Problem 21 - Advanced Engineering Mathematics

Question

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

Step-by-Step Solution

Answer

The solution is \( A(t) = 1000 - 1000e^{-t/100} \).

Step 1: Understanding the Problem

We need to start by understanding that the differential equation given is \( \frac{dA}{dt} = 10 - \frac{A}{100} \). We are also given its solution in the form \( A = 1000 + ce^{-t/100} \), and we need to find the value of \( c \) given the initial condition \( A(0) = 0 \).

Step 2: Applying the Initial Condition

Substitute the initial condition \( A(0) = 0 \) into the solution \( A = 1000 + ce^{-t/100} \). This gives us \( 0 = 1000 + ce^{0} \). Since \( e^{0} = 1 \), we have \( 0 = 1000 + c \).

Step 3: Solving for Constant c

Solve the equation for \( c \). From \( 0 = 1000 + c \), we find \( c = -1000 \). This means our particular solution becomes \( A(t) = 1000 - 1000 e^{-t/100} \).

Key Concepts

Initial Value Problem

An initial value problem in differential equations involves finding a specific solution to a differential equation that satisfies a given condition at a certain point. This condition is usually expressed as the value of the function at a particular time or position, known as the initial condition. Here’s how it works:

• You start with a differential equation, which is a mathematical equation that relates a function with its derivative.
• An initial condition is provided alongside this equation to help find a unique solution. For instance, in our case, we have A(0) = 0.
• Using the initial condition, you solve for any constants in the general solution to find the particular solution, which uniquely satisfies the differential equation at the given point.

The beauty of initial value problems is that they lead us to a specific answer, tailoring a general solution to meet specific needs at a given point. They are crucial in modeling real-world situations where initial conditions are known or can be measured.

Exponential Solutions

Exponential solutions emerge when solving certain differential equations, particularly those with terms involving exponential functions. Let’s delve into their properties:

• They are marked by solutions that include the exponential function e. These functions have a base of e, Euler's number, which is approximately 2.718.
• Exponential solutions often take the form ce^-kt in differential equations, where c is a constant and k is a rate constant.
• The exponential function has interesting properties, such as rapid growth or decay, which makes it highly applicable to natural phenomena like population growth, radioactive decay, and more.

In our context, the solution includes an exponential term ce^{-t/100}, reflecting decay as time increases. By substituting the initial condition, we found the constant c that fits our particular solution to the problem's constraints.

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear differential equations of the form \( \frac{dy}{dt} + P(t)y = Q(t) \). It simplifies these equations in a few systematic steps:

• The goal is to multiply the entire equation by an integrating factor \( \mu(t) \), which is generally \( e^{\int P(t) \, dt} \). This transforms the equation into one that can be rewritten in terms of a derivative of a product.
• The equation becomes \( \frac{d}{dt}(\mu(t) y) = \mu(t) Q(t) \). This crucial step turns the initial problem into an easier one to integrate.
• After this transformation, the equation can be integrated with respect to \( t \), allowing us to solve for \( y(t) \).

This method is beneficial when dealing with equations that fit the first-order linear type, leading to solutions that might include exponential terms as seen in our problem. It provides a structured path to arriving at the solution precisely and efficiently.