Problem 54 - Calculus for Biology and Medicine

Question

Find the general solution of the differential equation. $$ \frac{d y}{d t}=1-e^{-2 t}, t \geq 0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = t + \frac{1}{2}e^{-2t} + C \).

Step 1: Identify the Differential Equation Type

The given equation is \( \frac{dy}{dt} = 1 - e^{-2t} \). This is a simple first-order ordinary differential equation, which is in the form of \( \frac{dy}{dt} = f(t) \).

Step 2: Integrate Both Sides

To solve for \( y(t) \), integrate both sides of the equation with respect to \( t \). This results in the equation \( y(t) = \int (1 - e^{-2t}) \, dt \).

Step 3: Perform the Integration

Calculate the integral: \[ y(t) = \int 1 \, dt - \int e^{-2t} \, dt. \] The integral of 1 with respect to \( t \) is \( t \), and the integral of \( e^{-2t} \) is \( -\frac{1}{2}e^{-2t} \). Therefore, \[ y(t) = t + \frac{1}{2}e^{-2t} + C, \] where \( C \) is the constant of integration.

Step 4: State the General Solution

Combine the results to state the general solution to the differential equation. The general solution is \[ y(t) = t + \frac{1}{2}e^{-2t} + C. \]

Key Concepts

Integration

Integration is a key mathematical process which involves finding the anti-derivative or "integral" of a function. In other words, when you integrate a function, you are essentially reversing the process of differentiation. Integration helps us solve differential equations by allowing us to find a function, given its rate of change.

In the original exercise, we integrate both sides of the equation \( \frac{dy}{dt} = 1 - e^{-2t} \) with respect to the variable \( t \). This is represented as:

• \( y(t) = \int (1 - e^{-2t}) \, dt \)

This equation tells us what \( y(t) \) should be, based on integrating the right-hand side expression. Performing this integration leads to:

• \( y(t) = \int 1 \, dt - \int e^{-2t} \, dt \)
• Solving these integrals gives us \( y(t) = t + \frac{1}{2}e^{-2t} + C \)

The constant \( C \) represents the constant of integration, which appears because when we integrate, we retrieve a whole family of solutions, depending on the value of \( C \). This reflects the multitude of functions whose derivatives could equal the given rate of change.

First-Order Ordinary Differential Equation

A first-order ordinary differential equation (ODE) is an equation involving derivatives of a function concerning one variable. It usually has the form \( \frac{dy}{dt} = f(t, y) \). The primary characteristic is that it features only the first derivative of the unknown function.

In this exercise, the equation \( \frac{dy}{dt} = 1 - e^{-2t} \) is a first-order ODE because it involves the first derivative of the function \( y \) with respect to \( t \), the independent variable. The terms on the right do not involve \( y \), making it particularly straightforward to solve.

Differential equations like this model processes where change rates depend on the current state. They are crucial in fields such as physics, engineering, and economics to describe systems in motion or change.

General Solution

The "general solution" of a differential equation provides a comprehensive set of solutions representing all possible instances of the function that satisfy the equation. It usually includes an arbitrary constant, symbolized as \( C \), reflecting the unaccounted initial conditions.

For the equation \( \frac{dy}{dt} = 1 - e^{-2t} \), we find the general solution by integrating. Here’s a recap of that process:

• We integrated \( 1 - e^{-2t} \) to get \( y(t) = t + \frac{1}{2}e^{-2t} + C \)

This expression \( t + \frac{1}{2}e^{-2t} + C \) includes a constant \( C \), which is a defining feature of a general solution. The value of \( C \) can be determined if initial conditions (e.g., \( y(0) = y_0 \)) are provided, which would then lead to a "particular solution".

The central idea of a general solution is capturing the entire family of potential functions corresponding to the differential equation. This makes it immensely beneficial for modeling various scenarios within applied mathematics.