Problem 7 - Applied Calculus

Question

Find particular solutions \(\frac{d B}{d t}+2 B=50, \quad B(1)=100\)

Step-by-Step Solution

Answer

The particular solution is \( B(t) = 25 + 75e^{2}e^{-2t} \).

Step 1: Write down the differential equation

The given differential equation is \( \frac{dB}{dt} + 2B = 50 \). We also have an initial condition \( B(1) = 100 \).

Step 2: Find the general solution

This is a first order linear differential equation and can be solved using an integrating factor. The integrating factor \( \mu(t) \) is \( e^{\int 2 \, dt} = e^{2t} \). Multiply through by the integrating factor to get \( e^{2t} \frac{dB}{dt} + 2e^{2t}B = 50e^{2t} \).

Step 3: Simplify using the integrating factor

The left hand side can be rewritten as \( \frac{d}{dt}(e^{2t}B) = 50e^{2t} \). This simplifies the equation.

Step 4: Integrate both sides

Integrate both sides with respect to \( t \) to find \( e^{2t}B = \int 50e^{2t} \, dt \). The integral of \( 50e^{2t} \) is \( 25e^{2t} + C \) where \( C \) is the constant of integration.

Step 5: Solve for B(t)

We have \( e^{2t}B = 25e^{2t} + C \). Dividing through by \( e^{2t} \) gives \( B = 25 + Ce^{-2t} \).

Step 6: Use the initial condition

Plug the initial condition \( B(1) = 100 \) into the equation \( B = 25 + Ce^{-2t} \). This gives \( 100 = 25 + Ce^{-2(1)} \), simplifying to \( 100 = 25 + Ce^{-2} \).

Step 7: Solve for the constant C

Rearrange the equation to solve for \( C \). This gives \( 75 = Ce^{-2} \), so \( C = 75e^{2} \).

Step 8: Write the particular solution

Substitute \( C \) back into the equation from Step 5 to find the particular solution: \( B(t) = 25 + 75e^{2}e^{-2t} \).

Key Concepts

First Order Linear Differential Equations

First order linear differential equations are equations that involve the first derivative of a function and the function itself. They can be expressed in the general form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

Here, \( y \) is the dependent variable we're trying to find, and \( x \) is the independent variable. The functions \( P(x) \) and \( Q(x) \) are known, and they dictate how \( y \) and its derivative relate to each other. These equations are called "linear" because they involve a linear combination of \( y \) and its derivative.

This type of equation is prevalent in real-world applications, including physics, biology, and economics, as it often models processes that change over time in a linear fashion. Understanding how to solve these equations is crucial as it forms the basis for more complex differential equations encountered later in studies.

Integrating Factor Method

The integrating factor method is a powerful tool used to solve first order linear differential equations. This method makes solving these equations systematic and straightforward.

• Identify the differential equation in the form: \( \frac{dy}{dx} + P(x)y = Q(x) \).
• Find the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the entire equation by \( \mu(x) \) to simplify the left-hand side into the derivative of a product.

This transformation helps us express the equation as:

• \( \frac{d}{dx}(\mu(x)y) = \mu(x)Q(x) \)

The benefit is that we can integrate both sides of this equation directly. Once integrated, we solve for \( y(x) \) by isolating \( y \). This method effectively unravels the relationship encapsulated within the original equation, allowing for an explicit solution.

Initial Condition

In solving differential equations, initial conditions are critical because they allow us to find specific solutions rather than just a general solution. The initial condition provides a particular value of the dependent variable at a specific point, typically given as:

• \( y(x_0) = y_0 \)

Here, \( x_0 \) is the point where the initial condition is applied, and \( y_0 \) is the value of the function at that point.

When dealing with a differential equation, the general solution often contains arbitrary constants that arise from the integration process. By substituting the values from the initial condition into the general solution, we can determine these constants, giving us a unique solution known as the particular solution. This is crucial when the solution must align with real-world phenomena or specific experimental data.

Particular Solution

A particular solution of a differential equation is a solution that satisfies both the differential equation and the initial conditions. It is derived from the general solution by using the initial conditions to determine any constants present in the equation.

Using the initial condition provided, substitute this into the general solution:

• The general solution is of the form: \( y = g(x) + C \cdot h(x) \)

Here, \( C \) is the constant to be found, based on the initial condition. For example, if the initial condition is \( y(x_0) = y_0 \), apply these values:

• Plug \( x_0 \) and \( y_0 \) into the general equation to solve for \( C \).

Once \( C \) is known, put it back into the general solution to form the particular solution. This targeted solution is critical when modeling systems as it correctly aligns with specific starting conditions or observational data.