Problem 18 - Thomas Calculus

Question

Solve the initial value problems. \(\theta \frac{d y}{d \theta}-2 y=\theta^{3} \sec \theta \tan \theta, \quad \theta>0, \quad y(\pi / 3)=2\)

Step-by-Step Solution

Answer

The solution is \( y = \theta^2 \sec\theta + \frac{18 - 2\pi^2}{\pi^2} \theta^2 \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \theta \frac{dy}{d\theta} - 2y = \theta^3 \sec\theta \tan\theta \). In this form, it resembles a linear first-order differential equation. Identify the standard form, which is \( \frac{dy}{d\theta} + P(\theta)y = Q(\theta) \), and rewrite the given equation: \( \frac{dy}{d\theta} - \frac{2}{\theta}y = \theta^2 \sec\theta \tan\theta \). Here, \( P(\theta) = -\frac{2}{\theta} \) and \( Q(\theta) = \theta^2 \sec\theta \tan\theta \).

Step 2: Find the Integrating Factor

The integrating factor for a linear differential equation is given by \( \mu(\theta) = e^{\int P(\theta)d\theta} \). Compute the integrating factor: \( \mu(\theta) = e^{\int -\frac{2}{\theta}d\theta} = e^{-2\ln\theta} = \theta^{-2} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \theta^{-2} \): \( \theta^{-2} \frac{dy}{d\theta} - 2\theta^{-3}y = \sec\theta \tan\theta \). This can be rewritten as \( \frac{d}{d\theta}(\theta^{-2}y) = \sec\theta \tan\theta \).

Step 4: Integrate Both Sides

Integrate both sides with respect to \( \theta \): \( \int \frac{d}{d\theta}(\theta^{-2}y) \, d\theta = \int \sec\theta \tan\theta \, d\theta \). The left side simplifies to \( \theta^{-2}y \), and the integral on the right side gives \( \sec\theta + C \). Thus, \( \theta^{-2}y = \sec\theta + C \).

Step 5: Solve for y

Solve for \( y \) by multiplying both sides by \( \theta^2 \): \( y = \theta^2 \sec\theta + C\theta^2 \).

Step 6: Apply the Initial Condition

Use the initial condition \( y(\pi/3) = 2 \) to find the value of \( C \): \( 2 = (\pi/3)^2 \sec(\pi/3) + C(\pi/3)^2 \). With \( \sec(\pi/3) = 2 \), the equation becomes \( 2 = \frac{\pi^2}{9} \times 2 + C\frac{\pi^2}{9} \). Simplify to find \( C \): \( 2 = \frac{2\pi^2}{9} + C\frac{\pi^2}{9} \), so \( C = \frac{18 - 2\pi^2}{\pi^2} \).

Step 7: Final Expression for y

Insert the value of \( C \) back into the expression for \( y \): \( y = \theta^2 \sec\theta + \frac{18 - 2\pi^2}{\pi^2} \theta^2 \).

Key Concepts

Linear Differential Equations

Linear differential equations are essential in understanding a wide range of physical phenomena. They are expressed in the standard form:

• \( \frac{dy}{d\theta} + P(\theta)y = Q(\theta) \)

This signifies a relationship between the derivative \( \frac{dy}{d\theta} \) and the function \( y \) itself, with the terms \( P(\theta) \) and \( Q(\theta) \) representing functions of the variable \( \theta \). The goal is to solve for \( y \) given these expressions. Linear differential equations can generally be solved using methods like integrating factors, which you'll learn more about next. It's crucial to recognize the type of differential equation you're dealing with, to apply the correct solving method.

Integrating Factor

An integrating factor is a strategic multiplier used to solve linear differential equations. It transforms the differential equation into one that is easier to integrate. The magic lies in finding the integrating factor \( \mu(\theta) \), calculated as follows:

• \( \mu(\theta) = e^{\int P(\theta) d\theta} \)

Once the integrating factor is found, you multiply the entire differential equation by \( \mu(\theta) \). This step simplifies the equation into a form where the left-hand side is the derivative of a product, which directly leads to an easier integration. Thus, the integrating factor simplifies the original problem drastically, making it possible to solve for \( y \).

First-Order Differential Equations

First-order differential equations involve derivatives of the first order, meaning they contain only the first derivative, \( \frac{dy}{d\theta} \). They are common and useful in modeling dynamic systems. In solving these, one must first bring the equation to its standard form:

• For linear cases: \( \frac{dy}{d\theta} + P(\theta)y = Q(\theta) \)

The key difference of first-order equations is their simplicity, as they involve only the first derivative. Despite being simpler, first-order differential equations can describe sophisticated processes such as electrical circuits and population dynamics. The beauty of these equations lies in their ability to provide clear insights into how processes evolve over time.

Initial Conditions

Initial conditions are critical in obtaining a specific solution to a differential equation. They provide the necessary information to determine the constants that arise from integration. When solving differentials, the general solution includes arbitrary constants, usually denoted by \( C \). To find these exact constants, a known value of \( y \), called an initial condition, is used. For instance, if \( y(\pi/3) = 2 \), this initial condition helps to incorporate specific scenarios into the solution, thereby making it unique. Initial conditions are like boundary markers, grounding the mathematical abstraction in realistic contexts, ensuring that the derived solutions are meaningful and applicable to real-world phenomena.