Problem 5 - Applied Calculus

Question

Find particular solutions \(\frac{d Q}{d t}=0.3 Q-120, \quad Q=50\) when \(t=0\)

Step-by-Step Solution

Answer

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

Step 1: Recognize the Differential Equation Type

The given equation \( \frac{dQ}{dt} = 0.3Q - 120 \) is a first-order linear ordinary differential equation, and it can be rewritten as \( \frac{dQ}{dt} = aQ + b \), where \( a = 0.3 \) and \( b = -120 \).

Step 2: Find the Homogeneous Solution

To find the homogeneous solution, consider the equation \( \frac{dQ}{dt} = 0.3Q \). This is a separable differential equation, and its solution is \( Q_h(t) = Ce^{0.3t} \), where \( C \) is an arbitrary constant.

Step 3: Find the Particular Solution

We need a particular solution to \( \frac{dQ}{dt} = 0.3Q - 120 \). Assume a constant solution of the form \( Q = K \). Substituting into the differential equation, we get \( 0 = 0.3K - 120 \). Solving for \( K \), we have \( K = 400 \). Thus, \( Q_p = 400 \) is a particular solution.

Step 4: Write the General Solution

The general solution is the sum of the homogeneous and particular solutions: \( Q(t) = Ce^{0.3t} + 400 \).

Step 5: Apply the Initial Condition

Use the initial condition \( Q(0) = 50 \) to find \( C \). Substitute \( t = 0 \) and \( Q = 50 \) into the general solution: \( 50 = Ce^{0.3\times0} + 400 \). This simplifies to \( 50 = C + 400 \), leading to \( C = -350 \).

Step 6: Write the Particular Solution with Initial Condition

Substitute \( C = -350 \) back into the general solution to obtain the particular solution: \( Q(t) = -350e^{0.3t} + 400 \).

Key Concepts

First-order Linear Differential Equations

In calculus, first-order linear differential equations are essential for understanding change and growth over time. The general form is given by:
\[ \frac{dy}{dt} + P(t)y = Q(t) \]where \( P(t) \) and \( Q(t) \) are functions of \( t \), and \( y \) is the function we want to solve for.
These equations are called 'first-order' because they involve only the first derivative (rate of change) of the function \( y \).
Key aspects to remember:

• The coefficients (\( a \) and \( b \)), which are constants in many problems, define the behavior of the solution.
• An exponential component often arises in solutions, addressing growth or decay processes.

Recognizing the structure can simplify solving the equation, as it leads to standard methods and solutions.

Initial Conditions

Initial conditions in differential equations specify the value of the solution at a specific point, often at the start of the time period being considered. For example, in our problem:
\( Q(0) = 50 \)
This tells us where the process begins; it is also crucial for determining specific solutions from general solutions.
These conditions are essential for applying particular scenarios and ensuring solutions reflect real-world situations precisely. Without them, we'd only have a general family of solutions.

• Initial conditions are used to uniquely determine the constant in the general solution of a differential equation.
• They align the mathematical solution with practical real-life conditions.

Incorporating initial conditions is a step that tailors the solution to match actual situations.

Particular Solutions

A particular solution to a differential equation is a solution that satisfies the differential equation and any given initial conditions. It represents one specific instance among the family of possible solutions.
In our problem:
The particular solution arises from assuming the form \( Q = K \) and solving for \( K \).
This is often a constant solution used to adjust the general solution.

• Particular solutions can be constants, functions, or combinations thereof.
• They help 'anchor' the equation to specific situations or points.

Particularly in linear equations, this solution is added to the homogeneous solution to form the complete general solution. Considering the constant solutions helps in simplifying many real-life modeling scenarios.

Homogeneous Solutions

Homogeneous solutions address the part of the differential equation without the external, non-homogeneous component (in our case, the \(-120\)). By setting the external influence to zero, for instance:
\[ \frac{dQ}{dt} = 0.3Q \],
we solve the equation as if it's purely driven by its own dynamics.
The solution, typically in the form \( Q_h(t) = Ce^{at} \), arises from natural growth or decay principles.

• This form helps understand the natural or intrinsic behavior of the system.
• It allows the understanding of transient states before the external factors act.

Combining the homogeneous solution with the particular solution gives us the comprehensive behavior of the system.