Problem 4 - Applied Calculus

Question

Find particular solutions \(\frac{d B}{d t}=4 B-100, \quad B=20\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \(B = 25 - 5e^{4t}\).

Step 1: Identify the Type of Equation

The differential equation given is \(\frac{dB}{dt} = 4B - 100\). This is a first-order linear ordinary differential equation.

Step 2: Rewrite the Equation in Standard Form

To solve the differential equation, it can be rewritten in the standard form: \(\frac{dB}{dt} - 4B = -100\). This will help us use integrative techniques.

Step 3: Find the Integrating Factor

The integrating factor \(\mu(t)\) is calculated as \(e^{\int -4 \, dt} = e^{-4t}\). We will multiply through by this integrating factor to solve the equation.

Step 4: Solve Using the Integrating Factor

Multiply the entire differential equation by \(e^{-4t}\): \(e^{-4t}\frac{dB}{dt} - 4e^{-4t}B = -100e^{-4t}\). The left-hand side becomes the derivative of \((Be^{-4t})\).

Step 5: Integrate Both Sides

Integrate both sides of the equation: \(\int d(Be^{-4t}) = \int -100e^{-4t} \, dt\). This gives \(Be^{-4t} = 25e^{-4t} + C\).

Step 6: Solve for B

To find \(B\), multiply through by \(e^{4t}\): \(B = 25 + Ce^{4t}\). This is the general solution of the differential equation.

Step 7: Apply Initial Conditions

Substitute the initial condition \(B = 20\) when \(t = 0\): \(20 = 25 + C\cdot 1\). Solving for \(C\) gives \(C = -5\).

Step 8: Write the Particular Solution

Using the value of \(C\), the particular solution to the differential equation is \(B = 25 - 5e^{4t}\).

Key Concepts

First-Order Linear Differential Equation

First-order linear differential equations are a type of ordinary differential equation (ODE) that involve the first derivative of a function. These equations have the general form: \[ \frac{dy}{dt} + P(t) y = Q(t) \] Here, \( y \) is the dependent variable, \( t \) is the independent variable, \( P(t) \) and \( Q(t) \) are functions of \( t \) only. The term 'first-order' indicates that the equation includes only the first derivative, meaning it describes how a process changes at a particular moment. Such equations are quite common because they can model many real-life systems, like heat flow or population growth.

• This specific structure lets us use special techniques to find solutions.
• Recognizing the equation type is the first step in solving it effectively.

For the differential equation given: \( \frac{dB}{dt} = 4B - 100 \), it is rearranged to match the standard form and solutions are pursued.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. This method involves finding a multiplier, called the integrating factor, which simplifies the process of solving the equation. The key steps are:1. Identify the function \( P(t) \) in the standard form equation \( \frac{dy}{dt} + P(t) y = Q(t) \).2. Calculate the integrating factor \( \mu(t) \) using the formula \( e^{\int P(t) \, dt} \).Once the integrating factor is determined, multiply the entire differential equation by this factor. The purpose is to simplify the left-hand side into an exact derivative, which can be more straightforwardly integrated. In our problem scenario, for the equation: \[ \frac{dB}{dt} - 4B = -100 \]The integrating factor \( \mu(t) \) is calculated as: \[ \mu(t) = e^{-4t} \]This allows the simplified equation to be integrated easily, leading you towards solving for the function \( B(t) \). Using this method can seem tricky at first, but it is very systematic once you practice.

Initial Value Problem

An initial value problem involves finding a particular solution to a differential equation that meets a specific condition at an initial point. This is crucial because many real-world problems require not only the solution to a differential equation but one that fits a certain scenario at the start.For initial value problems, you are typically given:

• A differential equation.
• An initial condition, represented as a point such as \( y(t_0) = y_0 \).

In our exercise, we have the initial value condition given as \( B = 20 \) when \( t = 0 \). This means that the solution to the differential equation must not only fit the equation itself but must also pass through the point when \( B \) equals 20 at \( t = 0 \). Practically, this involves substituting the initial values into the general solution obtained to solve for the constant of integration, \( C \). When done correctly, this results in a particular solution that is tailored to the initial situation provided, as seen in the specific solution: \[ B = 25 - 5e^{4t} \]Thus, solving initial value problems is crucial in crafting solutions that are applicable and meaningful for specific scenarios.