Problem 28 - 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. \(\frac{d G}{d t}=0.75 G,\) where \(G(0)=2000\)

Step-by-Step Solution

Answer

The particular solution is \(G(t) = 2000e^{0.75t}\), and it satisfies the differential equation.

Step 1: Identify the Type of Differential Equation

The equation \(\frac{d G}{d t} = 0.75 G\) is a first-order linear differential equation. The form is \(\frac{dG}{dt} = kG\) where \(k = 0.75\).

Step 2: Solve the Differential Equation

For a first-order linear differential equation of the form \(\frac{dG}{dt} = kG\), the general solution is \(G(t) = Ce^{kt}\). Here, \(k=0.75\). So, the solution becomes \(G(t)=Ce^{0.75t}\).

Step 3: Apply the Initial Condition

Use the initial condition \(G(0) = 2000\) to find \(C\). Substitute \(t = 0\) and \(G(0) = 2000\) into \(G(t) = Ce^{0.75t}\):\[2000 = Ce^{0.75\cdot0} = Ce^{0} = C\]Thus, \(C = 2000\). The particular solution is \(G(t) = 2000e^{0.75t}\).

Step 4: Verify the Solution

Differentiate the particular solution \(G(t) = 2000e^{0.75t}\) with respect to \(t\):\[\frac{dG}{dt} = 2000 \cdot 0.75e^{0.75t} = 1500e^{0.75t}\]Now check that this expression satisfies the original differential equation by substituting:\[1500e^{0.75t} = 0.75 \cdot 2000e^{0.75t} = 1500e^{0.75t}\]Both sides are equal, thus the solution satisfies the differential equation.

Key Concepts

First-Order Linear Differential Equation

When tackling differential equations, it's crucial to identify the type of problem you are dealing with. A **first-order linear differential equation** is characterized by its specific structure. It typically takes the form of \( \frac{dG}{dt} = kG \), where the rate of change of a function \( G \) is proportional to the function itself. This simplicity makes them particularly accessible for many students.

• The equation involves the first derivative, hence the term *first-order*.
• The equation is *linear* because it can be expressed as a linear combination of the derivative of \( G \) and \( G \) itself.
• The constant \( k \) is the proportionality constant and can greatly affect the behavior of the solution.

In our example, \( \frac{dG}{dt} = 0.75G \), we identify \( k \) as 0.75. Recognizing these elements is the starting point for finding a solution.

Initial Condition

The **initial condition** is a vital part of solving differential equations because it helps us determine the specific behavior of the solution over time. It is typically given in the form of a value of the function at a particular point, simplifying the process of finding a particular solution.

• In our problem, the initial condition is \( G(0) = 2000 \).
• It provides a starting point or anchor for the solution curve, ensuring the solution satisfies not only the differential equation but specific criteria as well.
• By substituting \( t = 0 \) into the general solution, we apply the initial condition to solve for any unknown constants in the equation, in this case, \( C \).

Thus, the initial condition helps in tailoring the general solution to match real-world scenarios or specific requirements.

Particular Solution

Once we have the general solution to the differential equation, we use the initial condition to find the **particular solution**. This form of the solution reflects a specific answer that caters precisely to the initial conditions outlined.

• The general solution derived from our equation is \( G(t) = Ce^{0.75t} \).
• Using the initial condition \( G(0) = 2000 \), we find \( C = 2000 \), leading us to the particular solution: \( G(t) = 2000e^{0.75t} \).
• This process translates a broad, theoretical solution into a practical, applicable one by ensuring it starts with the given conditions.

Verifying the solution ensures that it not only meets the initial condition but also satisfies the original differential equation, confirming its correctness.