Problem 35 - Advanced Engineering Mathematics

Question

In Problems 33-38, solve the given differential equation subject to the indicated conditions. $$ y^{\prime \prime}-y=x+\sin x, y(0)=2, y^{\prime}(0)=3 $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{5}{2} e^{x} + \frac{1}{2} e^{-x} + x - 1 - \frac{1}{2} \sin x \).

Step 1: Solve the Homogeneous Equation

Begin by solving the homogeneous part of the differential equation: \( y'' - y = 0 \). Assume a solution of the form \( y_h = e^{rx} \). This gives \( r^2e^{rx} - e^{rx} = 0 \), resulting in the characteristic equation \( r^2 - 1 = 0 \). The solutions to this equation are \( r = 1 \) and \( r = -1 \), so the general solution of the homogeneous equation is \( y_h = C_1 e^{x} + C_2 e^{-x} \).

Step 2: Find a Particular Solution

To find a particular solution \( y_p \), consider a solution of the form \( y_p = Ax + B + C\sin x + D\cos x \). Substituting this into the original equation yields a system of equations for \( A, B, C, \) and \( D \). Solving these gives \( A = 1, B = -1, C = -\frac{1}{2}, D = 0 \). Thus, the particular solution is \( y_p = x - 1 - \frac{1}{2}\sin x \).

Step 3: Form the General Solution

The general solution of the differential equation combines the homogeneous version and the particular solution. So, \( y = y_h + y_p = C_1 e^{x} + C_2 e^{-x} + x - 1 - \frac{1}{2} \sin x \).

Step 4: Apply Initial Conditions

Use the conditions \( y(0) = 2 \) and \( y'(0) = 3 \) to determine \( C_1 \) and \( C_2 \). Substituting \( x = 0 \) into the general solution gives \( C_1 + C_2 - 1 = 2 \), so \( C_1 + C_2 = 3 \). Differentiating the general solution yields \( y' = C_1 e^{x} - C_2 e^{-x} + 1 - \frac{1}{2} \cos x \). Substituting \( x = 0 \) gives \( C_1 - C_2 + 1 = 3 \), or \( C_1 - C_2 = 2 \). Solving these two equations simultaneously yields \( C_1 = \frac{5}{2}, \) \( C_2 = \frac{1}{2} \).

Step 5: Write the Final Solution

Substitute \( C_1 = \frac{5}{2} \) and \( C_2 = \frac{1}{2} \) into the general solution: \( y = \frac{5}{2} e^{x} + \frac{1}{2} e^{-x} + x - 1 - \frac{1}{2} \sin x \). This is the solution to the differential equation with the specified initial conditions.

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions or "initial values". These conditions are used to find a particular solution to the differential equation. In the given problem, we have the second-order differential equation:

• \( y'' - y = x + \sin x \)

And the initial conditions are provided as:

• \( y(0) = 2 \)
• \( y'(0) = 3 \)

The goal is to find a function \( y(x) \) that satisfies both the differential equation and these initial conditions. By using the initial values, we can determine any arbitrary constants that appear in the general solution of the differential equation. This process ensures that the solution is unique and precisely fits the given scenario.
These initial conditions are crucial as they help in narrowing down the infinite possible solutions to pinpoint the exact one that matches the criteria of the problem.

Homogeneous Equation

A homogeneous equation is a fundamental piece when solving linear differential equations. For our case, the homogeneous equation is obtained by removing the non-homogeneous part from the differential equation, leading to:

• \( y'' - y = 0 \)

The solution involves assuming a solution of the form \( y_h = e^{rx} \). Substituting into the homogeneous equation helps us derive the characteristic equation:

• \( r^2 - 1 = 0 \)

Solving the characteristic equation tells us the type of solutions. Here, we have two distinct real roots \( r = 1 \) and \( r = -1 \), resulting in:

• \( y_h = C_1 e^{x} + C_2 e^{-x} \)

This general solution represents the family of all possible solutions that satisfy the homogeneous part of the differential equation.

Particular Solution

Finding a particular solution is necessary to handle the non-homogeneous part of a differential equation. In our exercise, this involves the right-hand side:

• \( x + \sin x \)

To solve for the particular solution \( y_p \), we assume a similar form of solution that matches the non-homogeneous terms. A suitable guess for this form is:

• \( y_p = Ax + B + C\sin x + D\cos x \)

By substituting \( y_p \) into the original differential equation and solving for constants \( A, B, C, \) and \( D \), we find:

• \( A = 1, B = -1, C = -\frac{1}{2}, D = 0 \)

This gives us the particular solution:

• \( y_p = x - 1 - \frac{1}{2}\sin x \)

The particular solution accounts for the specific non-homogeneous aspects of the differential equation.

Characteristic Equation

The characteristic equation is a pivotal tool in solving linear homogeneous differential equations with constant coefficients. When you assume a solution like \( y_h = e^{rx} \) and plug it into the homogeneous equation, you derive the characteristic equation. For our case, it is:

• \( r^2 - 1 = 0 \)

This equation determines the behavior of the solutions—whether they are real and distinct, real and repeated, or complex. Here, solving \( r^2 - 1 = 0 \) gives roots \( r = 1 \) and \( r = -1 \). This results in the general solution for the homogeneous equation:

• \( y_h = C_1 e^{x} + C_2 e^{-x} \)

Understanding and solving the characteristic equation allows us to grasp the foundational structure of the general solution for homogeneous differential equations.