Solve the initial value problems in Exercises 87 and 88.
88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1

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Identify the given differential equation and initial conditions: \( \frac{d^2y}{dx^2} = \sec^2 x \), with \( y(0) = 0 \) and \( y'(0) = 1 \).

Integrate the second derivative \( \frac{d^2y}{dx^2} \) once with respect to \( x \) to find the first derivative \( y'(x) \). This means computing \( y'(x) = \int \sec^2 x \, dx + C_1 \), where \( C_1 \) is a constant of integration.

Use the initial condition \( y'(0) = 1 \) to solve for the constant \( C_1 \) after finding the integral expression for \( y'(x) \).

Integrate \( y'(x) \) with respect to \( x \) to find \( y(x) \). This involves computing \( y(x) = \int y'(x) \, dx + C_2 \), where \( C_2 \) is another constant of integration.

Apply the initial condition \( y(0) = 0 \) to solve for the constant \( C_2 \), completing the solution for \( y(x) \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Second-Order Differential Equations

A second-order differential equation involves the second derivative of a function. Solving such equations typically requires integrating the given expression twice and applying initial conditions to find the constants of integration.

Integration of Trigonometric Functions

Integrating functions like sec²x is essential in solving the differential equation. Knowing that the integral of sec²x is tan x helps in finding the first derivative, which can then be integrated again to find the original function.

Initial Value Problems (IVP)

An initial value problem specifies values of the function and its derivatives at a particular point. These conditions allow us to determine the constants after integration, ensuring a unique solution that fits the given initial values.

Solve the initial value problems in Exercises 87 and 88.88. d²y/dx² = sec²x, y(0)=0 and y'(0)=1