Problem 7 - A First Course in Differential Equations with Modeling Applications

Question

Solve each differential equation by variation of parameters. $$y^{\prime \prime}-y=\cosh x$$

Step-by-Step Solution

Answer

The general solution is \( y = c_1 e^x + c_2 e^{-x} + \frac{x}{4} e^x - \frac{x}{4} e^{-x} \).

Step 1: Identify the Homogeneous Solution

First, solve the homogeneous equation associated with the given differential equation: \( y'' - y = 0 \). The characteristic equation is \( r^2 - 1 = 0 \), which factors to \( (r-1)(r+1) = 0 \). The roots are \( r = 1 \) and \( r = -1 \). Thus, the general solution to the homogeneous equation is \( y_h = c_1 e^x + c_2 e^{-x} \).

Step 2: Formulate Particular Solution Using Variation of Parameters

To find a particular solution \( y_p \) to the non-homogeneous differential equation, use variation of parameters. The solution takes the form: \[ y_p = u_1(x) e^x + u_2(x) e^{-x} \]where \( u_1' \) and \( u_2' \) are obtained by:\[ u_1' = - \frac{e^{-x} \cdot \, ext{cosh} \, x}{W(e^x, e^{-x})} \]\[ u_2' = \frac{e^x \cdot \, ext{cosh} \, x}{W(e^x, e^{-x})} \]The Wronskian \( W(e^x, e^{-x}) = -2. \)

Step 3: Calculate the Integrals for u₁′ and u₂′

Compute the integrals to find \( u_1(x) \) and \( u_2(x) \):\[ u_1(x) = -\int \frac{e^{-x} \cosh x}{-2} \, dx = \frac{1}{2}\int e^{-x} \cosh x \, dx \]The integral simplifies using the identity \( \cosh x = \frac{e^x + e^{-x}}{2} \):\[ u_1(x) = \frac{1}{2} \int e^{-x} \left(\frac{e^x + e^{-x}}{2}\right) \, dx = \frac{1}{4} \int (1 + e^{-2x}) \, dx = \frac{x}{4} + \frac{1}{8} e^{-2x}\]

Step 4: Compute the Integral for u₂

Now compute \( u_2(x) \):\[ u_2(x) = \int \frac{e^x \cosh x}{-2} \, dx = -\frac{1}{2}\int e^x \cosh x \, dx \]Simplify using \( \cosh x = \frac{e^x + e^{-x}}{2} \):\[ u_2(x) = -\frac{1}{2}\int \left( \frac{e^{2x} + 1}{2} \right) \, dx = -\frac{1}{4} \int (e^{2x} + 1) \, dx = -\frac{1}{8} e^{2x} - \frac{x}{4} \]

Step 5: Construct the Particular Solution

Substitute \( u_1(x) \) and \( u_2(x) \) back into the equation for \( y_p \):\[ y_p = \left( \frac{x}{4} + \frac{1}{8} e^{-2x} \right) e^x + \left( -\frac{1}{8} e^{2x} - \frac{x}{4} \right) e^{-x} \]Simplify:\[ y_p = \frac{x}{4} e^x + \frac{1}{8} - \frac{x}{4} e^{-x} - \frac{1}{8} = \frac{x}{4} e^x - \frac{x}{4} e^{-x} \]

Step 6: Write the General Solution

The general solution of the differential equation is the sum of the homogeneous and particular solutions:\[ y = c_1 e^x + c_2 e^{-x} + \frac{x}{4} e^x - \frac{x}{4} e^{-x} \]

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In simple terms, it describes how a quantity changes over time or space depending on its initial state and the rate of change. Differential equations are essential in modeling real-world phenomena, such as

• physical systems,
• biological processes, and
• economic trends.

When we talk about solving a differential equation, we want to find the function that satisfies this relationship. In our exercise, the differential equation is expressed as: \[ y'' - y = \cosh x \] Here, \( y'' \) represents the second derivative of function \( y \) with respect to \( x \), and \( \cosh x \) signifies the hyperbolic cosine of \( x \).
To solve this differential equation, we need to determine both the homogeneous solution and the particular solution, which we will explore further.

Homogeneous Solution

The homogeneous solution of a differential equation refers to the solution of the equation when all external forces or functions are absent, represented by the equation set equal to zero. For our specific problem, the corresponding homogeneous equation is: \[ y'' - y = 0 \]
This implies that we seek solutions only influenced by their intrinsic characteristics, without any external influence.

By solving the above equation, we derive a characteristic equation in terms of \( r \): \[ r^2 - 1 = 0 \] Solving \( (r-1)(r+1) = 0 \), yields roots \( r = 1 \) and \( r = -1 \). With these roots, the general solution to the homogeneous equation becomes: \[ y_h = c_1 e^x + c_2 e^{-x} \]
In this expression:

• \( c_1 \) and \( c_2 \) are constants that are defined by initial conditions,
• \( e^x \) and \( e^{-x} \) represent the exponential functions corresponding to the roots of the characteristic equation.

The homogeneous solution plays a crucial role when composing the overall solution of the differential equation.

Particular Solution

The particular solution is the part of a differential equation's solution that accounts for the non-homogeneous part, or external driving forces. In our equation, this is represented by the function \( \cosh x \). Variation of parameters is a method used to determine the particular solution.

The particular solution takes the form: \[ y_p = u_1(x) e^x + u_2(x) e^{-x} \]
Here, \( u_1(x) \) and \( u_2(x) \) are functions that we determine by solving two integrals derived from formulas using the Wronskian, resulting in:

• \( u_1'(x) = - \frac{e^{-x} \cosh x}{-2} \)
• \( u_2'(x) = \frac{e^x \cosh x}{-2} \)

By integrating, we find:

• \( u_1(x) = \frac{x}{4} + \frac{1}{8} e^{-2x} \)
• \( u_2(x) = -\frac{1}{8} e^{2x} - \frac{x}{4} \)

This results in the particular solution: \[ y_p = \frac{x}{4} e^x - \frac{x}{4} e^{-x} \]
The particular solution is essential as it fine-tunes the general solution to match the specific conditions imposed by the non-homogeneous element.

Wronskian

The Wronskian is a determinant used primarily in the study of differential equations to determine linear independence of solutions. For two functions, \( f(x) \) and \( g(x) \), the Wronskian \( W(f, g) \) is calculated as: \[ W(f, g) = \begin{vmatrix} f & g \ f' & g' \end{vmatrix} \]
In simpler terms, it is the determinant of a matrix that includes the functions and their first derivatives.

For this particular problem, the Wronskian of \( e^x \) and \( e^{-x} \) is calculated as follows: \[ W(e^x, e^{-x}) = \begin{vmatrix} e^x & e^{-x} \ e^x & -e^{-x} \end{vmatrix} \]
This simplifies to: \[ e^x(-e^{-x}) - (e^{-x}e^x) = -2 \]
The Wronskian is a crucial element in the variation of parameters method as it assists in forming the equations used to compute the particular solution.