Initial Value Problems
Solve the initial value problems in Exercises 89–92.
d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1

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Identify the given differential equation and initial conditions: \(\frac{d^3 r}{dt^3} = -\cos t\), with \(r''(0) = 0\), \(r'(0) = 0\), and \(r(0) = -1\).

Integrate the third derivative \(\frac{d^3 r}{dt^3} = -\cos t\) once with respect to \(t\) to find the second derivative \(r''(t)\). Remember to add an integration constant \(C_1\).

Integrate \(r''(t)\) to find the first derivative \(r'(t)\), adding another integration constant \(C_2\).

Integrate \(r'(t)\) to find the original function \(r(t)\), adding a third integration constant \(C_3\).

Use the initial conditions \(r''(0) = 0\), \(r'(0) = 0\), and \(r(0) = -1\) to set up equations and solve for the constants \(C_1\), \(C_2\), and \(C_3\).

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주요 개념

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Higher-Order Differential Equations

Higher-order differential equations involve derivatives of order two or more. In this problem, the third derivative of r with respect to t is given, requiring integration multiple times to find the original function r(t). Understanding how to reduce the order by successive integration is essential.

Initial Conditions in Differential Equations

Initial conditions specify the values of a function and its derivatives at a particular point, allowing for the determination of integration constants. Here, values of r(0), r'(0), and r''(0) are given, which help uniquely solve for the constants after integrating the differential equation.

Integration of Trigonometric Functions

Solving the differential equation requires integrating the function -cos(t) multiple times. Familiarity with the integrals of sine and cosine functions, including their signs and constants of integration, is crucial to correctly find r(t) from its third derivative.

Initial Value Problems Solve the initial value problems in Exercises 89–92. d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1