Problem 8 - 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? $$ \frac{d y}{d t}=-\frac{2}{3} y, \quad y=20 \text { when } t=0 $$

Step-by-Step Solution

Answer

The solution to the differential equation \(\frac{d y}{d t} = -\frac{2}{3} y\) with the initial condition \(y = 20\) when \(t = 0\) is \(y = 20e^ {-\frac{2}{3}t}\), which represents an exponential decay.

Step 1: Identify the given differential equation and initial condition

The given differential equation is \(\frac{d y}{d t} = -\frac{2}{3} y\), and the initial condition is \(y=20\) when \(t=0\). The minus sign in the equation indicates a negative rate of change, which ususally signifies a decay process.

Step 2: Solve the given differential equation

The given differential equation can be solved by separation of variables. Rearranging the equation gives \(\frac{1}{y} d y = - \frac{2}{3} dt\). Then, integrate both sides of the equation to get \(\int \frac{1}{y} d y = - \frac{2}{3} \int dt\), which results in \(ln |y| = - \frac{2}{3} t +C\). To rid of the natural logarithm, apply the exponential function to both sides to give \(|y| = e^ {- \frac{2}{3} t +C}\). The absolute value sign can be eliminated by introducing another constant \(K = \pm e^C\), to get \(y = Ke^{-\frac{2}{3}t}\).

Step 3: Use the initial condition to find the constant 'K'

Plug \(t=0\) and \(y = 20\) into the equation gives \(20 = Ke^{-\frac{2}{3}*0}\), which simplifies to \(20 = K\). Therefore, the solution of the differential equation is \(y = 20e^ {-\frac{2}{3}t}\).

Step 4: Confirm the result

To confirm the result, differentiate the solution \(y = 20e^{-\frac{2}{3} t}\) with respect to \(t\). This gives \(\frac{d y}{d t} = -\frac{2}{3}*20e^{-\frac{2}{3} t}\), which is equivalent to the original differential equation \(\frac{d y}{d t}=-\frac{2}{3} y\).

Step 5: Determine the nature of the function

Since the coefficient of \(t\) in the exponential function is negative, this indicates that the function represents exponential decay.

Key Concepts

Exponential Growth and Decay

Exponential growth and decay are key concepts when dealing with differential equations.
They describe how quantities change over time.
Simply put, these processes determine whether a quantity increases or decreases as time progresses.

• Exponential growth occurs when a quantity increases at a constant rate, often represented by a positive coefficient in an exponential function.
• Exponential decay, on the other hand, happens when a quantity decreases over time, indicated by a negative coefficient.

In the exercise, the differential equation is \(\frac{d y}{d t} = -\frac{2}{3} y\).Here, the negative sign in front of the fraction \(-\frac{2}{3}\) tells us we're dealing with exponential decay. This means, as time increases, the value of \(y\) decreases exponentially.
Decaying processes are common in real-world situations like radioactive decay, cooling liquids, or depreciation of assets. The solution to this equation is given as \(y = 20e^{-\frac{2}{3} t}\), a classic example of exponential decay. The function decreases as \(t\) increases, following the negative exponent.

Separation of Variables

Separation of variables is a technique used to solve differential equations, which breaks the problem into manageable parts. By rearranging the terms, we make it easier to integrate each side of the equation separately.
This strategy is particularly helpful when you have an equation involving the product of a function of \(y\) and a function of \(t\). In basic terms, we aim to get all \(y\)-related terms on one side and all \(t\)-related terms on the other.

• For this problem, starting with \(\frac{d y}{d t} = -\frac{2}{3} y\), the separation leads to \(\frac{1}{y} dy = -\frac{2}{3} dt\).
• We then integrate each side: \(\int \frac{1}{y} dy = -\int \frac{2}{3} dt\).

Integrating gives \(\ln |y| = -\frac{2}{3}t + C\).
Applying exponential functions eliminates the logarithm: \(|y| = e^{-\frac{2}{3}t + C}\).
To remove the absolute value, introduce a constant, \(K = \pm e^C\), leading to the solution \(y = Ke^{-\frac{2}{3}t}\).This method is powerful and widely used in solving various types of differential equations. The goal? Separate, integrate, and solve step-by-step.

Initial Conditions

Initial conditions provide specific values at a certain point, often used to find particular solutions to differential equations.
In other words, they help us in determining unknown constants such as \(K\) in a general solution.
The exercise gives us an initial condition: \(y = 20\) when \(t = 0\).

• By substituting these values into the general solution \(y = Ke^{-\frac{2}{3}t}\), we find \(K\).
• For \(t = 0\), the equation simplifies to \(20 = Ke^{0}\) (since \(e^0 = 1\)), leading to \(K = 20\).

Having this initial condition ensures that our solution fits the specific scenario we're examining. It confirms that our equation accurately predicts behaviors starting from the known state.Without initial conditions, we'd have a range of potential solutions, making it difficult to pinpoint the exact one that characterizes a specific situation. Initial conditions, therefore, are crucial in both real-world problems and mathematical scenarios.