Problem 97 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d s}{d t}=1+\cos t, \quad s(0)=4$$

Step-by-Step Solution

Answer

The solution is \( s(t) = t + \sin t + 4 \).

Step 1: Understand the Problem

We need to solve the differential equation \( \frac{d s}{d t}=1+\cos t \) with the initial condition \( s(0)=4 \). This means we have to find the function \( s(t) \) that satisfies both the differential equation and the initial condition.

Step 2: Integrate the Differential Equation

To solve for \( s(t) \), we integrate the right side of the equation with respect to \( t \). Therefore, we have: \[ s(t) = \int (1 + \cos t) \, dt \]The integral of \( 1 \) with respect to \( t \) is \( t \), and the integral of \( \cos t \) is \( \sin t \). So, \[ s(t) = t + \sin t + C \] where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

We use the initial condition \( s(0) = 4 \) to find \( C \). Substituting \( t = 0 \) into the equation for \( s(t) \), we have: \[ 4 = 0 + \sin(0) + C \] Since \( \sin(0) = 0 \), this simplifies to \[ 4 = C \] Thus, \( C = 4 \).

Step 4: Write the Solution

Now that we have the value of \( C \), we can express the solution to the differential equation: \[ s(t) = t + \sin t + 4 \] This is the function that satisfies both the differential equation and the initial condition.

Key Concepts

Differential Equation

A differential equation is an equation that involves an unknown function and its derivatives. In this case, the equation \( \frac{d s}{d t} = 1 + \cos t \) involves the derivative of the function \( s(t) \) with respect to \( t \). It tells us how the function \( s(t) \) changes as \( t \) changes. Solving a differential equation involves finding a function that satisfies the equation. This often requires the use of integration, which we'll touch on next.

This specific type of differential equation, where a derivative equals a combination of constants and other functions, is quite common in modeling situations where change is described in terms of both constant rates and periodic influences, like the reign of a pendulum or electrical currents.

Integration

Integration is a key technique used in solving differential equations. It is the process of finding a function whose derivative is the given function. In our problem, we integrate \( 1 + \cos t \) to find \( s(t) \). The integration of a constant \( 1 \) with respect to \( t \) is \( t \), and the integration of \( \cos t \) is \( \sin t \). Thus, when we integrate \( 1 + \cos t \), we obtain \( s(t) = t + \sin t + C \), where \( C \) is the constant of integration.

The constant \( C \) appears because of the indefinite integral, which represents a family of functions. Each value of \( C \) corresponds to a different function that still satisfies the original differential equation. Finding this constant is crucial to satisfy specific conditions.

Initial Condition

An initial condition is a value that allows us to determine the specific solution to a differential equation that meets certain criteria. It's like a piece of additional information that helps us narrow down a single function from a family of possible solutions.

In this exercise, the initial condition given is \( s(0) = 4 \), which means when \( t = 0 \), the value of \( s(t) \) should be 4. We use this condition to find the constant \( C \) from our integration result.

• Substitute \( t = 0 \) into the equation \( s(t) = t + \sin t + C \).
• Since \( \sin(0) = 0 \), the equation becomes \( 4 = 0 + 0 + C \), giving \( C = 4 \).

So, the initial condition helps finalize the solution, ensuring it meets the real-world criteria or initial circumstances described.

Function of a Variable

A function of a variable describes a relationship where each input (from a domain) relates to one and only one output (from a range). In the context of differential equations and initial value problems, our function \( s(t) \) defines how the variable \( s \) changes with respect to \( t \).

The ultimate goal in solving a differential equation is determining this kind of function. The solution \( s(t) = t + \sin t + 4 \) assigns a unique value of \( s \) for every \( t \), which embodies both the arbitrary changes from the sine component and the linear growth from the \( t \) component, refined by the constant from the initial condition.

• \( t \) represents the independent variable or input.
• \( s(t) \) would be the dependent variable or output.

Understanding functions in this way is critical to seeing the bigger picture in situations modeled by differential equations, from physics to economics.