91Ó°ÊÓ

Differential equations are mathematical equations that involve the rates of change of a quantity. They describe how a certain function changes over time or other variables, using derivatives. In our context, the differential equation \( \frac{d W}{d t} = e^{-3 t} \) describes how the function \( W(t) \) evolves as the variable \( t \) changes. This equation indicates that the rate of change of \( W \) with respect to \( t \) is given by \( e^{-3 t} \). Such equations are fundamental in mathematical modeling of physical phenomena, from the growth of populations to the decay of radioactive material. Recognizing and setting up a differential equation is a crucial first step in solving problems involving dynamic systems. Once a differential equation is established, we can move on to finding solutions, which often involves integration.

Integration is a fundamental mathematical process used to solve differential equations. It allows us to find the original function given its rate of change. When we integrate the function \( e^{-3t} \), we essentially "reverse" the differentiation process to recover \( W(t) \).The integration of both sides in the differential equation \( \int dW = \int e^{-3t} dt \) leads to:

• \( W(t) \) on the left side
• \( -\frac{1}{3}e^{-3t} + C \) on the right side

Here, the symbol \( \int \) denotes integration, and the result of the integral on the right-hand side includes a term \( C \), known as the constant of integration. Each step in integration brings us closer to solving for the function \( W(t) \).

The constant of integration, \( C \), appears in indefinite integrals because integration is the reverse of differentiation, and the differentiation of a constant is zero. If we only know the derivative of a function, the original function might contain an unknown constant. That's where \( C \) comes into play, representing any constant value that could have been present in the original function.In our solution, after integrating \( e^{-3t} \), the general solution includes the term \( C \). To find the exact form of \( W(t) \), and not just any solution but the one that fits specific conditions, we determine \( C \) using initial conditions. In our exercise, \( C \) was determined using the condition \( W(0) = \frac{2}{3} \). This particular step ensures that we derive a solution specific to the problem's given context.

A particular solution to a differential equation satisfies not only the equation itself but also a given set of initial conditions. The initial condition \( W(0) = \frac{2}{3} \) allows us to specify which of the infinitely many solutions the equation could have is the correct one for this problem.Inserting the initial condition into the general solution \( W(t) = -\frac{1}{3}e^{-3t} + C \), we calculate and find \( C = 1 \). Thus, the particular solution becomes \( W(t) = -\frac{1}{3}e^{-3t} + 1 \). This particular solution not only solves the differential equation \( \frac{d W}{d t} = e^{-3 t} \) but also satisfies the specific condition \( W(0) = \frac{2}{3} \), ensuring it is tailored to the problem's requirements.