Problem 73 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d s}{d t}=1+\cos t, \quad s(0)=4 $$

Step-by-Step Solution

Answer

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

Step 1: Integrate the Differential Equation

To solve the initial value problem, we first need to find the general solution of the differential equation \( \frac{d s}{d t} = 1 + \cos t \). This requires finding an antiderivative of the right-hand side. Integrating both sides with respect to \( t \), we obtain: \[ s(t) = \int (1 + \cos t) \, dt = \int 1 \, dt + \int \cos t \, dt = t + \sin t + C, \] where \( C \) is the constant of integration.

Step 2: Apply the Initial Condition

Now, use the initial condition \( s(0) = 4 \) to find the constant \( C \). Substitute \( t = 0 \) and \( s = 4 \) into the general solution:\[ 4 = 0 + \sin(0) + C. \] This simplifies to \( 4 = C \). Thus, the constant \( C = 4 \).

Step 3: Write the Particular Solution

Insert the value of \( C \) back into the general solution to get the particular solution to the initial value problem. Thus, \[ s(t) = t + \sin t + 4. \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve the rates of change of a quantity. They are essential in describing various phenomena in science and engineering. In this exercise, we deal with a differential equation of the form \( \frac{d s}{d t} = 1 + \cos t \). This represents the rate of change of \( s \) with respect to \( t \), where the right-hand side (\( 1 + \cos t \)) determines how \( s \) changes over time.
Differential equations can be complex, but breaking them down into simpler parts, like separating variables or finding antiderivatives, helps solve them. Solving a differential equation often involves integrating to find the unknown function, \( s \), in terms of the independent variable, \( t \).

Antiderivative

An antiderivative is a function whose derivative is the given function. It's the opposite of differentiation. In our problem, the antiderivative helps find the function \( s(t) \) such that \( \frac{d s}{d t} = 1 + \cos t \).
By integrating \( 1 + \cos t \), we find the antiderivative:

• The antiderivative of \( 1 \) is \( t \).
• The antiderivative of \( \cos t \) is \( \sin t \).

Combining these, the general antiderivative of \( 1 + \cos t \) is \( t + \sin t + C \), where \( C \) is a constant of integration to be determined by initial conditions.

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies given initial conditions. For our problem, the initial condition is \( s(0) = 4 \).
To find the particular solution, we substitute \( t = 0 \) into the antiderivative, \( s(t) = t + \sin t + C \), and solve for \( C \).

• Plugging in \( t = 0 \) and \( s = 4 \), we have: \( 4 = 0 + \sin(0) + C \).
• This simplifies to \( C = 4 \).

Substituting \( C = 4 \) back into our antiderivative gives us the particular solution: \( s(t) = t + \sin t + 4 \). This particular solution solves the differential equation while meeting the initial condition.

Integration

Integration is the process of finding the antiderivative of a function. In the context of differential equations, integration is used to reverse differentiation. This leads to finding the function that describes the original relationship.
In our exercise, we integrated \( 1 + \cos t \) to determine \( s(t) \), the function of \( t \). Integration performed is:

• \( \int 1 \, dt = t \)
• \( \int \cos t \, dt = \sin t \)

Putting these results together gives \( \int (1+\cos t) \, dt = t + \sin t + C \). This function \( s(t) \) is the general solution to the differential equation.

Constant of Integration

The constant of integration, symbolized by \( C \), appears in the context of indefinite integration. It represents the family of all possible functions that differentiate to a given function. Every integrated function has a constant because differentiation of a constant yields zero.
In solving initial value problems, the constant \( C \) is determined by applying the initial condition. In our example, we found \( C = 4 \) by using the condition \( s(0) = 4 \).

• This ensures the particular solution is not just any solution, but the one that solves the problem for specific initial values.
• Without this step, the solution would be incomplete.

Once \( C \) is known, the particular solution becomes \( s(t) = t + \sin t + 4 \), fully satisfying the differential equation and initial condition.