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Chapter 4: Problem 97

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

Short Answer

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

Step by step solution

01

Integrate the Differential Equation

The given differential equation is \( \frac{ds}{dt} = 1 + \cos t \). To integrate both sides with respect to \( t \), we integrate \( 1 + \cos t \) to get \( s(t) \). The integral of \( 1 \) is \( t \) and the integral of \( \cos t \) is \( \sin t \). Therefore, \[s(t) = \int (1 + \cos t) \, dt = t + \sin t + C\]where \( C \) is the constant of integration.
02

Apply the Initial Condition

We are given the initial condition \( s(0) = 4 \). Use this condition to solve for \( C \). Substituting \( t = 0 \) into our equation, we have:\[s(0) = 0 + \sin(0) + C = C = 4\]Thus, \( C = 4 \).
03

Write the Final Solution

Substitute \( C = 4 \) back into the expression we found for \( s(t) \). The final solution to the initial value problem is:\[s(t) = t + \sin t + 4\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation along with a specific value at a given point. This given value is called the initial condition. The purpose of an IVP is to find a particular solution, satisfying both the differential equation and the initial condition.
In our exercise, we are given the differential equation \( \frac{ds}{dt} = 1 + \cos t \) with the initial condition \( s(0) = 4 \).
Initial Value Problems are common in mathematics and physics as they often represent real-world phenomena where the initial state of a system is known.
  • The differential equation shows how a rate of change occurs.
  • The initial condition gives a starting point.
This combination allows us to find a unique solution tailored to the specific context of the problem.
Integration
Integration is a fundamental concept in calculus. It involves finding a function whose derivative is the given function. In other words, it is the reverse process of differentiation.
In solving our problem, we integrate the expression \( 1 + \cos t \). This turns the rate of change \( \frac{ds}{dt} \) into an actual function \( s(t) \).
Useful points to remember about integration:
  • The integral of a constant, like \( 1 \), with respect to \( t \) is simply \( t \).
  • The integral of \( \cos t \) is \( \sin t \).
  • Integration introduces a constant of integration, denoted by \( C \).
The constant \( C \) is vital as it accounts for any constant offset in the function, determined later through initial conditions.
Cosine Function
The cosine function, denoted as \( \cos t \), is a trigonometric function that describes the ratio of the adjacent side to the hypotenuse in a right-angled triangle. In the realm of calculus, it is often involved in integrals and derivatives.
For integration purposes, it's essential to know that:
  • The derivative of \( \sin t \) is \( \cos t \).
  • Thus, the integral of \( \cos t \) is \( \sin t \).
This fact is crucial in our problem, as integrating \( \cos t \) gives us part of the solution for \( s(t) \).
The cosine function is periodic, meaning it repeats its values in regular intervals of \( 2\pi \). This property can be useful in analyzing cyclical or oscillating behaviors in physics and engineering.

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Most popular questions from this chapter

Solve the initial value problems. $$y^{(4)}=-\sin t+\cos t;$$ $$y^{\prime \prime \prime}(0)=7, \quad y^{\prime \prime}(0)=y^{\prime}(0)=-1, \quad y(0)=0$$

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