Problem 68 - Calculus for Biology and Medicine

Question

Solve the initial-value problem. $$ \frac{d W}{d t}=e^{-5 t}, \text { for } t \geq 0 \text { with } W(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( W(t) = -\frac{1}{5} e^{-5t} + \frac{6}{5} \).

Step 1: Recognize the type of equation

The given equation \( \frac{d W}{d t} = e^{-5t} \) is a first-order ordinary differential equation of the form \( \frac{dW}{dt} = f(t) \). Our goal is to find \( W(t) \) that satisfies this equation and the initial condition \( W(0) = 1 \).

Step 2: Integrate both sides

Integrate both sides of the equation with respect to \( t \):\[W(t) = \int e^{-5t} \, dt.\]

Step 3: Solve the integral

The integral \( \int e^{-5t} \, dt \) can be solved using the substitution method or by recognizing the antiderivative. The antiderivative of \( e^{-5t} \) is:\[-\frac{1}{5} e^{-5t} + C,\]where \( C \) is the constant of integration.

Step 4: Apply the initial condition

Use the initial condition \( W(0) = 1 \) to solve for the constant \( C \). Substitute \( t = 0 \) into the general solution:\[W(0) = -\frac{1}{5} e^{0} + C = 1.\]Since \( e^{0} = 1 \), simplifying gives:\[-\frac{1}{5} + C = 1.\]

Step 5: Solve for Constant C

Rearrange the equation from Step 4 to solve for \( C \):\[C = 1 + \frac{1}{5} = \frac{6}{5}.\]

Step 6: Write the particular solution

Substitute the constant \( C = \frac{6}{5} \) back into the general solution:\[W(t) = -\frac{1}{5} e^{-5t} + \frac{6}{5}.\]This is the particular solution to the initial-value problem.

Key Concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving a function and its derivatives. They relate some function to its rate of change, which is expressed by the derivatives of that function. This makes them vital in modeling how things evolve over time, such as the change in populations or physical systems. In our problem, the equation \( \frac{d W}{d t} = e^{-5t} \) is a first-order ODE, meaning it involves the first derivative of function \( W(t) \). Since ODEs can often predict or explain physical phenomena or mathematical processes, they're crucial in both mathematics and applied fields. Our task is to determine the function \( W(t) \) that satisfies both the differential equation and the given initial condition \( W(0) = 1 \). This ensures the solution is precisely tailored to a specific scenario.

Integration Techniques

Integration techniques are key to solving ordinary differential equations, especially when dealing with continuous processes. In our case, the core technique is finding the antiderivative, which helps reverse the differentiation process. The function \( e^{-5t} \) needs to be integrated to find \( W(t) \). Integration can often involve methods like substitution or integration by parts. However, when dealing with exponentials like \( e^{-kt} \), recognizing patterns or using simple substitutions can directly lead to the antiderivative. By finding the antiderivative, we are essentially summing up all the infinitesimal changes dictated by the differential equation, thus resulting in the original function \( W(t) \). This operation provides us with a general form of the solution, which is vital before applying any particular conditions.

Antiderivatives

Antiderivatives represent the family of functions which, when differentiated, give back the original function. For example, the antiderivative of \( e^{-5t} \) is \(-\frac{1}{5} e^{-5t} + C\), where \( C \) is a constant. The process of finding antiderivatives is known as integration, and is crucial in solving differential equations. In the context of our problem, finding the antiderivative helps us determine the general solution of the ordinary differential equation. Each member of the antiderivative family can be differentiated to result in the same derivative, but fit different scenarios explained by initial or boundary conditions. This means, without further information, any curve in this family could fit the derivative information of our problem.

Constant of Integration

The constant of integration, often denoted as \( C \), plays a pivotal role when integrating. It represents an entire family of solutions to an indefinite integral, as integration underdetermines a solution by ignoring shifts up or down along the y-axis. In our initial-value problem, the constant \( C \) ensures the particular solution aligns with the initial condition \( W(0) = 1 \). This step bounds one solution from the many potential ones indicated by the general formula. By plugging the initial condition into the general antiderivative, one can solve for \( C \), which precisely tailors the outcome to meet the starting point of the scenario. Thus, it transforms a generalized solution into a specific one fitting the given conditions.