Problem 29 - 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. \(\frac{d S}{d t}=0.12 S,\) where \(S=750\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \( S(t) = 750e^{0.12t} \), and it satisfies the differential equation.

Step 1: Identify the type of differential equation

The differential equation given is \( \frac{dS}{dt} = 0.12 S \). This is a first-order linear differential equation commonly known as a separable differential equation.

Step 2: Separate variables

Rearrange the equation to separate the variables, resulting in \( \frac{dS}{S} = 0.12 \, dt \). This form allows us to integrate each side with respect to its respective variable.

Step 3: Integrate both sides

Integrate both sides of the equation. The left side becomes \( \int \frac{dS}{S} = \, \ln |S| \) and the right side is \( \int 0.12 \, dt = 0.12t + C \), where \( C \) is the constant of integration.

Step 4: Solve for \( S \)

Rewrite the equation from the integration step: \( \ln |S| = 0.12t + C \). Exponentiate both sides: \( |S| = e^{0.12t + C} \). Recognize that \( e^C \) is still a constant, rename it as \( C' \), therefore \( S = C' e^{0.12t} \).

Step 5: Apply initial condition

Use the initial condition \( S = 750 \) when \( t = 0 \). Substitute these values into the equation \( S = C' e^{0.12t} \) to find \( C' \): \( 750 = C' e^{0.12 \times 0} \). Since \( e^0 = 1 \), \( C' = 750 \). The particular solution is \( S(t) = 750e^{0.12t} \).

Step 6: Check the solution by substitution

Differentiate the solution \( S(t) = 750e^{0.12t} \) to find \( \frac{dS}{dt} \). This results in \( \frac{dS}{dt} = 750 \times 0.12 \times e^{0.12t} = 90e^{0.12t} \). Substitute \( S(t) = 750e^{0.12t} \) back into the original differential equation \( \frac{dS}{dt} = 0.12S \) to verify: \( 0.12 \times 750e^{0.12t} = 90e^{0.12t} \). Both sides are equal, confirming the solution is correct.

Key Concepts

Separable Differential Equations

Separable differential equations are a specific type of differential equations where the variables can be 'separated' onto different sides of the equation. This makes them easier to solve.
To identify a separable differential equation, look for an equation that can be restructured into the form \( \frac{dy}{dx} = g(y)h(x) \), meaning the derivative of \( y \) with respect to \( x \) can be expressed as a product of functions of \( y \) and \( x \).

• In our example, the given differential equation is \( \frac{dS}{dt} = 0.12 S \). You can see here that this equation can be expressed as \( \frac{1}{S} \frac{dS}{dt} = 0.12 \), effectively separating the variables \( S \) and \( t \).

This characteristic allows for straightforward integration with each variable integrated independently:- Integrate with respect to \( S \) on one side.- Integrate with respect to \( t \) on the other side.
Once separated and integrated, the solution is often found by connecting back the separated functions, as was shown in the original step-by-step solution.

Initial Value Problems

In differential equations, an initial value problem means you're not only asked to find the general solution of the equation, but also a particular solution that satisfies given initial conditions.
Initial conditions provide specific values for the variables at a certain point which allow us to find any constants of integration that arise during the process.

• In the given exercise, the initial condition is \( S = 750 \) when \( t = 0 \). This specifically instructs us to find the solution that matches this state, not just any solution.

Here's the process:1. Find the general solution of the differential equation. In the example, this was found to be \( S = Ce^{0.12t} \).2. Substitute the initial condition values into this general solution to solve for the constant \( C \). Here, substituting \( S = 750 \) when \( t = 0 \) led us to find that \( C = 750 \).3. Use this specific value of \( C \) to write out the particular solution that satisfies the initial condition, resulting in \( S(t) = 750e^{0.12t} \).

Solutions of Differential Equations

Finding solutions of differential equations involves determining functions that satisfy the given equations upon differentiation.
Differential equations can have general and particular solutions:

• The general solution includes all possible solutions and usually contains arbitrary constants (like \( C \) in our example).
• A particular solution meets specific conditions set by the problem, often initial conditions, to ensure it holds true in a specific scenario.

Here is how we approached solving:1. **Differentiate and Integrate:** Often begins by differentiating existing functions or integrating as needed, which helps isolate variables or constants. Here, the integration revealed \( S = Ce^{0.12t} \).
2. **Apply Initial Conditions:** Incorporating initial or specific values into the integrated solution narrows down the possibilities to a specific curve or line, in our case, giving \( S(t) = 750e^{0.12t} \).
3. **Validate Solution:** To verify, substitute back into the original differential equation to confirm both sides equal the same expression. Thus, both sides simplifying to \( 90e^{0.12t} \) proved our solution's correctness.Remember, knowing how to manipulate, solve, and validate solutions are crucial in handling any differential equations.