Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1

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Identify the given differential equation and initial conditions: \( \frac{d^2 y}{dx^2} = 2 - 6x \), with \( y'(0) = 4 \) and \( y(0) = 1 \).

Integrate the second derivative \( \frac{d^2 y}{dx^2} \) once with respect to \( x \) to find the first derivative \( y'(x) \). Remember to add an integration constant \( C_1 \): \[ y'(x) = \int (2 - 6x) \, dx = 2x - 3x^2 + C_1 \]

Use the initial condition \( y'(0) = 4 \) to solve for the constant \( C_1 \) by substituting \( x = 0 \) into the expression for \( y'(x) \).

Integrate \( y'(x) \) with respect to \( x \) to find \( y(x) \), adding another integration constant \( C_2 \): \[ y(x) = \int (2x - 3x^2 + C_1) \, dx = x^2 - x^3 + C_1 x + C_2 \]

Use the initial condition \( y(0) = 1 \) to solve for \( C_2 \) by substituting \( x = 0 \) into the expression 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 it requires integrating twice, often leading to a general solution with two arbitrary constants. Understanding how to handle these equations is essential for modeling systems with acceleration or curvature.

Initial Conditions

Initial conditions specify the values of a function and its derivatives at a particular point, allowing us to find the unique solution to a differential equation. For example, y(0) = 1 and y′(0) = 4 provide specific values to determine the constants after integration.

Integration Techniques

Solving the given differential equation requires integrating the right-hand side function step-by-step. Mastery of basic integration rules and applying them correctly to find the antiderivatives is crucial to progress from the differential equation to the explicit solution.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.d²y/dx² = 2 − 6x; y′(0) = 4, y(0) = 1