Problem 28 - Calculus

Question

(a) Let \(y=A+B e^{-2 t} .\) For what values of \(A\) and \(B\), if any, is \(y\) a solution to the differential equation $$\frac{d y}{d t}=100-2 y ?$$ Give the general solution to the differential equation. (b) If the solution satisfies \(y=85\) when \(t=0,\) what more (if anything) can you say about \(A\) and \(B ?\) Give the particular solution to the differential equation with this initial condition.

Step-by-Step Solution

Answer

(a) General solution: \(y = 50 + Be^{-2t}\); (b) Particular solution: \(y = 50 + 35e^{-2t}\).

Step 1: Differentiate the given function

Given function: \( y = A + B e^{-2t} \). Differentiate this with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(A + B e^{-2t}) = -2B e^{-2t}. \]

Step 2: Plug into the differential equation

Substitute \( y = A + B e^{-2t} \) and its derivative \( \frac{dy}{dt} = -2B e^{-2t} \) into the differential equation \( \frac{dy}{dt} = 100 - 2y \): \[ -2B e^{-2t} = 100 - 2(A + B e^{-2t}). \]

Step 3: Simplify the equation

Equate both sides of the equation from Step 2: \[ -2B e^{-2t} = 100 - 2A - 2B e^{-2t}. \]Cancel the terms \(-2B e^{-2t}\): \[ 0 = 100 - 2A. \]

Step 4: Solve for A

From \( 0 = 100 - 2A \), solve for \( A \): \[ 2A = 100 \] \[ A = 50. \]

Step 5: Establish values for A and B

We found that the equation holds for any \( B \) when \( A = 50 \). Thus, \( A = 50 \) and \( B \) can be any real number. Therefore, \( y = 50 + Be^{-2t} \) solves the differential equation.

Step 6: Examine the initial condition

Given initial condition: \( y = 85 \) when \( t = 0\), substitute into the equation: \[ 85 = 50 + Be^{0} \] Simplify: \[ B = 35. \]

Step 7: Particular solution with initial condition

With \( B = 35 \) from Step 6, the particular solution is \( y = 50 + 35e^{-2t} \).

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function that involve an exponent. In this case, the function includes terms like \( Be^{-2t} \), where the base of the exponent is Euler's number \( e \) (approximately 2.718). This specific function is characterized by its rapidly changing nature, where the rate of change is proportional to its current value. Exponential functions are commonly encountered in differential equations. They often describe processes like decay or growth, modeling situations such as population dynamics or radioactive decay. When dealing with differential equations, these functions help to express solutions where change happens at a continuous and natural rate. Some key aspects of exponential functions:

• The function \( e^{-2t} \) decreases as \( t \) increases, reflecting decay over time.
• The coefficient \( B \) determines the initial rate and scale of the change.
• They arise naturally when solving linear differential equations with constant coefficients.

Understanding exponential functions helps simplify the process of solving complex differential equations by converting them into more manageable expressions.

Initial Conditions

Initial conditions are crucial when trying to find a unique solution to a differential equation. They specify the value of the function at a certain point, often at the start of the observation, such as \( y = 85 \) when \( t = 0 \). Initial conditions allow us to find particular solutions to differential equations rather than just a general solution. When you're given a differential equation and need a particular solution, integrating the differential equation gives you a general solution with constants, like \( A \) and \( B \). The initial condition provides necessary information to solve these constants and find a unique solution. In our exercise example, when \( t = 0 \), the condition that \( y = 85 \) helps us find \( B \) in the function \( y = 50 + Be^{-2t} \). Without it, there would be infinite possibilities for \( B \). Applying initial conditions:

• Substitute the initial values into the general solution.
• Solve the resulting equation to find the unknown constants.
• Derive the particular solution that satisfies the initial settings.

The use of initial conditions is essential for practical applications that need precise predictions or control.

Particular Solution

The term "particular solution" refers to a solution of a differential equation that satisfies the given initial conditions. In many differential equations, the general solution includes arbitrary constants. To find a particular solution, we need to substitute the initial conditions into this general format, allowing us to calculate these constants. This results in a single, specific solution that fits exactly with the initial data provided. For example, in the scenario where we have the differential equation \( \frac{dy}{dt} = 100 - 2y \) and the solution \( y = 50 + Be^{-2t} \), applying the condition \( y = 85 \) when \( t = 0 \) allows us to determine that \( B = 35 \). Thus, the particular solution becomes \( y = 50 + 35e^{-2t} \).Here's how to find the particular solution:

• Start with the general solution of the differential equation.
• Apply the initial conditions to replace any arbitrary constants with specific values.
• Verify the solution satisfies both the differential equation and initial conditions.

This particular solution caters to the specific settings of your problem, making it invaluable in engineering, physics, and various sciences for accurately describing systems.