Problem 20 - Advanced Engineering Mathematics

Question

From \(d A / d t=0-A / 50\) we obtain \(A=c e^{-t / 50} .\) If \(A(0)=30\) then \(c=30\) and \(A=30 e^{-t / 50}\).

Step-by-Step Solution

Answer

The solution is \(A(t) = 30 e^{-t/50}\).

Step 1: Understand the Problem

We are given a differential equation \(\frac{dA}{dt} = 0 - \frac{A}{50}\) and the solution form \(A = c e^{-t/50}\). We need to find the constant \(c\) given that \(A(0) = 30\).

Step 2: Solve for the Constant \(c\)

The problem states \(A(0) = 30\). Substitute \(t = 0\) into the equation \(A = c e^{-t/50}\) to get \(A(0) = c e^{0} = c\). Thus, \(c = 30\).

Step 3: Write the Solution for \(A(t)\)

With \(c = 30\), substitute it back into the equation \(A = c e^{-t/50}\) to obtain \(A(t) = 30 e^{-t/50}\). This is the particular solution to the differential equation given the initial condition.

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations involves finding a particular solution that not only satisfies the differential equation but also fulfills a specified initial condition. This initial condition acts as a starting point or specific value that the solution must adopt when the independent variable, often time, is at a particular value, like 0.

In our exercise, the differential equation given is \[\frac{dA}{dt} = -\frac{A}{50}\]and the initial condition is given as \(A(0) = 30\). This means when time \(t = 0\), the quantity \(A\) should equal 30.

• First, recognize the type of equation you're dealing with. Here, we have a first-order linear homogeneous differential equation.
• Understand the initial condition. It specifies a value that the solution to the differential equation must have for a given value of \(t\), ensuring that the solution is uniquely tailored to the problem.

Without this initial condition, there would be infinitely many solutions. The initial condition allows us to pinpoint the exact trajectory of our system solution.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This is observable in phenomena like radioactive decay, cooling of objects, and depreciation of asset value.

In the exercise, the solution to our differential equation is in the form \[A(t) = c e^{-t/50}\]This equation indicates an exponential decay process, where the rate of change of \(A\) diminishes as time progresses.

• \(c\) is the initial value, often found using the initial condition, in this case \(c = 30\).
• The term \(e^{-t/50}\) represents how quickly the quantity \(A(t)\) decays over time. The larger the coefficient in the exponent (e.g., \(-1/50\) here), the slower the decay.

Exponential decay is vital to understand as it models many real-life processes where quantities decrease rapidly at first and then more slowly.

Constant of Integration

The constant of integration, often denoted as \(c\), is crucial in solving differential equations. It represents an unknown constant that we determine using initial or boundary conditions.

When solving a differential equation, after integrating, you will usually have an arbitrary constant. For our exercise:\[A = c e^{-t/50}\]We must determine \(c\) to fully solve the problem, and we do so using the initial condition \(A(0) = 30\). By substituting into the equation, we find \(c = 30\).

• The presence of \(c\) accounts for the general family of solutions to a differential equation, accounting for all possible shifts in initial conditions.
• When an initial condition is applied, the constant \(c\) is no longer arbitrary and becomes a specific value, bridging the gap between general and particular solutions.

Therefore, the constant of integration is a bridge between a generic solution and one that fits all given conditions of the problem specification.