Problem 20 - 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. $$y^{\prime \prime}+2 y^{\prime}-24 y=16-(x+2) e^{4 x}$$

Step-by-Step Solution

Answer

The solution to the differential equation is \( y = c_1 e^{4x} + c_2 e^{-6x} - \frac{1}{5}x e^{4x} + \frac{2}{5} \).

Step 1: Identify the Homogeneous Equation

First, solve the homogeneous part of the equation, which is \( y'' + 2y' - 24y = 0 \). This is done by finding the characteristic equation related to it.

Step 2: Solve the Characteristic Equation

The characteristic equation for \( y'' + 2y' - 24y = 0 \) is \( r^2 + 2r - 24 = 0 \). Solving this quadratic equation using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a=1, b=2, c=-24 \). Calculating gives roots: \( r_1 = 4 \), \( r_2 = -6 \).

Step 3: Write the General Solution to the Homogeneous Equation

The solution to the homogeneous equation is given by \( y_h = c_1 e^{4x} + c_2 e^{-6x} \), where \( c_1 \) and \( c_2 \) are arbitrary constants.

Step 4: Propose a Particular Solution

For the non-homogeneous term \( 16 - (x+2) e^{4x} \), propose the particular solution. Since \( e^{4x} \) is part of \( y_h \), use \( y_p = Ax e^{4x} + Bx + C \) as a trial solution for the particular solution.

Step 5: Calculate Derivatives for the Particular Solution

Compute the first and second derivatives: \( y_p' = A(4x + 1)e^{4x} + B \) and \( y_p'' = A(4x + 2)e^{4x} + (4A) e^{4x} \).

Step 6: Substitute Particular Solution into Differential Equation

Substitute \( y_p, y_p' \), and \( y_p'' \) into the original equation \( y'' + 2y' - 24y \) to form an equation involving \( A, B, \) and \( C \). This results in collecting coefficients for \( e^{4x} \), \( x e^{4x} \), and constants.

Step 7: Solve for the Coefficients

Equate coefficients from the left-hand side and right-hand side of the substituted equation: Solve for \( A \), \( B \), and \( C \) based on the respective coefficients. From solving, \( A = -\frac{1}{5} \), \( B = 0 \), and \( C = \frac{2}{5} \).

Step 8: Write the General Solution to the Non-Homogeneous Equation

Combine the homogeneous and particular solutions to get \( y = y_h + y_p = c_1 e^{4x} + c_2 e^{-6x} - \frac{1}{5}x e^{4x} + \frac{2}{5} \).

Key Concepts

Differential Equation

A differential equation is an equation that involves an unknown function and its derivatives. In simpler terms, it is a mathematical tool used to describe how things change. Think about how your position changes over time when you're walking—that's a kind of differential equation!
In the given exercise, the differential equation translates to finding a function, denoted as \( y \), which satisfies both the presence and change expressions described in the mathematical relation:

• \( y'' + 2y' - 24y = 16-(x+2) e^{4x} \)

Here, \( y'' \) denotes the second derivative of \( y \), which reflects the rate of change of the first derivative, \( y' \), showcasing the acceleration or deceleration of the function.
The equation is a combination of two parts: 'homogeneous' and 'non-homogeneous.' Understanding this distinction invites us to explore the solutions in systematically methodical ways.

Homogeneous Solution

In the context of differential equations, the homogeneous solution addresses the equation where the non-variable term is zero. This part of the equation focuses on the intrinsic dynamics of the system, free from external influences.
For the given problem, the homogeneous differential equation is:

• \( y'' + 2y' - 24y = 0 \)

The aim here is to find what is called the 'general solution' to this equation. We achieve this by first determining the characteristic equation from it, predicate on the assumption that potential solutions have the form \( e^{rx} \). The solutions to this characteristic equation not only help us find the roots but also form the basis of the homogeneous solution.
In our worked example, solving the characteristic equation yields roots \( r_1 = 4 \) and \( r_2 = -6 \). Therefore, the homogeneous solution, \( y_h \), is:

• \( y_h = c_1 e^{4x} + c_2 e^{-6x} \)

where \( c_1 \) and \( c_2 \) are constants determined by initial conditions of the problem.

Particular Solution

Contrasted with the homogeneous solution, the particular solution manages the specific adjustments requisite for accommodating the non-homogeneous aspect of the differential equation. It focuses on the influence of external sources added onto the system's natural behavior.
For our differential equation, the non-homogeneous part is:

• \( 16 - (x+2) e^{4x} \)

To construct a particular solution, we employ the method of undetermined coefficients. This requires us to guess a form for the solution based on the structure of the non-homogeneous term. In this case, since \( e^{4x} \) already appears as part of the homogeneous solution, we enhance our guess, leading to the trial solution:

• \( y_p = Ax e^{4x} + Bx + C \)

By substituting \( y_p \) into the original differential equation and solving, we determine the coefficients, which makes \( y_p \) fit the equation. Calculations conclude:

• \( A = -\frac{1}{5} \)
• \( B = 0 \)
• \( C = \frac{2}{5} \)

Thus, the particular solution is finalized.

Characteristic Equation

The characteristic equation is a key tool when solving linear homogeneous differential equations. It helps us move from a problem involving derivatives, to a polynomial one, where algebraic methods can be applied.
In constructing the characteristic equation for our homogeneous differential equation, \( y'' + 2y' - 24y = 0 \), the first step is to assume a solution like \( y = e^{rx} \). By substituting this into the differential equation, we obtain a quadratic in terms of \( r \):

• \( r^2 + 2r - 24 = 0 \)

This is the characteristic equation. Solving it reveals the 'r' values or roots which correspond to the natural exponential solutions that make up the homogeneous solution. By using the quadratic formula:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

with \( a=1 \), \( b=2 \), and \( c=-24 \), we find:

• \( r_1 = 4 \)
• \( r_2 = -6 \)

The roots determine the form of the homogeneous solution, \( y_h = c_1 e^{4x} + c_2 e^{-6x} \), providing foundational insights into the equation's intrinsic behavior.