Problem 13 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation by undetermined coefficients.In Problems \(1-26\) solve the given differential equation by undetermined coefficients. $$y^{\prime \prime}-16 y=2 e^{4 x}$$

Step-by-Step Solution

Answer

The solution is: \( y = C_1e^{4x} + C_2e^{-4x} + \frac{1}{4}xe^{4x} \).

Step 1: Write the Differential Equation

The given differential equation is \( y'' - 16y = 2e^{4x} \). Our task is to solve it by using the method of undetermined coefficients.

Step 2: Solve the Homogeneous Equation

First, we solve the homogeneous equation \( y'' - 16y = 0 \). The characteristic equation is \( r^2 - 16 = 0 \), which factors as \( (r - 4)(r + 4) = 0 \). Thus, the roots are \( r = 4 \) and \( r = -4 \). The solution to the homogeneous equation is \( y_h = C_1e^{4x} + C_2e^{-4x} \), where \( C_1 \) and \( C_2 \) are constants.

Step 3: Determine the Particular Solution Form

The right-hand side of the equation is \( 2e^{4x} \). Since \( e^{4x} \) is already a solution to the homogeneous equation, we use the form \( y_p = Ax e^{4x} \) for the particular solution, where \( A \) is a constant to be determined.

Step 4: Compute Derivatives of the Particular Solution

Calculate the first and second derivatives of \( y_p = Ax e^{4x} \). First derivative: \( y_p' = A(e^{4x} + 4xe^{4x}) = Ae^{4x} + 4Axe^{4x} \). Second derivative: \( y_p'' = A(4e^{4x} + 16xe^{4x}) + 4A(4xe^{4x}) = 16Axe^{4x} + 8Ae^{4x} \).

Step 5: Substitute Particular Solution into the Differential Equation

Substitute \( y_p \) into the original differential equation to find \( A \). We have: \( 16Axe^{4x} + 8Ae^{4x} - 16(Axe^{4x}) = 2e^{4x} \), which simplifies to \( 8Ae^{4x} = 2e^{4x} \).

Step 6: Solve for Coefficient A

Equalizing the coefficients gives \( 8A = 2 \), so \( A = \frac{1}{4} \). The particular solution is \( y_p = \frac{1}{4}xe^{4x} \).

Step 7: Write the General Solution

Combine the homogeneous and particular solutions to get the general solution: \( y = y_h + y_p = C_1e^{4x} + C_2e^{-4x} + \frac{1}{4}xe^{4x} \).

Key Concepts

Undetermined Coefficients

The method of undetermined coefficients is a popular technique for finding particular solutions to non-homogeneous linear differential equations. It works best when the non-homogeneous term on the right side of the equation is a simple function like polynomials, exponentials, or sines and cosines. To use this method, we assume a particular form for the solution based on the type of function. For example:

• If the right-hand side is a polynomial, the particular solution form is also a polynomial of the same degree.
• If it is an exponential, like in our problem, the particular solution also has an exponential term, possibly multiplied by an additional polynomial if the term already solves the homogeneous equation.

Before moving on, ensure the assumed solution form is valid by checking whether it conflicts with solutions of the homogeneous equation. If it does, modify it by multiplying with an appropriate power of x.

Homogeneous Equation

A homogeneous equation in the context of differential equations is one without a non-homogeneous term on the right side, or basically where the right side equals zero. In our given equation, the homogeneous part is obtained by setting the equation equal to zero: \[ y'' - 16y = 0 \] To solve this, we look for solutions of the form \( y = e^{rx} \), which leads directly to the characteristic equation. Solving this equation provides solutions that typically include exponential terms. In this equation, our characteristic equation is \( r^2 - 16 = 0 \). Solving it, we have roots \( r = 4 \) and \( r = -4 \), which means our general solution to the homogeneous equation is a combination:

• \( y_h = C_1 e^{4x} + C_2 e^{-4x} \)

These solutions form the complementary part of the overall solution.

Particular Solution

The particular solution represents a specific function that satisfies the non-homogeneous differential equation. Here, our goal is to find a particular solution, \( y_p \), which solves the equation: \[ y'' - 16y = 2e^{4x} \] To find \( y_p \), first analyze the form of the non-homogeneous term \( 2e^{4x} \) and guess a form for \( y_p \). However, because \( e^{4x} \) is already part of the solution to the homogeneous equation, we use \( y_p = Ax e^{4x} \) to ensure that the particular solution is independent of the homogeneous solution. Find the derivatives, substitute back, and equate coefficients to solve for \( A \). Here, this method gives us \( A = \frac{1}{4} \). Thus, our particular solution is:

• \( y_p = \frac{1}{4} xe^{4x} \)

Characteristic Equation

To solve a homogeneous linear differential equation, we use a mathematical tool called the characteristic equation. This equation helps us find the values of \( r \) (roots) that satisfy the homogeneous part of our differential equation. For our differential equation:

• \( y'' - 16y = 0 \)
• The characteristic equation is formed by replacing derivatives with the notation \( r \). Hence, we have \( r^2 - 16 = 0 \).

Solving the characteristic equation is straightforward: its roots, \( r = 4 \) and \( r = -4 \), determine the solution for the homogeneous equation. These roots are used to form the complementary function:

• \( y_h = C_1 e^{4x} + C_2 e^{-4x} \)

By finding the characteristic roots, we can write the homogeneous solution, which, when combined with the particular solution, gives the general solution of the differential equation.