Question

Solve the differential equation. $$ f^{\prime \prime}(x)=\sin x, f^{\prime}(0)=1, f(0)=6 $$

Short answer

The solution to the differential equation is \( f(x) = -\sin x + 2x + 6 \).

Step-by-step solution

Integration Step 1

To solve the differential equation, we start by integrating the equation \( f^{\prime \prime}(x) = \sin x \) with respect to \( x \). The integral of \( \sin x \) with respect to \( x \) is \( -\cos x \) plus a constant \( C_1 \). Therefore, \( f^{\prime}(x) = -\cos x + C_1 \).

Integration Step 2

Next, integrate the equation \( f^{\prime}(x) = -\cos x + C_1 \) with respect to \( x \) to find \( f(x) \). The integral of \( -\cos x \) with respect to \( x \) is \( -\sin x \) and the integral of \( C_1 \) is \( C_1x \). Therefore, \( f(x) = -\sin x + C_1x + C_2 \), where \( C_2 \) is another constant.

Applying Initial Conditions

Use the initial conditions \( f'(0) = 1 \) and \( f(0) = 6 \) to solve for the constants. Substituting into the equation \( f^{\prime}(x) = -\cos x + C_1 \) yields \( 1 = -\cos 0 + C_1 \) or \( C_1 = 1 + \cos 0 = 2 \). Substituting into the equation \( f(x) = -\sin x + C_1x + C_2 \) yields \( 6 = -\sin 0+ 2*0 + C_2 \) or \( C_2 = 6 \). Therefore, the solution to the differential equation is \( f(x) = -\sin x + 2x + 6 \).

Key concepts

Integration of Functions

Integration is a fundamental technique in calculus used for finding the antiderivatives of functions. The process of integration, in essence, is the reverse operation to differentiation. When given a differential equation, such as the one in our exercise \( f^{\prime \prime}(x) = \sin x \), the goal is to find a function \( f(x) \) whose second derivative is \( \sin x \).

To integrate a function like \( \sin x \), we rely on known integrals and apply them step by step. The first integral gives us \( f^\prime(x) \), which represents the velocity if \( f(x) \) is the position. In our case, integrating \( \sin x \) with respect to \( x \) gives us \( -\cos x + C_1 \), where \( C_1 \) is the integration constant. This constant represents the indefinite nature of the integration process, since an infinite number of antiderivatives exist, differing only by a constant.

It's important for students to recognize the pattern of integration and to remember the fundamental integrals, such as the integral of \( \sin x \), to solve differential equations efficiently.

Initial Conditions

Initial conditions are specific values given for a function and/or its derivatives at a particular point. These conditions allow us to find the exact solution to a differential equation by determining the integration constants.

In our exercise, the initial conditions are provided as \( f^\prime(0) = 1 \) and \( f(0) = 6 \). Think of the initial conditions as a 'starting point' for the problem. By applying the first initial condition, \( f^\prime(0) = 1 \), we find the value of \( C_1 \). Similarly, by applying the second condition, \( f(0) = 6 \), we determine the value of \( C_2 \). These values guide us towards the unique solution that satisfies both the differential equation and the initial conditions.

Understanding and utilizing initial conditions are crucial for solving real-world problems modeled by differential equations, where the conditions represent the known state at the start of a scenario.

Integration Constants

Integration constants, often denoted as \( C \) or \( C_1, C_2, ... \) depending on the number of times integration is performed, emerge naturally during the process of indefinite integration. These constants account for all the possible functions that have the same derivative.

As seen in the solution steps for our example, the integration of \( f^\prime(x) \) produced a constant \( C_1 \) and subsequent integration to find \( f(x) \) introduced another constant \( C_2 \). The presence of these constants means that without additional information, we cannot deduce a single, unique antiderivative.

This is where initial conditions play a role; they allow us to solve for the values of these constants, providing a complete and unique solution to the differential equation. Students should remember that every time an indefinite integration is performed, an integration constant must be included, and its value can be determined if appropriate initial conditions are given.