions that affect the system described by the equation.

Characteristic Equation

The characteristic equation is a crucial component in solving homogeneous linear differential equations with constant coefficients. To find it, we first translate the differential equation into an algebraic equation by assuming solutions of the form \( e^{rt} \), where \( r \) is a constant. This yields an auxiliary equation that is polynomial in nature.
For example, consider the homogeneous differential equation \( y'' + y = 0 \). By substituting \( y = e^{rt} \), we form the characteristic equation \( r^2 + 1 = 0 \).
Solutions to this equation provide the roots, which in turn help us construct the general solution \( y_h \). Here, the roots are \( r = \pm i \), signifying complex roots. As a result, the general solution assumes the form:

• \( y_h = c_1 \cos x + c_2 \sin x \) for complex roots.

This demonstrates how various root types (e.g., real, repeated, complex) guide the formulation of the solution, ensuring it fits the behavior dictated by the differential equation.

Particular Solution

The particular solution \( y_p \) of a non-homogeneous differential equation addresses the component \( g(x) \). It can be found using several methods, but for equations with constant coefficients, the method of undetermined coefficients is commonly used. This involves assuming a form for \( y_p \) that matches the nature of \( g(x) \).
In the exercise, \( g(x) = 2x \sin x \), which suggests a combination of trigonometric and polynomial components.

• Assume \( y_p = (Ax + B)x \cos x + (Cx + D)x \sin x \), which accounts for both \( x \sin x \) and potential overlap with the homogeneous solution.

This method necessitates substituting the assumed \( y_p \) and its derivatives into the differential equation, allowing the determination of coefficients (A, B, C, and D) by comparing terms.
This ensures the particular solution satisfies the differential equation, providing the part of the solution influenced by \( g(x) \).

Second-Order Differential Equation

A second-order differential equation involves a second derivative, indicating the rate of change of the rate of change of the function in question. Its general form is given by \( a y'' + b y' + c y = g(x) \), where \( a \), \( b \), and \( c \) are constants or functions of \( x \), and \( g(x) \) is either zero (making it homogeneous) or non-zero.
This type of equation can model various physical phenomena, such as oscillations, electrical circuits, and mechanical vibrations. The solutions to these equations are called general solutions and encompass both homogeneous and particular parts.

• The homogeneous part captures the natural behavior of the system.
• The particular part accounts for external influences defined by \( g(x) \).

Understanding second-order differential equations is essential for analyzing systems where acceleration or curvature plays a major role in the system's behavior.