Problem 84 - Thomas Calculus in SI Units

Question

Solve the initial value problems $$\frac{d^{2} y}{d x^{2}}=\sec ^{2} x, \quad y(0)=0 \quad \text { and } \quad y^{\prime}(0)=1$$

Step-by-Step Solution

Answer

The solution is \( y = -\ln |\cos x| + x \).

Step 1: Integrate the Second Derivative

Given the differential equation \( \frac{d^{2}y}{dx^{2}} = \sec^2 x \), we first integrate with respect to \( x \) to find the first derivative \( \frac{dy}{dx} \). The integral of \( \sec^2 x \) is \( \tan x \). Thus, \( \frac{dy}{dx} = \tan x + C_1 \), where \( C_1 \) is a constant of integration.

Step 2: Use Initial Condition for First Derivative

We use the initial condition \( y'(0) = 1 \) to find \( C_1 \). Substitute \( x = 0 \) and \( \frac{dy}{dx} = 1 \) into \( \frac{dy}{dx} = \tan x + C_1 \). Since \( \tan(0) = 0 \), we get \( 1 = 0 + C_1 \), so \( C_1 = 1 \). The equation for the first derivative is now \( \frac{dy}{dx} = \tan x + 1 \).

Step 3: Integrate the First Derivative

Next, we integrate \( \frac{dy}{dx} = \tan x + 1 \) to find \( y \). The integral of \( \tan x \) is \( -\ln |\cos x| \), and the integral of 1 is \( x \). Therefore, \( y = -\ln |\cos x| + x + C_2 \), where \( C_2 \) is another constant of integration.

Step 4: Use Initial Condition for the Function

Utilize the initial condition \( y(0) = 0 \) to solve for \( C_2 \). Substitute \( x = 0 \) into \( y = -\ln |\cos x| + x + C_2 \). Since \( \cos(0) = 1 \), the equation simplifies to \( 0 = -\ln 1 + 0 + C_2 \). Given that \( \ln 1 = 0 \), we find \( C_2 = 0 \). Thus, \( y = -\ln |\cos x| + x \).

Step 5: Conclusion

The solution to the initial value problem is \( y(x) = -\ln |\cos x| + x \). This satisfies both initial conditions: \( y(0) = 0 \) and \( y'(0) = 1 \).

Key Concepts

Second-Order Differential Equation

A second-order differential equation is a type of differential equation that involves the second derivative of a function. This means it includes terms like \( \frac{d^2y}{dx^2} \), which is the second derivative of \( y \) with respect to \( x \). These equations are fundamental in modeling phenomena where the rate of change of a system's state is itself changing over time.
When working on second-order differential equations, the goal is often to find the original function \( y \) that satisfies given conditions. Such functions may describe a variety of physical situations, such as oscillations or waves, depending on the nature of the equation.
Solving these requires not only finding an expression for \( y \) but also considering initial conditions, like initial position or speed. Initial conditions help us solve each constant of integration step-by-step, ensuring the solution uniquely fits the specifics of the problem.
In the given exercise, the equation \( \frac{d^2y}{dx^2} = \sec^2 x \) models the change in a function derived from a trigonometric context.

Trigonometric Integration

Trigonometric integration is important when solving equations that involve trigonometric functions. This type of integration helps us find anti-derivatives for functions like \( \sin x \), \( \cos x \), and \( \sec x \). The special case we encounter in the exercise is the integration of \( \sec^2 x \).
When integrating \( \sec^2 x \) with respect to \( x \), we find that it equals \( \tan x \). Understanding this integral is crucial in advancing our work on second-order equations that include trigonometric terms.
Some tips on integrating trigonometric functions include knowing common derivatives and integrals, such as:

• \( \int \sec^2 x \, dx = \tan x + C \)
• \( \int \tan x \, dx = -\ln |\cos x| + C \)
• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)

These identities are the backbone of manipulating trigonometric integrals, which crop up again and again in calculus-related problems.

Integration Constants

Integration constants are values added during the indefinite integration process. They reflect the fact that an indefinite integral can represent an infinite family of functions. Each constant corresponds to the need for specific initial conditions to determine a unique solution.
When we perform an indefinite integration, we account for unknown starting points or shifts by adding a constant, often denoted as \( C_1, C_2, \) etc. In solving initial value problems, these constants are what we focus on determining from given initial conditions.
To understand their role, consider the function \( y = f(x) + C \) resulting from an integration. The constant \( C \) shifts or translates the entire function graph, which can significantly affect function behavior and ensure the solution fits real-world data.
In our example, two constants were determined using initial conditions: \( y'(0) = 1 \) gave us \( C_1 = 1 \), and \( y(0) = 0 \) resulted in \( C_2 = 0 \). Solving these constants transforms potential family of solutions into a single, unique answer that meets all initial requirements.