Problem 65 - Calculus for Biology and Medicine

Question

Solve the initial-value problem. $$ \frac{d W}{d t}=e^{t}, \text { for } t \geq 0 \text { with } W(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( W(t) = e^t \).

Step 1: Integrate the Differential Equation

First, we need to integrate the differential equation \( \frac{dW}{dt} = e^t \) with respect to \( t \). This gives us \( W(t) = \int e^t \, dt = e^t + C \), where \( C \) is the constant of integration.

Step 2: Apply the Initial Condition

We are given the initial condition \( W(0) = 1 \). Substitute \( t = 0 \) and \( W(0) = 1 \) into the equation \( W(t) = e^t + C \) to solve for \( C \). Thus, \( 1 = e^0 + C \), which simplifies to \( 1 = 1 + C \). Therefore, \( C = 0 \).

Step 3: Write the Particular Solution

Substitute the value of \( C = 0 \) back into the general solution to find the particular solution to the initial-value problem. So, \( W(t) = e^t + 0 \) or simply \( W(t) = e^t \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with an additional condition imposed at the start of the problem. This extra piece of information is known as the "initial condition." It specifies the value of the function at a particular point, often at time zero or the start of observations. For example, in the problem given, the condition is, \( W(0) = 1 \). This means that at time \( t = 0 \), the value of \( W \) is 1.

Initial value problems are crucial because they ensure there is enough information to find a unique solution to a differential equation. Without an initial condition, we might only find a general solution family depending on an arbitrary constant. Applying the initial condition allows us to determine the exact value of this constant, providing a specific solution known as the "particular solution."

Understanding initial value problems is essential in many fields such as physics, engineering, and economics, where real-world systems are often modeled with differential equations having known starting conditions.

Integration Techniques

Integration is a powerful mathematical tool used to find accumulation of quantities. It reverses the process of differentiation. In the context of differential equations, it is used to solve equations involving derivatives by finding the original function.

When faced with a differential equation like \( \frac{dW}{dt} = e^t \), we integrate \( e^t \) with respect to \( t \) to find \( W(t) \). The integral \( \int e^t \, dt \) results in \( e^t + C \), where \( C \) is a constant of integration.

The process of solving differential equations using integration consists of:

• Identifying the differential equation to be solved.
• Applying appropriate integration techniques to evaluate the integral.
• Including a constant of integration to account for the family of possible solutions.
• Using boundaries or initial conditions to solve for any unknown constants.

Integration techniques not only help solve initial value problems but are also applicable in calculating areas under curves, volumes, and other quantities in calculus.

Exponential Functions

Exponential functions are a fundamental mathematical concept characterized by their constant rate of growth or decay. The basic form is \( f(t) = a \cdot e^{rt} \), where \( e \) is the base of the natural logarithm, and \( r \) is the rate of growth.

In the given initial value problem, the function \( W(t) = e^t \) is an example of an exponential function. Exponential functions have notable properties, such as:

• The function grows or decays proportionally to its current value.
• The derivative \( \frac{d}{dt}(e^t) = e^t \) is the same as the function, indicating a constant rate of growth.

Their importance is far-reaching, appearing in various physical processes like population growth, radioactive decay, and compound interest.

In solving an initial value problem involving an exponential function, recognizing its form enables us to easily find the particular solution once the initial condition is applied.