Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

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Recognize that the given differential equation is a third-order ordinary differential equation: \(\frac{d^{3}y}{dx^{3}} = 6\). Our goal is to find the function \(y(x)\) that satisfies this equation along with the initial conditions.

Integrate the differential equation once with respect to \(x\) to find the second derivative \(y''(x)\). Since \(\frac{d^{3}y}{dx^{3}} = 6\), integrating gives \(y''(x) = 6x + C_1\), where \(C_1\) is an integration constant.

Use the initial condition \(y''(0) = -8\) to solve for \(C_1\). Substitute \(x=0\) into \(y''(x) = 6x + C_1\) and set it equal to \(-8\) to find \(C_1\).

Integrate \(y''(x)\) to find the first derivative \(y'(x)\). This gives \(y'(x) = \int y''(x) \, dx = \int (6x + C_1) \, dx = 3x^{2} + C_1 x + C_2\), where \(C_2\) is another integration constant.

Use the initial condition \(y'(0) = 0\) to solve for \(C_2\). Substitute \(x=0\) into \(y'(x)\) and set it equal to \(0\) to find \(C_2\). Then integrate \(y'(x)\) once more to find \(y(x)\), and use the initial condition \(y(0) = 5\) to solve for the last constant.

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주요 개념

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

Higher-Order Derivatives and Integration

This involves understanding derivatives beyond the first order, such as the third derivative given here. To solve the differential equation, you integrate the third derivative step-by-step to find the second, first derivatives, and finally the original function, adding constants of integration at each step.

Initial Conditions in Differential Equations

Initial conditions specify the values of the function and its derivatives at a particular point, allowing us to determine the constants of integration uniquely. Here, values for y(0), y′(0), and y″(0) are given, which help solve for the constants after integration.

Solving Initial Value Problems (IVPs)

An IVP requires finding a function that satisfies a differential equation and meets given initial conditions. The process involves integrating the differential equation and applying the initial values to find the specific solution that fits the problem.

Initial Value ProblemsSolve the initial value problems in Exercises 71–90.d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5