Problem 56 - Calculus for Biology and Medicine

Question

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 is \( \frac{d y}{d s}=\cos (2 \pi s)\). This is a first-order ordinary differential equation, where the derivative of \(y\) with respect to \(s\) is given.

Step 2: Set Up the Integral

To find \(y(s)\), we need to integrate the right-hand side of the equation: \( \int \cos (2 \pi s) \, ds \). This integral will help us find the expression for \(y(s)\).

Step 3: Solve the Integral

The integral \( \int \cos (2 \pi s) \, ds \) is a standard integral. To solve it, recall that the integral of \(\cos(ks)\) is \(\frac{1}{k}\sin(ks)\). Here, \(k = 2\pi\). So, the integral becomes \(\frac{1}{2\pi}\sin(2\pi s) + C\), where \(C\) is the constant of integration.

Step 4: Write the General Solution

The general solution to the differential equation is \(y(s) = \frac{1}{2\pi} \sin(2\pi s) + C\). This expression includes the constant \(C\), representing the family of solutions based on the initial conditions if they were provided.

Key Concepts

First-order differential equation

A first-order differential equation involves derivatives of a function with respect to only one variable and the order of derivative considered is the first. The general form is:

• \( \frac{dy}{dx} = f(x, y) \)

In this form, \( x \) is the independent variable, \( y \) is a function of \( x \), and \( f(x, y) \) represents a function involving \( x \), \( y \), or both. The key here is to understand that we are looking at how quickly \( y \) changes as \( x \) changes.
In our given exercise, \( \frac{dy}{ds} = \cos(2 \pi s) \), the function \( f(s) \) only involves the variable \( s \) and not \( y \). This makes it an explicit first-order ordinary differential equation.
These types of equations are common in modeling processes with constant rates of change, such as physics problems involving motion or wave functions.

Integration techniques

Integration techniques are fundamental in solving differential equations, as they allow us to reverse the process of differentiation to find the original function. In our context, it involves finding the antiderivative.
Consider the integral in the exercise:

• \( \int \cos(2 \pi s) \, ds \)

This is a standard integral where the function \( \cos(ks) \) has an antiderivative \( \frac{1}{k}\sin(ks) \). Recognizing this pattern is critical for efficiently solving integrals of trigonometric functions.
In this instance:

• Let \( k = 2\pi \).
• Thus, \( \int \cos(2 \pi s) \, ds = \frac{1}{2 \pi} \sin(2\pi s) + C \).

The constant \( C \) is crucial as it accounts for all possible vertical shifts of the function on a graph, representing the family of solutions.

General solution

The general solution of a differential equation includes the function found after integration plus a constant of integration, usually denoted as \( C \). The purpose of the general solution is to represent all possible solutions that satisfy the differential equation.
For the equation \( \frac{dy}{ds} = \cos(2 \pi s) \) after solving its integral, the general solution is:

• \( y(s) = \frac{1}{2\pi} \sin(2\pi s) + C \)

• \( C \) is an arbitrary constant that can be determined if additional conditions, such as initial values, are provided.

This is important in practical applications, as conditions may dictate specific solutions. Without further context, \( C \) remains unspecified, indicating all potential function shifts can be solutions. This flexibility aligns with the real-world phenomena where starting conditions might differ.