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

Particular solution: \(B = 25 + 75e^{2-2t}\).

Step 1: Identify the Type of Differential Equation

This equation, \(\frac{dB}{dt} + 2B = 50\), is a first-order linear differential equation. It fits the standard form \(\frac{dy}{dt} + P(t)y = Q(t)\), where \(P(t) = 2\) and \(Q(t) = 50\).

Step 2: Find the Integrating Factor

The integrating factor, \(\mu(t)\), is given by the expression \(\mu(t) = e^{\int P(t) \, dt} = e^{\int 2 \, dt} = e^{2t}\).

Step 3: Multiply through by the Integrating Factor

Multiply the entire differential equation by the integrating factor: \(e^{2t} \frac{dB}{dt} + 2Be^{2t} = 50e^{2t}\). This simplifies to \(\frac{d}{dt}(Be^{2t}) = 50e^{2t}\).

Step 4: Integrate Both Sides

Integrate both sides with respect to \(t\). For the left side: \(\int \frac{d}{dt}(Be^{2t}) \, dt = Be^{2t}\). For the right side: \(\int 50e^{2t} \, dt = 25e^{2t} + C\), where \(C\) is the constant of integration.

Step 5: Solve for \(B\)

Equating both integrated results gives \(Be^{2t} = 25e^{2t} + C\). Dividing by \(e^{2t}\), we find \(B = 25 + Ce^{-2t}\).

Step 6: Apply Initial Condition

Use the initial condition \(B(1) = 100\) to find \(C\). Substitute \(t = 1\) and \(B = 100\) into the equation \(100 = 25 + Ce^{-2(1)}\). This simplifies to \(100 = 25 + Ce^{-2}\). Solving for \(C\), we get \(C = 75e^2\).

Step 7: Write the Particular Solution

Substitute \(C = 75e^2\) back into \(B = 25 + Ce^{-2t}\) to get the particular solution: \(B = 25 + 75e^{2-2t}\).

Key Concepts

Integrating Factor

The integrating factor is a powerful technique used in solving first-order linear differential equations, and it's very straightforward once you understand the concept. In our problem, we had the differential equation \( \frac{dB}{dt} + 2B = 50 \). This is written in the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \), where \( P(t) = 2 \) and \( Q(t) = 50 \).
To solve such an equation, we need what's called an "integrating factor," typically denoted as \( \mu(t) \). This factor transforms the left side of the equation into the derivative of a product, making integration straightforward.

• Calculate \( \mu(t) \) using \( \mu(t) = e^{\int P(t) \, dt} \).
• Here, \( P(t) = 2 \), so \( \mu(t) = e^{\int 2 \, dt} = e^{2t} \).
• Multiplying the entire equation by \( e^{2t} \) transforms our differential equation into a new form that is easier to integrate: \( \frac{d}{dt}(Be^{2t}) = 50e^{2t} \).

The integrating factor simplifies the equation, turning it into something we can solve by simple integration on both sides.

Initial Conditions

Initial conditions are essential whenever we seek a particular solution to a differential equation. In our exercise, we were given the initial condition \( B(1) = 100 \). This means that when \( t = 1 \), the value of \( B \) is 100.
Initial conditions help us determine any constants that appear in the general solution of the differential equation.

• Once we have a solution, often it contains a constant of integration, \( C \), because integration can produce such constants.
• To find \( C \), we plug the initial conditions into the equation. Here, that means substituting \( t = 1 \) and \( B = 100 \) into \( B = 25 + Ce^{-2t} \).
• From this equation, \( 100 = 25 + Ce^{-2} \) allowed us to isolate \( C \), giving a specific value, \( C = 75e^2 \).

This provides us with an exact solution that fits not only the differential equation but also satisfies the given initial condition.

Particular Solution

The goal of solving a differential equation with an initial condition is ultimately to find a particular solution. A "particular solution" is one that not only solves the differential equation but also meets specific criteria set by initial conditions.In our case, we worked through the equation \( \frac{dB}{dt} + 2B = 50 \) using the integrating factor method, arriving at a general solution \( B = 25 + Ce^{-2t} \).
However, since we know \( B(1) = 100 \), we used this information to solve for \( C \), finding it to be \( 75e^2 \).

• By substituting \( C = 75e^2 \) back into our general solution, we achieved the particular solution: \( B = 25 + 75e^{2-2t} \).
• This solution fits the basic differential equation and fulfills the initial condition that \( B = 100 \) when \( t = 1 \).

Thus, through the process of applying the integrating factor and initial conditions, the particular solution gives the precise behavior of the system as described by the original problem.