Problem 17 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Write a second order homogeneous differential equation that is satisfied by \(y(t)=\) \(e^{t} \sin t .\) (The answer is not unique.)

Step-by-Step Solution

Answer

So, the second order homogeneous differential equation of the given solution \(y(t)=e^{t}\sin t\) can be written as \(a(2e^{t} cos(t)+2e^{t}sin(t)) + b(e^{t} cos(t) + e^t sin(t)) + c(e^t sin(t))=0\). Several combinations of coefficients a, b, and c can satisfy this equation, thus rendering it not unique.

Step 1: Obtain the First Derivative

The first derivative \(y'(t)\) of \(y(t)=e^{t} \sin t\) can be computed using the product rule. The product rule is given as \((uv)'=u'v+uv',\) where \(u=e^t\) and \(v=\sin t\). Therefore, applying the product rule gives \(y'(t)=e^{t} \cos t+e^{t} \sin t.\)

Step 2: Obtain the Second Derivative

Next, apply the product rule again to compute the second derivative \(y''(t)\) of the first derivative \(y'(t)=e^{t} \cos t+e^{t} \sin t.\) This leads to \(y''(t)=e^{t} cos(t) + e^{t} cos(t) + e^{t} sin(t) + e^{t} sin(t) = 2e^{t} \cos t+2e^{t}\sin t.\)

Step 3: Form the Second Order Homogeneous Differential Equation

A second order homogeneous differential equation is generally in the form \(ay''+by'+cy=0\). Substituting \(y, y', y''\) into the equation gives \(a(2e^{t} \cos t+2e^{t}\sin t) + b(e^{t} \cos t + e^t sin t) + c(e^t sin t)=0.\) The given solutions can satisfy this equation for certain values of a, b, and c. In fact, the solution to the equation is not unique. There are multiple sets of (a, b, c) that can satisfy the equation.

Key Concepts

Understanding Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it describes a relationship between a changing quantity, the function, and the rate at which it changes, the derivative. Differential equations are fundamental in the modeling of physical systems, where something's rate of change is proportional to the state of the system itself.

For example, the population growth of a species may be modeled by a differential equation where the rate of population change (the derivative) depends on the current population size (the function itself). The differential equation given in the exercise is a second order homogeneous equation, which means it contains derivatives up to the second order and all terms involve the unknown function or its derivatives.

Applying the Product Rule

The product rule is a vital tool in calculus used to find the derivative of the product of two functions. It states that the derivative of a product u(t)v(t) is given by u'(t)v(t) + u(t)v'(t), where u'(t) and v'(t) are the derivatives of u(t) and v(t) respectively.

When you come across a function that is the product of two simpler functions, like the one in our problem, using the product rule allows you to break down the problem into more manageable parts. The proper application of the product rule is essentiel in both obtaining the first and the second derivative of the given function.

Finding the First Derivative

The first derivative of a function represents the rate at which the function value changes with respect to its variable. For the function y(t) in the exercise, obtaining the first derivative required using the product rule. It is the first step to understanding how the function's rate of change behaves over time.

The computation of y'(t) from y(t) reveals how quickly the function is changing at any point t, providing insights into the behavior of the system it represents. In this context, the resulting first derivative encompasses both the exponential growth from the term e^t and the oscillatory behavior of the sine function.

Calculating the Second Derivative

The second derivative, y''(t), gives us information about the acceleration, or the rate of change of the rate of change, of the function. In many physical contexts, this would correspond to the acceleration of an object if y(t) was its position. In the exercise, we apply the product rule a second time to the first derivative to find the second derivative.

This process further reveals the nature of the function's change over time and is a critical step toward forming a second order homogeneous differential equation. The second derivative introduces a new level of complexity and depth to our understanding of the function's overall behavior.