Problem 29 - Advanced Engineering Mathematics

Question

Assume \(R d q / d t+(1 / C) q=E(t), R=200, C=10^{-4},\) and \(E(t)=100\) so that \(q=1 / 100+c e^{-50 t} .\) If \(q(0)=0\) then \(c=-1 / 100\) and \(i=\frac{1}{2} e^{-50 t}\).

Step-by-Step Solution

Answer

The solution is consistent with \( c = -\frac{1}{100} \) and \( i(t) = \frac{1}{2} e^{-50t} \).

Step 1: Identify the Given Equation and Parameters

We are given the differential equation \( R \frac{dq}{dt} + \frac{1}{C} q = E(t) \) with parameters \( R = 200 \), \( C = 10^{-4} \), and \( E(t) = 100 \). We also have the solution form \( q = \frac{1}{100} + c e^{-50t} \).

Step 2: Determine the Value of c Using Initial Condition

Since \( q(0) = 0 \), we substitute this into the solution form: \( 0 = \frac{1}{100} + c e^{0} \). This simplifies to \( 0 = \frac{1}{100} + c \), leading to \( c = -\frac{1}{100} \).

Step 3: Confirm Charge Function q(t)

Substitute \( c = -\frac{1}{100} \) back into the general solution to get \( q(t) = \frac{1}{100} - \frac{1}{100} e^{-50t} \). This satisfies the initial condition \( q(0) = 0 \).

Step 4: Define and Confirm Current Function i(t)

Based on the solution, the current \( i(t) = \frac{dq}{dt} \) is given as \( i = \frac{1}{2} e^{-50t} \). We differentiate \( q(t) \): \( \frac{d}{dt} \left( \frac{1}{100} - \frac{1}{100} e^{-50t} \right) = 50 \times \frac{1}{100} e^{-50t} = \frac{1}{2} e^{-50t} \). This confirms \( i(t) \) as computed.

Key Concepts

First-order differential equations

A first-order differential equation involves derivatives of the first degree. In these equations, the solution depends on the value of one derivative.
These types of equations model a variety of real-world phenomena such as electrical circuits, population growth, and fluid movement. They are typically written in the form:

• \( \frac{dy}{dt} = f(t, y) \)

In our specific exercise, the given first-order differential equation is:

• \( R \frac{dq}{dt} + \frac{1}{C} q = E(t) \)

Here, \( R \) and \( C \) are constants, \( E(t) \) represents an external input, often referred to as a driving function, and \( q \) represents the charge.
Solving such equations often involves identifying the dependent and independent variables, and using initial conditions to determine the particular solution.

Initial value problem

An initial value problem (IVP) in differential equations emphasizes finding a solution when some initial conditions are given.
It is crucial because a differential equation can have many possible solutions; the initial conditions help pinpoint a unique one. In our context, the initial condition states that the charge \( q(0) = 0 \) when \( t = 0 \).
This condition is essential to determining the constant \( c \) in the equation \( q = \frac{1}{100} + c e^{-50t} \). By substituting \( q(0) = 0 \) into the expression, we find that:

• \( 0 = \frac{1}{100} + c \times e^{0} \)
• This simplifies to \( c = -\frac{1}{100} \)

Thus, the initial condition helps us adjust our equation to accurately reflect the starting scenario of our model.

Exponential solutions

In many differential equations, exponential functions emerge as part of the solution due to their unique derivatives.

• The exponential function is often in the form \( e^{kt} \), where \( k \) is a constant.

In our case, the solution possesses an exponential component \( c e^{-50t} \).
The properties of the exponential function are useful because they model decay or growth processes efficiently. Here, the exponential term represents how the charge \( q(t) \) changes over time, decaying as time progresses.
Exponential solutions are significant in real life for processes that involve rapid change or decay, such as radioactive decay, cooling, and electrical discharge in circuits. By understanding the rate of change determined by the exponent, one can predict future behavior more accurately.