Problem 19 - Advanced Engineering Mathematics

Question

From \(d A / d t=4-A / 50\) we obtain \(A=200+c e^{-t / 50} .\) If \(A(0)=30\) then \(c=-170\) and \(A=200-170 e^{-t / 50}\).

Step-by-Step Solution

Answer

The expression for \( A(t) \) is \( A(t) = 200 - 170 e^{-t/50} \).

Step 1: Understand the Problem

We need to verify the function \( A(t) = 200 + c e^{-t/50} \) given the differential equation \( \frac{dA}{dt} = 4 - \frac{A}{50} \) and initial condition \( A(0) = 30 \). Additionally, we need to find the constant \( c \) when \( A(0) = 30 \).

Step 2: Verify the Solution Form

Given the differential equation \( \frac{dA}{dt} = 4 - \frac{A}{50} \), we found the solution in form \( A(t) = 200 + c e^{-t/50} \). This form includes a particular solution (the constant 200) and a general solution with an arbitrary constant \( c \).

Step 3: Apply the Initial Condition

We substitute \( t = 0 \) into \( A(t) = 200 + c e^{-t/50} \) to find the constant \( c \). The condition provides \( A(0) = 30 \), leading to:\[30 = 200 + c e^{0} \Rightarrow 30 = 200 + c.\]

Step 4: Solve for Constant \( c \)

From the equation \( 30 = 200 + c \), solve for \( c \):\[c = 30 - 200 = -170.\]Thus, \( c = -170 \).

Step 5: Final Expression for \( A(t) \)

Substitute \( c = -170 \) back into the general solution expression:\[A(t) = 200 - 170 e^{-t/50}.\]This is the required solution for the differential equation with the given initial condition.

Key Concepts

Initial Conditions

Understanding initial conditions is essential when working with differential equations. This is because initial conditions allow us to determine a unique solution from a family of possible solutions. In our example, the differential equation given is \(\frac{dA}{dt} = 4 - \frac{A}{50}\), and we have an initial condition of \(A(0) = 30\).

To apply the initial condition, we substitute the known values into the solution. Here, the solution is of the form \(A(t) = 200 + c e^{-t/50} \). By substituting \(t = 0\), we can find the constant \(c\) that fits the specific condition set for the problem.

This process reveals that our function must satisfy \(A(0) = 30 = 200 + c\), leading us to determine that \(c = -170\). With this information, we can derive the particular solution specific to the initial condition provided.

Particular and General Solutions

Differential equations often supply solutions that include both a particular solution and a general solution. The particular solution satisfies a specific set of conditions, while the general solution encompasses a family of functions that might solve the same differential equation without additional constraints.

In the equation giving us \(A(t) = 200 + c e^{-t/50}\), the constant \(200\) represents a particular solution because it provides a steady-state solution when the effect of \(e^{-t/50}\) fades away. In contrast, the term \(c e^{-t/50}\) is the part of the general solution that accounts for variability and needs to be adjusted by initial conditions.

• Particular solution: Constant term determined by stabilizing terms
• General solution: Contains arbitrary constants fitted by initial conditions

This blending of particular and general solutions allows us to satisfy different scenarios, with initial conditions helping us pinpoint the exact behavior needed for our problem.

Exponential Functions

Exponential functions often appear in the solutions of differential equations, reflecting the dynamic nature of changing systems. They are particularly helpful in modeling growth and decay processes. In this exercise, the exponential term \(e^{-t/50}\) is central to our understanding of the solution's behavior.

The term \(e^{-t/50}\) originates from solving linear differential equations and represents the decaying influence over time. In our equation \(A(t) = 200 - 170 e^{-t/50}\), this term signifies how the initial conditions affect the rate of change and the eventual approach to the steady-state solution, represented by \(200\).

• Exponential decay \(e^{-t/50}\): Describes the reducing impact over time
• Steady state: Value the function approaches as time progresses

By understanding this relationship, we can predict how the function will evolve as time increases, revealing the trajectory of growth or decay in various contexts.