Problem 32 - A First Course in Differential Equations with Modeling Applications

Question

$$y^{\Solve the given initial-value problem. $$y^{\prime \prime}-y=\cosh x, \quad y(0)=2, y^{\prime}(0)=12$$

Step-by-Step Solution

Answer

The solution is \( y(x) = 7e^x - 5e^{-x} + \frac{1}{2}x^2 \).

Step 1: Find the homogeneous solution

First, solve the associated homogeneous differential equation, which is \[ y'' - y = 0. \] The characteristic equation is \[ r^2 - 1 = 0 \] which factors as \[ (r-1)(r+1) = 0. \] Thus, the roots are \( r = 1 \) and \( r = -1 \). The general solution for the homogeneous equation is \[ y_h(x) = C_1 e^x + C_2 e^{-x}, \] where \( C_1 \) and \( C_2 \) are constants.

Step 2: Find the particular solution

To solve the non-homogeneous differential equation \[ y'' - y = ext{cosh} \, x, \] we use the method of undetermined coefficients. Assume a particular solution of the form \[ y_p(x) = Ax^2 + Bx + C. \] Calculate the derivatives: \[ y'_p(x) = 2Ax + B \] \[ y''_p(x) = 2A. \] Substitute into the differential equation: \[ 2A - (Ax^2 + Bx + C) = ext{cosh} \, x. \] Compare coefficients to match \[ 2A = 1, \] \[ A = 1, \] \[ B = 0, \] and \[ C = 0. \] Thus, the particular solution is \[ y_p(x) = rac{1}{2}x^2. \]

Step 3: Write the general solution

Combine the homogeneous and particular solutions to obtain the general solution of the differential equation: \[ y(x) = y_h(x) + y_p(x) = C_1 e^x + C_2 e^{-x} + rac{1}{2}x^2. \]

Step 4: Use initial conditions to find constants

Apply the initial conditions \( y(0) = 2 \) and \( y'(0) = 12 \). First, using \( y(0) = 2 \): \[ C_1 + C_2 = 2. \] Now calculate \( y'(x) = C_1 e^x - C_2 e^{-x} + x. \) Using \( y'(0) = 12 \): \[ C_1 - C_2 = 12. \] Solve the system of equations: 1. \( C_1 + C_2 = 2 \) 2. \( C_1 - C_2 = 12 \) Add the equations to find \( 2C_1 = 14 \) so \( C_1 = 7 \). Substitute back into the first equation to get \( 7 + C_2 = 2 \) so \( C_2 = -5 \).

Step 5: Write the specific solution

Substitute \( C_1 = 7 \) and \( C_2 = -5 \) back into the general solution: \[ y(x) = 7 e^x - 5 e^{-x} + rac{1}{2}x^2 \] This is the specific solution to the initial-value problem.

Key Concepts

Differential equations

Differential equations are mathematical equations that relate some function with its derivatives. They play a crucial role in describing various physical phenomena like heat, sound, electricity, and even motion. A differential equation involves an unknown function whose derivatives are given as part of the equation. In simpler terms, it's a way to find out how a quantity changes when other specific parameters change.

• They can be first-order or higher-order depending on the highest derivative present.
• Initial-value problems involve solving these equations with initial conditions provided.

For example, in the given exercise, the differential equation is \(y'' - y = \cosh x\) with initial conditions \( y(0) = 2 \) and \( y'(0) = 12 \). Here, we have a second-order differential equation where the derivatives \( y' \) and \( y'' \) indicate the rate of change of \( y \) over time.

Homogeneous solution

The homogeneous solution involves solving a differential equation where the right side of the equation is equal to zero. For the exercise, the homogeneous form is \( y'' - y = 0 \). This means we only consider the terms with the derivative and the function itself, ignoring any external functions like \( \cosh x \).

• These solutions describe the natural behavior of the system without external forces.
• The solution is expressed in terms of exponential functions based on the roots of the characteristic equation.

In our problem, the homogeneous solution uses the exponential functions \( C_1 e^x + C_2 e^{-x} \), where \( C_1 \) and \( C_2 \) are constants to be determined by initial conditions.

Method of undetermined coefficients

This method is a way to find particular solutions to non-homogeneous linear differential equations. In our exercise, the differential equation \( y'' - y = \cosh x \) features an external function, \( \cosh x \), that disrupts the homogeneous equilibrium.

• Assume a form for the particular solution based on the form of the non-homogeneous term, \( \cosh x \) in this case.
• Derive this assumed form and substitute back into the original equation to determine unknown coefficients.

By matching coefficients from both sides of the equation, we determine the constants for the particular solution. Here, the assumed particular solution is \( y_p(x) = \frac{1}{2}x^2 \), achieved by equating coefficients appropriately to solve the problem.

Characteristic equation

The characteristic equation is a pivotal step in solving linear homogeneous differential equations. This equation is derived by substituting an exponential function \( e^{rx} \) into the differential equation. It helps us find solutions that capture the natural response of the system.

• The characteristic equation is obtained by treating \( y'' - y = 0 \) like an algebraic equation.
• For our problem, the characteristic equation \( r^2 - 1 = 0 \) was solved to give roots \( r = 1 \) and \( r = -1 \).

These roots indicate the type of exponential solution we use for the homogeneous part of the solution, which led to the terms \( e^x \) and \( e^{-x} \). Solving this equation correctly is paramount to establishing the homogeneous solution structure.