Problem 65 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$y^{\prime \prime}-64 y=16, \quad y(0)=1, y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The particular solution is \( y(x) = \frac{5}{8} e^{8x} + \frac{5}{8} e^{-8x} - \frac{1}{4} \).

Step 1: Write the given differential equation

We are given the differential equation \( y'' - 64y = 16 \). This is a linear second-order differential equation with constant coefficients.

Step 2: Find the homogeneous solution

First, solve the homogeneous equation \( y'' - 64y = 0 \). Assume a solution of the form \( y_h = e^{rx} \). Find the characteristic equation: \( r^2 - 64 = 0 \), which factors to \( (r - 8)(r + 8) = 0 \). Thus, the roots are \( r = 8 \) and \( r = -8 \). The general solution of the homogeneous equation is: \( y_h = C_1 e^{8x} + C_2 e^{-8x} \).

Step 3: Find the particular solution

To find a particular solution \( y_p \), we note that the non-homogeneous term is a constant (16). Assume a particular solution of the form \( y_p = A \). Substitute into the differential equation: \( 0 - 64A = 16 \), which gives \( A = -\frac{1}{4} \). Thus, the particular solution is \( y_p = -\frac{1}{4} \).

Step 4: Write the general solution of the differential equation

The general solution of the differential equation is:\[ y(x) = y_h + y_p = C_1 e^{8x} + C_2 e^{-8x} - \frac{1}{4} \].

Step 5: Apply initial conditions to find constants

Use the initial conditions to find \( C_1 \) and \( C_2 \). Using \( y(0) = 1 \): \[ C_1 + C_2 - \frac{1}{4} = 1 \quad \Rightarrow \quad C_1 + C_2 = \frac{5}{4} \].Using \( y'(0) = 0 \), first find \( y'(x) = 8C_1 e^{8x} - 8C_2 e^{-8x} \).Substitute \( x = 0 \): \[ 8C_1 - 8C_2 = 0 \quad \Rightarrow \quad C_1 = C_2 \].

Step 6: Solve the system of equations for constants

We have two equations: 1. \( C_1 + C_2 = \frac{5}{4} \) 2. \( C_1 = C_2 \) Substituting \( C_1 = C_2 \) into the first equation gives:\( 2C_1 = \frac{5}{4} \) Thus, \( C_1 = C_2 = \frac{5}{8} \).

Step 7: Write the particular solution with constants

Substitute \( C_1 \) and \( C_2 \) back into the general solution:\[ y(x) = \frac{5}{8} e^{8x} + \frac{5}{8} e^{-8x} - \frac{1}{4} \].

Key Concepts

Initial Value Problems

Initial value problems involve finding a solution to a differential equation that satisfies given conditions at the start. These initial conditions determine the specific constants in the general solution of the differential equation. In our case, the initial-value problem is:

• The differential equation: \( y'' - 64y = 16 \).
• Initial conditions: \( y(0) = 1 \) and \( y'(0) = 0 \).

The initial conditions facilitate finding the unique solution by providing specific information about the function and its derivative at a certain point. By applying these conditions, we solve for any unknown constants in the general solution, resulting in a particular solution that fits all parts of the problem.

Homogeneous and Particular Solutions

Differential equations often consist of two key parts: the homogeneous solution and the particular solution.

• The homogeneous solution \( y_h \) is found by setting the non-homogeneous part (right-hand side of the equation) to zero. For our problem \( y'' - 64y = 0 \), we use the characteristic equation to find \( y_h \) as \( C_1 e^{8x} + C_2 e^{-8x} \).
• The particular solution \( y_p \) addresses the non-homogeneous part of the differential equation. When the non-homogeneous part is a constant, like 16, we assume \( y_p = A \) and find \( A \) to be \(-\frac{1}{4}\) by substituting back into the original equation.

Combining these solutions gives the general solution to the differential equation.

Characteristic Equation

The characteristic equation is a pivotal step in solving linear differential equations with constant coefficients. It emerges from considering a solution of the form \( y = e^{rx} \) for the homogeneous part. For \( y'' - 64y = 0 \), substituting \( y = e^{rx} \) leads to the characteristic equation: \[ r^2 - 64 = 0 \] Factoring this gives: \[ (r - 8)(r + 8) = 0 \] Which then results in roots \( r = 8 \) and \( r = -8 \). The general solution of the homogeneous equation, \( y_h = C_1 e^{8x} + C_2 e^{-8x} \), is thus derived from these roots. These roots indicate the exponential terms present in the homogeneous solution.

Constant Coefficients

In the realm of differential equations, constant coefficients simplify the process of finding solutions. They signify that terms like \( a \), \( b \), and \( c \) in a differential equation like \( ay'' + by' + cy = g(x) \) remain unchanged with respect to \( x \). For the given problem, the differential equation\( y'' - 64y = 16 \) has constant coefficients \( a = 1 \), \( b = 0 \), and \( c = -64 \). These constants facilitate the identification and solving of the characteristic equation, as the process for these operations is straightforward due to their simplicity and uniformity. Did you know? Constant coefficients help in yielding solutions that involve basic exponential functions rather than more complex, varying functions.