Problem 36 - A First Course in Differential Equations with Modeling Applications

Question

(a) \(\mathrm{By}\) inspection find a particular solution of\\[y^{\prime \prime}+2 y=10\\] (b) By inspection find a particular solution of \(y^{\prime \prime}+2 y=-4 x\). (c) Find a particular solution of \(y^{\prime \prime}+2 y=-4 x+10\). (d) Find a particular solution of \(y^{\prime \prime}+2 y=8 x+5\).

Step-by-Step Solution

Answer

(a) 5, (b) -2x, (c) -2x+5, (d) 4x+5/2

Step 1: Understand the General Form

The differential equation given is of the form \( y'' + 2y = g(x) \). We're tasked with finding particular solutions for different \( g(x) \). The aim is to find a function \( y_p(x) \) whose second derivative and twice its value add up to yield \( g(x) \).

Step 2: Solve for Part (a)

For part (a), \( g(x) = 10 \). A constant function \( y_p(x) = c \) might work. The derivative \( y'' \) is 0, so we have \( 0 + 2c = 10 \). Solving for \( c \), we get \( c = 5 \). Thus, \( y_p(x) = 5 \) is a particular solution.

Step 3: Solve for Part (b)

For part (b), \( g(x) = -4x \). Assume \( y_p(x) = Ax + B \). So, \( y'' = 0 \) and \( 2y_p = 2Ax + 2B \). We want \( 2Ax + 2B = -4x \). Therefore, \( A = -2 \) and \( B = 0 \). Hence, \( y_p(x) = -2x \) is a particular solution.

Step 4: Solve for Part (c)

For part (c), \( g(x) = -4x + 10 \). Use a combination of solutions from parts (a) and (b): \( y_p(x) = -2x + 5 \). Since \( y'' = 0 \), \( 2y_p = 2(-2x + 5) = -4x + 10 \), which matches \( g(x) \). Thus, \( y_p(x) = -2x + 5 \) is correct.

Step 5: Solve for Part (d)

For part (d), \( g(x) = 8x + 5 \). Assume \( y = Ax + B \). \( y'' = 0 \), thus \( 2y = 2Ax + 2B \). Set \( 2Ax + 2B = 8x + 5 \). Solving gives \( A = 4 \) and \( B = rac{5}{2} \). Therefore, the particular solution is \( y_p(x) = 4x + rac{5}{2} \).

Key Concepts

Particular Solutions

When solving differential equations, finding a particular solution is a key task. A particular solution, denoted as \( y_p(x) \), specifically satisfies the given non-homogeneous differential equation, unlike the general solution which includes all possible solutions. It is essential for determining how external forces or inputs, represented by \( g(x) \), affect the system.

In the context of second-order linear differential equations, a particular solution helps represent the specific behavior dictated by a particular function \( g(x) \). For instance, in the exercise given, the task is to find \( y_p(x) \) for different \( g(x) \). The method aims to find a simple function whose derivative properties make it fit exactly with the requirements of the modified equation.

Particular solutions are crucial in applications, where they can be adjusted or combined with the general solution of the homogeneous equation to form the complete solution to the boundary problem. This allows one to model real-life scenarios effectively.

Inspection Method

The inspection method is a straightforward and intuitive technique for finding particular solutions of differential equations, especially when \( g(x) \) is simple, like polynomials or constants. In this approach, the solver relies on educated guesses based on the form of \( g(x) \). If the differential equation is \( y'' + 2y = g(x) \), the key is to choose \( y_p(x) \) such that its second derivative and its twice addition gives the \( g(x) \).

For example, when \( g(x) = 10 \) (a constant), you might guess \( y_p(x) = c \) because its simplicity makes the arithmetic work out neatly. Similarly, for a polynomial like \( g(x) = -4x \), a linear function \( y_p(x) = Ax + B \) is a choice that allows us to adjust \( A \) and \( B \) appropriately. Inspection relies heavily on the solver being able to see potential matches visually and then verifying them mathematically.

Using the inspection method can save time and help understand the relationship between solution forms and their constituents. It works well in setups where the right-hand side function is straightforward and predictable.

Second-Order Linear Differential Equations

A second-order linear differential equation typically involves derivatives up to the second order and can be expressed in the form \( ay'' + by' + cy = g(x) \). These equations are ubiquitous in the modeling of physical phenomena, ranging from mechanical vibrations to electrical circuits.

The given equation \( y'' + 2y = g(x) \) is one example, where the coefficient of \( y'' \) is 1, and that of \( y \) is 2. Understanding the structure helps clarify how the equation models changes. The term \( y'' \) represents the acceleration (response to change), and \( 2y \) acts as a restorative force (like a spring).
Second-order equations include both homogeneous (\( g(x) = 0 \)) and non-homogeneous forms. Solving the homogeneous form provides the complementary solution, while the non-homogeneous solution is obtained by finding particular solutions, which are combined for the general solution.

Their solutions are essential for capturing the dynamics or transient responses in systems influenced by external inputs.

Modeling Applications

Differential equations, especially second-order linear ones, serve as foundational models in numerous scientific and engineering applications. They let us describe dynamic systems such as oscillations, heat conduction, and wave propagation.

In practical terms, a particular solution helps fine-tune models to fit real-world conditions. Imagine a bridge subject to oscillations; here, the external force is modeled as \( g(x) \). Through particular solutions, engineers can predict how the structure will react over time, contributing to safer designs.

Whether modeling the decline of a pharmaceutical substance in biology, the stress on building materials, or the vibration of mechanical parts, differential equations offer a precise language. By using them, scientists and engineers can simulate complex interactions and reactions accurately, ensuring that any modeled system behaves as expected in practice.