Problem 6 - Calculus for Biology and Medicine

Question

\text { In Problems , solve each pure-time differential equation. } $$ \frac{d x}{d t}=\sin (2 \pi(t+3)), \text { where } x(3)=1 $$

Step-by-Step Solution

Answer

The solution is \( x(t) = 1 + \frac{1}{2\pi}(1 - \cos(2\pi(t+3))) \).

Step 1: Understanding the Differential Equation

The given differential equation is \( \frac{dx}{dt} = \sin(2\pi(t+3)) \). We are asked to find the function \( x(t) \) that satisfies this equation. Also, an initial condition is given: \( x(3) = 1 \). This means that when \( t = 3 \), \( x = 1 \).

Step 2: Integration of Both Sides

To solve the differential equation \( \frac{dx}{dt} = \sin(2\pi(t+3)) \), we need to integrate both sides with respect to \( t \). This gives:\[x(t) = \int \sin(2\pi(t+3))\, dt + C\]where \( C \) is the constant of integration.

Step 3: Calculating the Integral

Calculate the integral to find \( x(t) \). The integration of \( \sin(2\pi(t+3)) \) is:\[\int \sin(2\pi(t+3))\, dt = -\frac{1}{2\pi} \cos(2\pi(t+3)) + C\]So, we have:\[ x(t) = -\frac{1}{2\pi} \cos(2\pi(t+3)) + C \]

Step 4: Applying the Initial Condition

Use the initial condition \( x(3) = 1 \) to find the constant \( C \). Substitute \( t = 3 \) and \( x = 1 \) in the equation:\[ 1 = -\frac{1}{2\pi} \cos(2\pi(3+3)) + C \]Simplify:\[ 1 = -\frac{1}{2\pi} \cos(12\pi) + C \]Since \( \cos(12\pi) = 1 \),\[ 1 = -\frac{1}{2\pi} + C \]Solving for \( C \),\[ C = 1 + \frac{1}{2\pi} \]

Step 5: Final Solution of the Differential Equation

Substitute \( C = 1 + \frac{1}{2\pi} \) back into the equation:\[ x(t) = -\frac{1}{2\pi} \cos(2\pi(t+3)) + 1 + \frac{1}{2\pi} \]Simplify (optional):\[ x(t) = 1 + \frac{1}{2\pi}(1 - \cos(2\pi(t+3))) \]

Key Concepts

Pure-time Differential Equation

A pure-time differential equation is a specific type of differential equation where the rate of change of a function is expressed explicitly in terms of time ( \( t \) ) alone. In contrast to other differential equations that may have dependencies on both the function and its derivatives, pure-time differential equations focus purely on how the function's rate of change depends on time.
This specific form often appears as:
\[ \frac{dy}{dt} = f(t) \]
where \( f(t) \) is a function of \( t \) only. In these equations, the dependent variable, often denoted as \( y \) or \( x(t) \), is not present in the right-hand side expression.

• These equations are generally easier to solve since they require only integration with respect to time.
• Understanding and identifying a pure-time differential equation is essential because it guides us on the method to solve it.

Thus, the primary focus is solving the integration to find the function \( y(t) \) or \( x(t) \) as shown in the exercise problem.

Integration

Integration is the process of finding the antiderivative of a function. In the context of differential equations, it helps in reversing the differentiation process to find a function where its derivative is known.
When given a differential equation like \( \frac{dx}{dt} = \sin(2\pi(t+3)) \), integration serves as the tool to discover \( x(t) \).
Here are some useful points about integration in differential equations:

• Identify the function inside the derivative: If \( f(t) = \sin(2\pi(t+3)) \), the task of integration is to find what \( x(t) \) would differentiate to \( f(t) \).
• Upon integration, always remember to add the constant of integration, \( C \), because integration can yield families of solutions.
• In the provided example, integrating \( \sin(2\pi(t+3)) \) gives \( -\frac{1}{2\pi}\cos(2\pi(t+3)) + C \).

This process is crucial for converting the rate of change expression into the function we need.

Initial Condition

An initial condition is an additional piece of information provided in a differential equation problem that helps find a unique solution from the general solution. It specifies the value of the unknown function, typically at a specific point in time.
For instance, in the exercise problem, the initial condition is \( x(3) = 1 \).

• This means that when \( t = 3 \), the value of the function should be 1.
• Initial conditions are crucial because they allow us to solve for the constant of integration, \( C \), making the solution specific to the problem context.
• Applying initial conditions effectively "anchor" the general solution, conferring it a particular real-world behavior.

In the worked example, substituting \( t = 3 \) and \( x = 1 \) into the integrated solution enables solving for \( C \). This ensures that our function \( x(t) \) bears the exact characteristics described by the condition.

Constant of Integration

The constant of integration, typically denoted as \( C \), arises from the integral operation. It represents the arbitrary constant added to the antiderivative due to the indefinite nature of integration.
Without the constant, the integration would not account for all possible antiderivatives of the function, thus missing potential solutions. Upon integrating \( \sin(2\pi(t+3)) \), the general solution becomes \( -\frac{1}{2\pi} \cos(2\pi(t+3)) + C \).

• The constant of integration ensures the solution encompasses all possible states the function could represent.
• The value of \( C \) is determined using the provided initial condition.
• For the problem example, substitution of the initial condition \( x(3) = 1 \) reveals that \( C = 1 + \frac{1}{2\pi} \).

Thus, the constant of integration transforms the general solution to a specific one, accurately reflecting the scenario defined by the initial condition.