Problem 30 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(\frac{d L}{d t}=0.68 L,\) where \(L=1200\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \(L(t) = 1200e^{0.68t}\). It satisfies the differential equation when verified.

Step 1: Identify the Differential Equation Type

The given differential equation is \(\frac{dL}{dt} = 0.68L\). This is a first-order linear homogeneous differential equation.

Step 2: Solve the Differential Equation

A first-order linear homogeneous differential equation of the form \(\frac{dL}{dt} = kL\), where \(k\) is a constant, has the general solution \(L(t) = Ce^{kt}\). Here, \(k = 0.68\), so the general solution is \(L(t) = Ce^{0.68t}\).

Step 3: Apply the Initial Condition

The initial condition given is \(L = 1200\) when \(t = 0\). Substituting these values into the general solution gives \(1200 = Ce^{0.68 \times 0}\). This results in \(1200 = C\), so the particular solution is \(L(t) = 1200e^{0.68t}\).

Step 4: Verify the Solution

To check, differentiate the particular solution \(L(t) = 1200e^{0.68t}\) with respect to \(t\). \(\frac{dL}{dt} = 1200 \times 0.68 e^{0.68t} = 816e^{0.68t}\). Substitute this back into the differential equation \(\frac{dL}{dt} = 0.68L\), getting \(816e^{0.68t} = 0.68 \times 1200e^{0.68t}\), which verifies the solution as both sides are equal.

Key Concepts

First-order Linear Equations

Understanding first-order linear equations is crucial when studying differential equations. These types of equations involve derivatives of an unknown function and are called 'first-order' because they involve only the first derivative. They generally look like this: \(\frac{dy}{dt} + P(t)y = Q(t)\), where \(y\) represents the unknown function, and \(t\) is the independent variable. In our exercise, the equation was simplified to \(\frac{dL}{dt} = 0.68L\), making it both linear and homogeneous. Linear equations often have certain properties that make them predictable and solvable through well-established methods.

• Standard Format: Expressed as \(a(t)y' + b(t)y = c(t)\).
• Solution Method: Using integrating factors or recognizing a pattern, as with exponential functions in our example.

By using these characteristics, solving first-order linear differential equations becomes a procedural task.

Homogeneous Equations

Homogeneous equations are a subtype of differential equations where every term is a function of the unknown variable and its derivatives. This typically means each term can be factored by the unknown function. In our example, \(\frac{dL}{dt} = 0.68L\) is homogeneous because both sides can be expressed in terms of \(L\).
When dealing with linear homogeneous equations, a general solution can often be found using exponential functions because they inherently satisfy the structure of these equations. For instance:

• The general solution form \(L(t) = Ce^{0.68t}\).
• This exponential property arises due to the nature of constant coefficients, a core characteristic of linear homogeneous equations.

The term homogeneous might sound complex, but it essentially reflects the internal consistency within the equation that allows us to use exponential functions to find solutions.

Initial Conditions

Initial conditions are values given at the starting point of the problem, which are critical to finding the particular solution. They allow us to narrow down from the general solution to a solution that specifically fits the given situation. In our case, the initial condition was \(L = 1200\) when \(t = 0\).
Applying the initial conditions involves plugging these values into the general solution to solve for constants. Here:

• Substitute to get \(1200 = Ce^{0}\).
• Simplifying gives the constant \(C = 1200\).

Initial conditions are what make differential equations so powerful for modeling real-world scenarios, as they reflect specific moments or states that the solution must adhere to.

Verification of Solutions

Verifying solutions is a way to confirm that the function we found actually satisfies the original differential equation. This step checks our computations and ensures our solution is mathematically consistent. To verify, we differentiate our particular solution \(L(t) = 1200e^{0.68t}\), resulting in:

• \(\frac{dL}{dt} = 1200 \times 0.68 e^{0.68t} = 816e^{0.68t}\)

Next, we substitute this back into the original differential equation:

• Confirm that \(816e^{0.68t} = 0.68 \times 1200e^{0.68t}\).

This holds true, proving the particular solution is correct. Verification reinforces confidence in our solution and helps identify errors if any discrepancies arise. It demonstrates the consistency of mathematical results with the posed problem.