Problem 7 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, use the given information to write an equation for \(y\). Confirm your result analytically by showing that the function satisfies the equation \(d y / d t=C y .\) Does the function represent exponential growth or exponential decay? $$ \frac{d y}{d t}=2 y, \quad y=10 \text { when } t=0 $$

Step-by-Step Solution

Answer

The equation for \( y \) is \( y = 10 * e^{2t}\). This function satisfies the given differential equation because differentiation results in \( \frac{d y}{d t} = 2y = 20 * e^{2t}\). Since the constant \( C = 2 \) is positive, it represents exponential growth.

Step 1: Solve the differential equation

The provided differential equation is \( \frac{d y}{d t}=2 y \). This resembles the form \( \frac{d y}{d t}=C y \) that we need to confirm, and in this case, \( C \) is equal to 2. This is a standard differential equation that can be solved using the method of separation of variables. The solution of this equation is \( y = e^{2t} \).

Step 2: Use the initial condition to solve for the constant

The given initial condition is \( y=10 \) when \( t=0 \). Substituting these values into the equation gives \( 10 = e^{0} .\). Simplifying it results in: \(10 = 1 \). This means that the complete solution to the differential equation will be \( y = 10 * e^{2t}\).

Step 3: Confirm whether the function satisfies the differential equation

Differentiating the function \( y = 10 * e^{2t}\) results in \( \frac{d y}{d t} = 20 * e^{2t}\), which is equal to \( 2y \). Thus, the function satisfies the differential equation.

Step 4: Determine whether the function represents exponential growth or decay

Since the constant \( C = 2 \) is positive, it indicates that the function represents exponential growth.

Key Concepts

Exponential Growth

Understanding exponential growth is crucial when dealing with differential equations like the one in this exercise. Exponential growth occurs when the increase in a quantity is proportional to the current amount. This means as the quantity grows, the rate of growth also increases.
In the differential equation given in the exercise, \( \frac{d y}{d t} = 2y \), the coefficient \( 2 \) is the proportionality constant, indicating growth. Since this value is positive, it confirms that the function is growing exponentially rather than decaying.
Exponential growth is commonly seen in real-life scenarios, such as population growth, compound interest, or radioactive decay. The standard form of an exponentially growing function is \( y = y_0 e^{Ct} \), where \( y_0 \) is the initial quantity, and \( C \) is the growth rate, which in this case is \( 2 \).
For easy recall:

• If the constant \( C > 0 \), you have exponential growth.
• If the constant \( C < 0 \), it represents exponential decay.

Separation of Variables

Separation of variables is a handy method for solving differential equations, especially those like \( \frac{d y}{d t}=2 y \). This method revolves around rewriting the differential equation so that each side depends on a different variable.
Here's a simple breakdown of how it works:

• First, you divide both sides by \( y \), getting \( \frac{1}{y} \frac{d y}{d t} = 2 \).
• Then, you multiply by \( dt \) to separate the variables: \( \frac{1}{y} dy = 2 dt \).
• Integrate both sides: \( \int \frac{1}{y} dy = \int 2 dt \).
• This results in \( \ln|y| = 2t + C \), where \( C \) is the constant of integration.

After solving and simplifying, you find \( y = e^{2t} \), which mirrors our solution perfectly. This equation represents the core idea that guided the solution in our exercise; recognizing this separation as the method confirms our approach.

Initial Conditions

Initial conditions are essential in uniquely determining the solution to a differential equation. They provide specific starting points, or moments, that allow us to solve for constants of integration, leading to a fully determined solution.
In our example, the initial condition is given as \( y = 10 \) when \( t = 0 \). Substituting this initial condition into our general solution, \( y = 10 e^{2t} \), helped us to ascertain and confirm the accurate form of our solution.
Why is this important?

• Initial conditions tailor the general solutions to a unique context or scenario, reflecting specific real-world elements.
• Using them early on ensures the solutions are relevant and applicable to the questions' requirements.

Initial conditions are a crucial resource for solving all types of equations, ensuring that mathematical solutions reflect practical scenarios accurately.