Problem 19 - Calculus and its applications

Question

(a) find the general solution of each differential equation, and (b) check the solution by substituting into the differential equation. \(\frac{d G}{d t}=0.005 G\)

Step-by-Step Solution

Answer

The general solution is \( G(t) = C e^{0.005t} \).

Step 1: Identify the type of differential equation

The differential equation given is \( \frac{dG}{dt} = 0.005G \). It is a first-order linear ordinary differential equation. This type of equation can be solved using the method of separation of variables.

Step 2: Separate the variables

Rearrange the equation to isolate terms involving \( G \) and \( t \). Divide both sides by \( G \) and multiply by \( dt \):\[ \frac{1}{G} \, dG = 0.005 \, dt \]

Step 3: Integrate both sides

Integrate the separated equation on both sides: \[ \int \frac{1}{G} \, dG = \int 0.005 \, dt \]The left integral becomes \( \ln |G| \) and the right integral becomes \( 0.005t + C \), where \( C \) is the integration constant.

Step 4: Solve for \( G \)

Exponentiate both sides to solve for \( G \). This gives:\[ e^{\ln |G|} = e^{0.005t + C} \]\[ |G| = e^C e^{0.005t} \]Let \( C_1 = e^C \) (still a constant), so \( G = C_1 e^{0.005t} \). Since \( C_1 \) can be any real number, write \( G(t) = C e^{0.005t} \), covering both positive and negative \( G \).

Step 5: Verify by substituting back into the differential equation

Take the derivative of the solution \( G(t) = C e^{0.005t} \): \( \frac{dG}{dt} = C \cdot 0.005 e^{0.005t} = 0.005G \), which matches the original differential equation. Therefore, the solution satisfies the differential equation.

Key Concepts

First-Order Linear Differential Equations

A first-order linear differential equation is a common type of differential equation often encountered in various scientific fields. These equations can be identified by their structure, which involves the first derivative of a function and the function itself, typically in the form of \( \frac{dy}{dx} + P(x)y = Q(x) \). In our original exercise, we have \( \frac{dG}{dt} = 0.005G \), a simplified case where only the term involving \( G \) appears on one side of the equation. These equations are termed 'linear' because they involve no powers or products of the dependent variable and its derivatives.First-order linear differential equations are characterized by:

• Having the highest derivative as the first derivative.
• Being linear in terms of the dependent variable and its derivatives.

Understanding the structure of these equations is crucial because it guides us in finding solutions, such as using the method of separation of variables. Recognizing the type lets us know how to proceed effectively with solving them.

Separation of Variables

Separation of variables is a straightforward yet powerful method used to solve first-order linear differential equations. This technique involves rearranging the equation to isolate terms involving each variable on opposite sides of the equation.In the exercise given, \( \frac{dG}{dt} = 0.005G \), we can separate the variables by dividing both sides by \( G \) and multiplying by \( dt \), leading us to \[ \frac{1}{G} \, dG = 0.005 \, dt \]. This step effectively isolates the variables, allowing us to handle one variable at a time.Once separated, integration is performed on both sides:

• The left-hand side, \( \int \frac{1}{G} \, dG \), integrates to \( \ln |G| \).
• The right-hand side, \( \int 0.005 \, dt \), integrates to \( 0.005t + C \), where \( C \) is the integration constant.

This approach transforms the differential equation into an algebraic equation, which can be further solved to express the general solution for \( G \). Separation of variables makes solving differential equations more approachable and reveals the relationship between the variables.

General Solution

The general solution of a differential equation provides a comprehensive representation of all possible solutions that satisfy the equation. For the equation \( \frac{dG}{dt} = 0.005G \), the general solution is derived after integrating both sides and including the integration constant, \( C \).Once integrated, we find that \( \ln |G| = 0.005t + C \). By exponentiating both sides, we solve for \( G \): \[ e^{\ln |G|} = e^{0.005t + C} \]which simplifies to:\[ |G| = e^C e^{0.005t} \]Letting \( C_1 = e^C \) leads us to \( G = C_1 e^{0.005t} \). Since \( C_1 \) represents any constant value, we denote it as \( C \). Thus, the general solution is \( G(t) = C e^{0.005t} \).This solution is termed 'general' because it encompasses all possible specific solutions for different initial conditions. Whether \( G \) is positive or negative, the constant \( C \) adjusts accordingly. Checking this solution involves substituting back into the original differential equation to verify correctness.