Problem 15 - Calculus: Early Transcendentals

Question

A function \(y(t)\) is a solution of \(y^{\prime}+k y=0 .\) Suppose that \(y(0)=100\) and \(y(2)=4 .\) Find \(k\) and find \(y(t)\).

Step-by-Step Solution

Answer

The constant \(k \approx 1.60944\) and \(y(t) = 100 e^{-1.60944 t}\).

Step 1: Recognize the Differential Equation Form

We are given the differential equation \(y^{\prime} + k y = 0\). This is a first-order linear differential equation of the form \(y^{\prime} + P(t) y = Q(t)\) with \(P(t) = k\) and \(Q(t) = 0\).

Step 2: Solve the Homogeneous Equation

For a differential equation \(y^{\prime} + k y = 0\), the solution has the form:\[ y(t) = C e^{-kt} \]where \(C\) is a constant determined by initial conditions.

Step 3: Apply Initial Condition to Find Constant C

Using the initial condition \(y(0) = 100\), substitute into the solution formula:\[ 100 = C e^{-k \cdot 0} = C \]Thus, \(C = 100\).

Step 4: Use Second Condition to Find k

We know \(y(2) = 4\), so substitute \(t = 2\) into the solution:\[ 4 = 100 e^{-2k} \]Solving for \(e^{-2k}\), we get:\[ e^{-2k} = \frac{4}{100} = 0.04 \]

Step 5: Solve for k

Take the natural logarithm on both sides to solve for \(k\):\[ -2k = \ln(0.04) \]\[ k = -\frac{1}{2} \ln(0.04) \]

Step 6: Simplify the Expression for k

Calculating further:\[ \ln(0.04) = \ln\left(\frac{1}{25}\right) = -\ln(25) \]\[ k = \frac{\ln(25)}{2} \approx \frac{3.21888}{2} \approx 1.60944 \]

Step 7: Finalize Solution for y(t)

Now, substitute \(C\) and \(k\) back into the exponential model:\[ y(t) = 100 e^{-1.60944 t} \]

Key Concepts

Initial Value Problem

An initial value problem involves a differential equation along with specific values, called initial conditions, that the solution must satisfy. The initial condition typically provides the value of the function at a certain point. In the problem given, we have the differential equation \( y'(t) + ky = 0 \) with two initial conditions: \( y(0) = 100 \) and \( y(2) = 4 \). These conditions are essential, as they allow us to determine the specific solution to the differential equation.
To solve an initial value problem effectively:

• Identify the mode of the differential equation.
• Use the given initial condition to find any constants in the solution.

This process ensures the solution not only satisfies the differential equation but also matches the real-world situation described by the initial conditions.

Homogeneous Equations

A homogeneous equation has the form \( y' + ky = 0 \), where the right-hand side is zero. This indicates that the rate of change of the function is directly proportional (and opposite in sign) to the current value of the function.
The term 'homogeneous' means that all terms involving the function equal zero when the function is equal to zero, highlighting the intrinsic symmetry and balance in such equations. The general solution for a homogeneous first-order linear equation is:\[ y(t) = Ce^{-kt} \]

• \( C \) is a constant determined by initial conditions.
• The exponential term \( e^{-kt} \) represents specific behaviors like decay or growth.

This solution showcases why homogeneous equations are pivotal—because of their simple form, solutions are easy to derive and interpret in physical and mathematical contexts.

Exponential Decay

Exponential decay occurs when quantities decrease at a rate proportional to their current value. It is characterized by the presence of an exponential function with a negative exponent, as seen in the equation \( y(t) = 100 e^{-1.60944 t} \). This type of decay is common in natural processes like radioactive decay and cooling of hot objects. Key characteristics of exponential decay include:

• The rate of decrease is swift initially but slows over time.
• No matter the time, the value of the quantity never reaches zero.

Understanding exponential decay is crucial for interpreting situations where resources diminish, or populations decrease over time, providing a realistic model of how systems respond to various conditions.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It is essential in solving equations involving exponential growth or decay, especially when finding the rate constant \( k \). In the given problem, we applied \( \ln \) to find \( k \) as follows:

• Rearrange the exponential equation: \( 4 = 100 e^{-2k} \).
• Solving gives \( e^{-2k} = 0.04 \).
• Taking \( \ln(0.04) \) revealed the value of \( k \).

The concept of a natural logarithm is fundamental in calculus and analytic geometry, seamlessly converting multiplicative processes into additive ones, which are often easier to manage.