Problem 77 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d s}{d t}=1+\cos t, \quad s(0)=4$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

We are given the differential equation \( \frac{d s}{d t} = 1 + \cos t \). Our task is to integrate this expression with respect to \( t \).

Step 2: Integrate Both Sides

Integrate both sides with respect to \( t \):\[s = \int (1 + \cos t) \, dt.\]This can be split into two separate integrals:\[s = \int 1 \, dt + \int \cos t \, dt.\]

Step 3: Solve Each Integral

1. The integral of 1 with respect to \( t \) is: \[ \int 1 \, dt = t. \]2. The integral of \( \cos t \) with respect to \( t \) is: \[ \int \cos t \, dt = \sin t. \]Combine the results:\[s = t + \sin t + C,\]where \( C \) is the constant of integration.

Step 4: Use the Initial Condition

We need to find the constant \( C \) using the initial condition \( s(0) = 4 \). Substitute \( t = 0 \) and \( s = 4 \) into our expression:\[ 4 = 0 + \sin(0) + C. \]Since \( \sin(0) = 0 \), this simplifies to:\[ 4 = C. \]

Step 5: Write the Specific Solution

Substitute \( C = 4 \) back into the general solution:\[s = t + \sin t + 4.\] This is the specific solution.

Key Concepts

Differential Equations

Differential equations serve as the cornerstone in understanding how different variables change in relation to one another. In our exercise, we encounter the differential equation \( \frac{d s}{d t} = 1 + \cos t \). Here, the equation describes how the function \( s(t) \) changes as a result of changes in \( t \).A differential equation involves derivatives, expressing how a function's rate of change is related to other functions. Whenever you come across such an equation, your main goal is to find the function that satisfies this relationship, often referred to as the "solution of the differential equation." This solution provides crucial insights, allowing us to predict behavior over time, which is incredibly useful in fields like physics, engineering, and economics. In simpler terms, think of a differential equation as a math sentence that tells how something is changing. To solve the equation, we need to "reverse" the differentiation, usually by integrating.

Integration

Integration is like finding the reverse of differentiation. When we integrate an equation, we're trying to discover the original function that was differentiated to yield the given derivative. In the problem at hand, we're tasked to integrate \(\frac{d s}{d t} = 1 + \cos t\).To integrate, we perform the integral of both sides with respect to \( t \):

• The integral of \(1\) with respect to \( t \) results in \( t \).
• The integral of \( \cos t \) is \( \sin t \), since the derivative of \( \sin t \) is \( \cos t \).

Putting these together, the integration of our differential equation looks like this: \[ s = t + \sin t + C \] Here, \( C \) represents the constant of integration, which emerges because the process of differentiation could have hidden any constant originally present.

Initial Conditions

Initial conditions are pivotal for finding specific solutions to differential equations. They provide exact values that the solution must satisfy at a certain point. In our exercise, the initial condition is given as \( s(0) = 4 \).To apply this, we substitute \( t = 0 \) into the general solution \( s = t + \sin t + C \), resulting in: \[ 4 = 0 + \sin(0) + C \]With \( \sin(0) = 0 \), this simplifies to \( 4 = C \). By finding this constant, we can fine-tune our general solution to fit this starting point exactly.Thus, the specific solution becomes \( s = t + \sin t + 4 \). Initial conditions transform the broad, general solution into something uniquely tailored, solving the problem precisely.

Constant of Integration

When you integrate a function, the result includes an arbitrary constant, known as the constant of integration \( C \). This constant arises because differentiation "loses" constant values; they become invisible as they turn into zeros.In our integrated equation \( s = t + \sin t + C \), \( C \) represents these potential missed constants. Without specified conditions, \( C \) could be numerous values, each one representing a different potential solution.To make this solution specific, initial conditions are used to identify the precise value of \( C \). In this exercise, the use of \( s(0) = 4 \) allows us to pinpoint \( C = 4 \), providing a tailored solution that fits the unique scenario of the problem.