Problem 56 - Calculus for Biology and Medicine

Question

In Problems 47-58, find the general solution of the differential equation. $$ \frac{d y}{d s}=\cos (2 \pi s), 0 \leq s \leq 1 $$

Step-by-Step Solution

Answer

The general solution is \( y(s) = \frac{1}{2\pi} \sin(2\pi s) + C \).

Step 1: Identify the type of differential equation

The given differential equation \( \frac{dy}{ds} = \cos(2\pi s) \) is a first-order linear ordinary differential equation. Our objective is to find a function \( y(s) \) whose derivative with respect to \( s \) is \( \cos(2\pi s) \).

Step 2: Integrate both sides with respect to \(s\)

To solve for \( y(s) \), we need to integrate both sides of the equation with respect to \( s \). This gives: \( \int \frac{dy}{ds} \, ds = \int \cos(2\pi s) \, ds \).

Step 3: Perform the integration

The left side simplifies to \( y(s) \), as it is the antiderivative of \( \frac{dy}{ds} \). For the right side, use the substitution for the integral of a cosine function: \( \int \cos(2\pi s) \, ds = \frac{1}{2\pi} \sin(2\pi s) \).

Step 4: Add the constant of integration

Since the integration on the right side can have an arbitrary constant, the general solution to the equation is: \( y(s) = \frac{1}{2\pi} \sin(2\pi s) + C \), where \( C \) is the constant of integration.

Key Concepts

First-Order Linear Equations

Differential equations come in various forms, and understanding their type helps in solving them efficiently. One common type is the first-order linear equation. This form looks like \( \frac{dy}{ds} + P(s)y = Q(s) \), where \( P(s) \) and \( Q(s) \) are functions of \( s \). However, the equation given in our problem is a simplified version since it does not include a \( P(s)y \) term. Instead, it directly relates the derivative of \( y \) with \( s \) to a given function, in this case, \( \cos(2\pi s) \). This type of equation is the easiest to work with as we primarily need to integrate to find the solution. Recognizing such structures aids in determining how to solve these equations effectively.

Integration Techniques

Integration is a crucial step when solving differential equations, particularly first-order ones. In our problem, we integrated both sides of the equation \( \frac{dy}{ds} = \cos(2\pi s) \) with respect to \( s \). This is a standard technique to find \( y(s) \).
For integration, knowing antiderivatives is essential. Here, the integration of \( \cos(2\pi s) \) involves a substitution to simplify the process. Recall that integrating \( \cos(kx) \), generally provides \( \frac{1}{k} \sin(kx) \). This leads to \( \int \cos(2\pi s) \, ds = \frac{1}{2\pi} \sin(2\pi s) \). This procedure showcases the power of antiderivatives and highlights how a small internalization of these formulae can simplify our work through seemingly complex integrals.

General Solution of Differential Equations

Once integration is completed, adding an arbitrary constant represents the general solution. This constant of integration, denoted as \( C \), is essential because it accounts for all potential vertical shifts of the function \( y(s) \) that satisfy the differential equation.
In our example, the solution \( y(s) = \frac{1}{2\pi} \sin(2\pi s) + C \) shows how an integration process leads to a family of solutions. Each specific value of \( C \) defines a unique solution, reflecting different initial conditions or particular solutions. Thus, finding the general solution implies summarizing all possibilities for \( y \), based on different starting points or conditions, making it critical in understanding the complete set of solutions for differential equations.