/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Euler Equations. An equation of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Euler Equations. An equation of the form $$ t^{2} y^{\prime \prime}+\alpha t y^{\prime}+\beta y=0, \quad t>0 $$ where \(\alpha\) and \(\beta\) are real constants, is called an Euler equation. Show that the substitution \(x=\ln t\) transforms an Euler equation into an equation with constant coefficients. Euler equations are discussed in detail in Section \(5.5 .\)

Short Answer

Expert verified
Answer: Yes, the substitution \(x = \ln t\) transforms an Euler equation into an equation with constant coefficients.

Step by step solution

01

Understand the Euler Equation and Substitution

The Euler equation is given as follows: $$ t^{2} y^{\prime \prime}+\alpha t y^{\prime}+\beta y=0, \quad t>0 $$ We have the substitution \(x = \ln t\). Note that taking the exponent on both sides gives us \(t = e^x\). We will now find the derivatives of \(y(t)\) with respect to \(t\) using this substitution and then substitute these derivatives into the Euler equation.
02

Find the Derivatives of y with respect to t

We will find \(y'(t)\) and \(y''(t)\) using the chain rule and the given substitution: Let \(y(t) = g(x)\), so \(y(x) = g(\ln t)\). 1. Find \(y'(t)\): $$ y'(t) = \frac{d g(x)}{dt} = \frac{d g(x)}{dx} \cdot \frac{dx}{dt} $$ Since \(x = \ln t\), we have \(\frac{dx}{dt} = \frac{1}{t}\). Therefore, $$ y'(t) = g'(x) \cdot \frac{1}{t} $$ 2. Find \(y''(t)\): $$ y''(t) = \frac{d^2 g(x)}{dt^2} = \frac{d g'(x)}{dt} = \frac{d g'(x)}{dx} \cdot \frac{dx}{dt} $$ Since \(g'(x) = y'(t)t\), we have \(g''(x) = y''(t)t^2 + y'(t)t\). Therefore, $$ y''(t) = \frac{1}{t^2}(g''(x) - g'(x) t) $$
03

Substitute the Derivatives into the Euler Equation

Now, we'll plug these derivatives into the Euler equation: $$ t^2 y''(t) + \alpha t y'(t) + \beta y = 0 $$ $$ t^2 \left(\frac{1}{t^2}(g''(x) - g'(x) t)\right) + \alpha t\left(g'(x) \cdot \frac{1}{t}\right) + \beta g(x) = 0 $$
04

Simplify the Equation

Simplify the equation by canceling out the terms with \(t\): $$ (g''(x) - g'(x)t) + \alpha g'(x) + \beta g(x) = 0 $$ Now, substitute back \(y(t) = g(x)\), \(y'(t)t = g'(x)\), and \(y''(t)t^2 = g''(x)\): $$ y''(t)t^2 - y'(t)t + \alpha y'(t)t + \beta y(t) = 0 $$ The equation has constant coefficients as required, so the substitution \(x = \ln t\) transforms an Euler equation into an equation with constant coefficients.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differential Equations
Differential equations are mathematical tools used to describe the relationship between functions and their derivatives, representing changes in physical quantities.

Take our Euler Equation for instance:
\[ t^{2} y^{\prime \prime} + \alpha t y^{\prime} + \beta y = 0, \quad t > 0\].This is a second-order linear homogeneous differential equation, implying that the highest derivative is second-order and no term is independent of the function y(t). Here, homogeneous indicates that the equation equals zero when y is absent, and linear means the equation does not contain any powers or functions of y or its derivatives other than the first degree.

Solving such an equation frequently involves finding a function y(t) that satisfies this relation for all values of the independent variable t within a certain domain (in this case, for all t > 0), making it an indispensable tool in both mathematics and applied sciences.
Chain Rule
In calculus, the chain rule is a formula for finding the derivative of the composition of two or more functions. Imagine you need to find the rate at which one quantity changes with respect to another, and the quantities are related through a third variable. This is exactly what we encounter with the substitution in our Euler Equation problem.

With the given substitution x = ln t, we can approach the derivatives of y with respect to t by also considering how x changes with t, or formally: \( y'(t) \) and \( y''(t) \).
1. For \( y'(t) \), we find:
\[y'(t) = g'(x) \cdot \frac{1}{t}\].
2. And for \( y''(t) \), the chain rule tells us:
\[y''(t) = \frac{1}{t^2}(g''(x) - g'(x) t)\].

By applying the chain rule, we can successfully transform the original differential equation into one where derivatives are expressed in terms of x, leading to the transformation into an equation with constant coefficients.
Equations with Constant Coefficients
When dealing with equations that have constant coefficients, like the transformed version of our Euler equation, we are simplifying computational complexity. This specific category of differential equations has the advantage of being more manageable due to the constants that do not change with the independent variable, in our case, x.

The transformation through the substitution x = ln t strips away the variable coefficients, reducing it to the linear form with constant coefficients as follows:
\[ g''(x) - g'(x)t + \alpha g'(x) + \beta g(x) = 0\],where \( \alpha \) and \( \beta \) remain fixed and do not vary with x. This allows us to apply theory and methods specifically developed for constant coefficient equations, such as characteristic equations and exponential solutions, to gain insights into the behavior of solutions and, in some cases, to find explicit solutions.

Through this lesson, we've seen how a clever substitution guided by the chain rule can transform an intimidating differential equation into one that's far more approachable by imposing constant coefficients, greatly simplifying our problem-solving journey.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mass of \(20 \mathrm{g}\) stretches a spring \(5 \mathrm{cm}\). Suppose that the mass is also attached to a viscous damper with a damping constant of \(400 \mathrm{dyne}\) -sec/cm. If the mass is pulled down an additional \(2 \mathrm{cm}\) and then released, find its position \(u\) at any time \(t .\) Plot \(u\) versus \(t .\) Determine the quasi frequency and the quasi period. Determine the ratio of the quasi period to the period of the corresponding undamped motion. Also find the time \(\tau\) such that \(|u(t)|<0.05\) \(\mathrm{cm}\) for all \(t>\tau\)

Use the method of Problem 32 to solve the given differential $$ 2 y^{\prime \prime}+3 y^{\prime}+y=t^{2}+3 \sin t \quad(\text { see Problem } 7) $$

(a) Show that the phase of the forced response of Eq. ( 1) satisfies tan \(\delta=\gamma \omega / m\left(\omega_{0}^{2}-\omega^{2}\right)\) (b) Plot the phase \(\delta\) as a function of the forcing frequency \(\omega\) for the forced response of \(u^{\prime \prime}+0.125 u^{\prime}+u=3 \cos \omega t\)

Find the general solution of the given differential equation. In Problems 11 and \(12 \mathrm{g}\) is an arbitrary continuous function. $$ y^{\prime \prime}-5 y^{\prime}+6 y=g(t) $$

A spring-mass system with a hardening spring (Problem 32 of Section 3.8 ) is acted on by a periodic external force. In the absence of damping, suppose that the displacement of the mass satisfies the initial value problem $$ u^{\prime \prime}+u+\frac{1}{5} u^{3}=\cos \omega t, \quad u(0)=0, \quad u^{\prime}(0)=0 $$ (a) Let \(\omega=1\) and plot a computer-generated solution of the given problem. Does the system exhibit a beat? (b) Plot the solution for several values of \(\omega\) between \(1 / 2\) and \(2 .\) Describe how the solution changes as \(\omega\) increases.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.