Problem 6 - Applied Calculus

Question

Find particular solutions. $$\frac{d Q}{d t}=0.3 Q-120, \quad Q=50 \text { when } t=0$$

Step-by-Step Solution

Answer

The particular solution is \( Q(t) = 400 - 350e^{0.3t} \).

Step 1: Set Up the Differential Equation

The given differential equation is \( \frac{dQ}{dt} = 0.3Q - 120 \). This is a first-order linear ordinary differential equation. We need to find a particular solution using the initial condition \( Q(0) = 50 \).

Step 2: Find the General Solution

First, rewrite the equation in standard linear form: \( \frac{dQ}{dt} - 0.3Q = -120 \). The solution involves finding the integrating factor \( \mu(t) = e^{\int -0.3 \, dt} = e^{-0.3t} \). Multiply the whole equation by this integrating factor: \( e^{-0.3t} \frac{dQ}{dt} - 0.3e^{-0.3t}Q = -120e^{-0.3t} \).

Step 3: Solve Using Integrating Factor

Recognize that the left side is the derivative of \( Qe^{-0.3t} \). Thus, integrate both sides: \( \int d(Qe^{-0.3t}) = \int -120e^{-0.3t} \, dt \). This simplifies to \( Qe^{-0.3t} = \int -120e^{-0.3t} \, dt \).

Step 4: Perform Integration

Calculating the integral \( \int -120e^{-0.3t} \, dt \), we get \( 400e^{-0.3t} + C \), where \( C \) is an integration constant. This makes \( Qe^{-0.3t} = 400e^{-0.3t} + C \).

Step 5: Solve for \( Q \)

Multiply both sides by \( e^{0.3t} \) to solve for \( Q \): \( Q = 400 + Ce^{0.3t} \).

Step 6: Apply Initial Condition

Use the initial condition \( Q(0) = 50 \) to find \( C \): \( 50 = 400 + Ce^{0} \) simplifies to \( C = 50 - 400 = -350 \). Thus, the particular solution is \( Q(t) = 400 - 350e^{0.3t} \).

Key Concepts

Integrating Factor

The integrating factor is a crucial tool when solving linear ordinary differential equations. It's a function used to make the equation easier to integrate. In our differential equation from the original exercise, transforming it into the standard linear form, we have:

• Original form: \(\frac{dQ}{dt} = 0.3Q - 120\)
• Standard form: \(\frac{dQ}{dt} - 0.3Q = -120\)

In this form, we recognize that we can use an integrating factor. The formula for the integrating factor \( \mu(t) \) is:\[ \mu(t) = e^{\int -0.3 \, dt} = e^{-0.3t} \]This function, when multiplied through the entire equation, allows the left side of the equation to be expressed as the derivative of a product. Specifically:\[ e^{-0.3t} \frac{dQ}{dt} - 0.3e^{-0.3t}Q = -120e^{-0.3t} \]This simplifies to \( \frac{d}{dt}(Qe^{-0.3t}) = -120e^{-0.3t} \), making it straightforward to integrate both sides and solve for \( Q \). This reduction in complexity is why an integrating factor is such a powerful technique in solving linear differential equations.

Initial Condition

Initial conditions are specifications needed to find particular solutions to differential equations. They provide extra information about the solution at a specific point, which helps to pin down specific constants that appear in the solution. In our exercise, the initial condition is specified as:

• \( Q(0) = 50 \)

This means at time \( t = 0 \), the quantity \( Q \) has the value 50. Applying this condition is done after finding the general solution of the differential equation. The general solution derived before applying the initial condition is:\[ Q = 400 + Ce^{0.3t} \]Here, \( C \) is an arbitrary constant that arises from integration. To find \( C \), substitute \( t = 0 \) and \( Q = 50 \):

• \( 50 = 400 + Ce^{0} \)

Since \( e^{0} \) equals 1, it follows that:

• \( C = 50 - 400 = -350 \)

This step confirms the specific value of \( C \) and leads to the particular solution for the differential equation.

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies both the differential equation and any initial conditions. Given the linear ordinary differential equation in our exercise, after applying the integrating factor and integrating, the general solution is obtained:\[ Q(t) = 400 + Ce^{0.3t} \]Utilizing the initial condition \( Q(0) = 50 \), we ascertain the particular value of \( C \). This gives us the particular solution:\[ Q(t) = 400 - 350e^{0.3t} \]In this context, the particular solution is unique to the conditions provided in the problem. A particular solution provides the precise behavior of \( Q \) over time, given the initial scenario. As \( t \) increases, \( e^{0.3t} \) grows, causing \( -350e^{0.3t} \) to decrease the overall value of \( Q \) from its initial value at \( t=0 \). Ultimately, the particular solution tells us exactly how the quantity \( Q \) changes with time from its starting point.