Problem 67 - Calculus for Biology and Medicine

Question

Solve the initial-value problem. $$ \frac{d W}{d t}=\exp (t+1), \text { for } t \geq 0 \text { with } W(0)=2 / 3 $$

Step-by-Step Solution

Answer

The solution is \( W(t) = e^{t+1} - e + \frac{2}{3} \).

Step 1: Identify the Differential Equation

We start with the given differential equation: \( \frac{dW}{dt} = \exp(t+1) \). This tells us that the derivative of the function \( W(t) \) with respect to \( t \) is equal to \( e^{t+1} \).

Step 2: Set Up the Integral for Solution

To find \( W(t) \), we need to integrate the right-hand side \( e^{t+1} \) with respect to \( t \). This gives us \( W(t) = \int e^{t+1} \, dt \).

Step 3: Perform the Integration

The integral of \( e^{t+1} \) can be found using the substitution method. Let \( u = t+1 \), then \( du = dt \). The integral transforms to \( \int e^u \, du = e^u + C \). Re-substituting \( u = t+1 \), we obtain \( e^{t+1} + C \). Thus, \( W(t) = e^{t+1} + C \).

Step 4: Apply the Initial Condition

We use the given initial condition \( W(0) = \frac{2}{3} \) to solve for \( C \). Substitute \( t = 0 \) into the equation \( W(t) = e^{t+1} + C \): \( \frac{2}{3} = e^{0+1} + C \). This simplifies to \( \frac{2}{3} = e + C \). Solve for \( C \): \( C = \frac{2}{3} - e \).

Step 5: Substitute Back to Find W(t)

Substitute the value of \( C \) back into the expression for \( W(t) \): \( W(t) = e^{t+1} + \frac{2}{3} - e \). Simplify to obtain \( W(t) = e^{t+1} - e + \frac{2}{3} \).

Step 6: Simplify the Expression

Since \( e^{t+1} = e \cdot e^t \), the final solution simplified is \( W(t) = e \cdot e^t - e + \frac{2}{3} \).

Key Concepts

Understanding Initial-Value Problems

An initial-value problem consists of a differential equation together with a specific condition, termed the initial condition. For example, in our exercise: \( \frac{dW}{dt} = \exp(t+1) \) with \( W(0) = \frac{2}{3} \). The objective is to find a specific function, in this case \( W(t) \), that satisfies both the differential equation and the initial condition.
This ensures that at \( t = 0 \), our function \( W(t) \) has the value \( \frac{2}{3} \). Initial-value problems are vital in modeling real-world scenarios where future states depend not only on evolving dynamics but also on initial conditions. By solving these problems, we establish a complete function that describes the phenomenon at any later time.
These steps are crucial:

• Identify the differential equation.
• Perform integration to find a general solution.
• Apply initial conditions to determine specific constants.

Integration Essentials

Integration is the process of finding the original function from its derivative. In solving \( \frac{dW}{dt} = e^{t+1} \), we performed integration to find \( W(t) \). The integral operation is the reverse of differentiation, essentially summing up tiny changes to yield the overall accumulation.

The formula for integration simplifies expressions and circumstances where we need complete information about a function based on its rate of change. Besides, integrating \( e^{t+1} \) involves shifting and scaling, needing special techniques like substitution to tackle.
When integrating, remember:

• The integral of \( e^x \) is \( e^x + C \), where \( C \) is the integration constant.
• Substitution may simplify the integral by changing variables to a more manageable form.

Solving with the Substitution Method

The substitution method is a technique used to simplify integration, especially when dealing with composite functions. For our differential equation, integrating \( e^{t+1} \) directly can be complex. By substituting \( u = t+1 \), the expression becomes simpler to handle.
Here's how substitution works:

• Substitute a part of an integral to make it simpler. Example: Let \( u = t + 1 \), then \( du = dt \). It simplifies the calculations.
• Change variables, thus transforming the integral into \( \int e^u \, du \), which is easier to compute.
• After finding the integral in terms of the new variable \( u \), re-substitute back to the original variable \( t \).

Substitution can significantly reduce the complexity of finding integrals, turning daunting computations into manageable steps. By mastering this method, tackling varied and complex integrals becomes much simpler.