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Chapter 31: Problem 17

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

Short Answer

Expert verified
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 by step solution

01

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 \(y'(t)=e^{t} \cos t+e^{t} \sin t.\)
02

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 \(y''(t)=e^{t} cos(t) + e^{t} cos(t) + e^{t} sin(t) + e^{t} sin(t) = 2e^{t} \cos t+2e^{t}\sin t.\)
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

For what value(s) of \(\beta\), if any, is (a) \(y=C_{1} \sin \beta t\) a solution to \(y^{\prime \prime}=16 y\) ? (b) \(y=C_{2} \cos \beta t\) a solution to \(y^{\prime \prime}=16 y\) ? (c) \(y=C_{3} e^{\beta t}\) a solution to \(y^{\prime \prime}=16 y\) ?

(a) You plan to save money starting today at a rate of \(\$ 4000\) per year over the next 30 years. You will deposit this money at a nearly continuous rate (a constant amount each day) into a bank account that earns \(5 \%\) interest compounded continuously. Let \(B(t)\) be the balance of money in the account \(t\) years from now, where \(0 \leq t \leq 30\) i. Write a differential equation whose solution is \(B(t)\). ii. Write an integral that is equal to \(B(30)\), the amount in the account at the end of 30 years. (b) Now assume that instead of making deposits continuously, you decide to make a deposit of \(\$ 4000\) once a year, starting today and continuing until you have made a total of 30 deposits. Suppose the bank account pays \(5 \%\) interest compounded annually. i. Write a geometric sum equal to the balance immediately after the final deposit. ii. Find a closed form expression (no \(+\cdots+\), no summation notation) for this sum.

When a population has unlimited resources and is free from disease and strife, the rate at which the population grows in often modeled as being proportional to the population. Assume that both the bee and the mosquito populations described below behave according to this model. In both scenarios described below you are given enough information to nd the proportionality constant \(k\). In one case the information allows you to nd \(k\) solely using the differential equation, without requiring that you solve it. In the other scenario you must actually solve the differential equation in order to nd \(k\). (a) Let \(M=M(t)\) be the mosquito population at time \(t, t\) in weeks. At \(t=0\) there are 1000 mosquitoes. Suppose that when there are 5000 mosquitoes the population is growing at a rate of 250 mosquitoes per week. Write a differential equation re ecting the situation. Include a value for \(k\), the proportionality constant. (b) Let \(B=B(t)\) be the bee population at time \(t, t\) in weeks. At \(t=0\) there are 600 bees. When \(t=10\) there are 800 bees. Write a differential equation re ecting the situation. Include a value for \(k\), the proportionality constant.

The population in a certain country grows at a rate proportional to the population at time \(t\), with a proportionality constant of \(0.03 .\) Due to political turmoil, people are leaving the country at a constant rate of 6000 people per year. Assume that there is no immigration into the country. Let \(P=P(t)\) denote the population at time \(t\). (a) Write a differential equation re ecting the situation. (b) Solve the differential equation for \(P(t)\) given the information that at time \(t=0\) there are 3 million people in the country. In other words, nd \(P(t)\), the number of people in the country at time \(t\).

For each system of differential equations, nd the nullclines and identify the equilibrium solutions. $$ \left\\{\begin{array}{l} \frac{d x}{d t}=0.02 x-0.001 x^{2}+0.002 x y \\ \frac{d y}{d t}=0.03 y-0.006 y^{2}+0.001 x y \end{array}\right. $$
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