Problem 20 - Advanced Engineering Mathematics

Question

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

Solve the homogeneous equation, then determine the particular solution using undetermined coefficients; combine for the general solution.

Step 1: Identify the Differential Equation

The given differential equation is a second-order linear non-homogeneous differential equation: \[y'' + 2y' - 24y = 16 - (x+2)e^{4x}.\]

Step 2: Solve the Homogeneous Equation

Find the complementary solution by solving the homogeneous equation:\[y'' + 2y' - 24y = 0.\]The characteristic equation is \(r^2 + 2r - 24 = 0\). Solving this quadratic equation using the quadratic formula yields the roots \(r = 4\) and \(r = -6\). The general solution of the homogeneous equation is:\[y_h = C_1e^{4x} + C_2e^{-6x}.\]

Step 3: Set up the Particular Solution

The non-homogeneous part of the differential equation is \[16 - (x+2)e^{4x}.\]A reasonable choice for the undetermined coefficients is:\[y_p = A + (Bx + C)x e^{4x}.\]This accounts for the polynomial and exponential components.

Step 4: Differentiate the Particular Solution

Differentiate \(y_p = A + (Bx + C)x e^{4x}\):1. First derivative: Use the product rule \[y_p' = (B + (B'x + C)e^{4x}) + (Bx + C)4e^{4x}.\] Simplify to get \[y_p' = B + Bx e^{4x} + C e^{4x} + 4BX e^{4x} + 4C e^{4x}.\]2. Second derivative: Differentiate again \[y_p'' = B'e^{4x} + (2Bxe^{4x} + 2Ce^{4x} + 4Bx e^{4x} + 4Ce^{4x}).\]

Step 5: Substitute Back and Solve

Substitute \(y_p\), \(y_p'\), and \(y_p''\) back into the differential equation:\[y_p'' + 2y_p' - 24y_p = 16 - (x+2)e^{4x}.\]Solve for \(A\), \(B\), and \(C\) by equating coefficients on both sides of the equation. Proceed to solve for these coefficients, ensuring each term on both sides matches in power and type.

Step 6: Combine Solutions

The general solution to the differential equation is the sum of the complementary and particular solutions:\[y = C_1e^{4x} + C_2e^{-6x} + A + (Bx + C)x e^{4x}.\]This expression encompasses both the homogeneous and non-homogeneous parts.

Key Concepts

Second-Order Differential Equations

A second-order differential equation is an equation that involves the second derivative of a function. In mathematical terms, it is often expressed as:\[ y'' + p(x)y' + q(x)y = g(x) \]where \( y'' \) denotes the second derivative of \( y \) with respect to \( x \), \( p(x) \) and \( q(x) \) are functions of \( x \), and \( g(x) \) is a given function. These equations are crucial for modeling various physical phenomena, including oscillations, forces, and circuit dynamics.

One key feature of second-order differential equations is their order, which reflects the highest derivative present in the equation. Solving such equations often involves finding a function \( y(x) \) that satisfies the equation. This solution can include both a complementary function and a particular function, representing the equation’s general behavior and specific scenarios, respectively.

Understanding these equations requires a grasp of calculus and algebraic manipulation, as they often necessitate techniques like undetermined coefficients or variation of parameters to solve.

Homogeneous Equation

A homogeneous equation in the context of differential equations refers to an equation of the form:\[ y'' + p(x)y' + q(x)y = 0 \]In this form, the function \( g(x) \) from the typical non-homogeneous equation is zero, making the equation dependent solely on the unknown function and its derivatives.

Solving a homogeneous equation involves finding the complementary solution \( y_h \), which satisfies the equation. This task usually starts with forming and solving the characteristic equation: a polynomial resulting from assuming a trial solution of the form \( e^{rx} \). For example, for the homogeneous equation \( y'' + 2y' - 24y = 0 \), the characteristic equation is \( r^2 + 2r - 24 = 0 \).

The roots of the characteristic equation, in this case, \( r = 4 \) and \( r = -6 \), lead to the complementary solution:

• \( y_h = C_1e^{4x} + C_2e^{-6x} \)

This solution represents the behavior of the system without any external input, reflecting natural oscillations or decay.

Particular Solution

In a second-order differential equation, the particular solution is a specific solution to the non-homogeneous equation. It reflects the system's response to an external input, described by the non-zero function \( g(x) \) on the right side of the equation.

To find the particular solution, often the method of undetermined coefficients is employed. This approach involves proposing a trial solution, typically mimicking the form of \( g(x) \). For the given differential equation, where the non-homogeneous term is \( 16 - (x+2)e^{4x} \), a corresponding trial solution could be:

• \( y_p = A + (Bx + C)x e^{4x} \)

Here, \( A \), \( B \), and \( C \) are coefficients to be determined.

By differentiating this trial solution and substituting it back into the original equation, we equate coefficients to solve for \( A \), \( B \), and \( C \). This tailored solution complements the general solution derived from the homogeneous equation.

Characteristic Equation

The characteristic equation is a crucial step in solving homogeneous second-order differential equations. It stems from substituting a trial exponential solution \( y = e^{rx} \) into the differential equation. Upon substitution, the terms of the equation involving \( y'' \), \( y' \), and \( y \) transform into a quadratic equation in terms of \( r \).

For a typical equation like \( y'' + 2y' - 24y = 0 \), the characteristic equation becomes:

• \( r^2 + 2r - 24 = 0 \)

Solving this quadratic equation gives us the roots \( r_1 \) and \( r_2 \), which are essential for constructing the complementary solution. The nature of these roots—real and distinct, real and repeated, or complex conjugates—determines the form of the complementary solution.

For real and distinct roots like \( r = 4 \) and \( r = -6 \), the complementary solution is a linear combination of exponential functions:

• \( y_h = C_1e^{4x} + C_2e^{-6x} \)

This solution encompasses the system's inherent response, built from the fundamental behaviors of the equation’s derivative terms.