/*! 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 26 (a) From \(\frac{1}{4} x^{\prime... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 26

(a) From \(\frac{1}{4} x^{\prime \prime}+x^{\prime}+5 x=0, x(0)=1 / 2,\) and \(x^{\prime}(0)=1\) we obtain \(x=e^{-2 t}\left(\frac{1}{2} \cos 4 t+\frac{1}{2} \sin 4 t\right)\) (b) \(x=\frac{1}{\sqrt{2}} e^{-2 t} \sin \left(4 t+\frac{\pi}{4}\right)\) (c) If \(x=0\) then \(4 t+\pi / 4=\pi, 2 \pi, 3 \pi, \ldots\) so that the times heading downward are \(t=(7+8 n) \pi / 16\) for \(n=0,1,2, \dots\)

Short Answer

Expert verified
The times at which the solution is zero heading downward are \( t = \frac{(7 + 8n)\pi}{16} \) for \( n=0,1,2, \ldots \).

Step by step solution

01

Verify Part (a) Solution

The original differential equation is \( \frac{1}{4} x'' + x' + 5x = 0 \) with initial conditions \( x(0) = \frac{1}{2} \) and \( x'(0) = 1 \). The given solution is \( x = e^{-2t} \left( \frac{1}{2} \cos 4t + \frac{1}{2} \sin 4t \right) \). Start by substituting \( x(t) \) into the differential equation and verify it satisfies both the equation and initial conditions. The solution appears to satisfy the differential equation and given initial conditions upon verification.
02

Verify Part (b) Solution

The proposed solution for part (b) is \( x = \frac{1}{\sqrt{2}} e^{-2t} \sin (4t + \frac{\pi}{4}) \). Use trigonometric identities to express \( \sin(a+b) \) as \( \sin a \cos b + \cos a \sin b \) and simplify to check if this matches the form found in part (a). Substituting \( a = 4t \) and \( b = \frac{\pi}{4} \), match it to the given solution by expressing it in terms of \( \cos 4t \) and \( \sin 4t \). This satisfies the differential equation when checked.
03

Solve Part (c) to Find Points Where Solution Equals Zero

To find when \( x = 0 \) from part (b), set \( \frac{1}{\sqrt{2}} e^{-2t} \sin (4t + \frac{\pi}{4}) = 0 \). Since \( e^{-2t} eq 0 \), \( \sin (4t + \frac{\pi}{4}) = 0 \) must be satisfied. The general solution for \( \sin \theta = 0 \) is \( \theta = n\pi \) for \( n \) an integer. Substitute \( 4t + \frac{\pi}{4} = n\pi \) and solve for \( t \) to get \( t = \frac{n\pi}{4} - \frac{\pi}{16} \). To ensure heading downward, adjust to the form \( t = \frac{(7 + 8n)\pi}{16} \). This gives us the specific values of \( t \) for which \( x = 0 \) at intervals.

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 91影视!

Key Concepts

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

Initial Value Problem
An initial value problem in differential equations involves finding a solution to the equation that satisfies certain given initial conditions. In this exercise, the differential equation is \( \frac{1}{4} x'' + x' + 5x = 0 \) with initial conditions \( x(0) = \frac{1}{2} \) and \( x'(0) = 1 \). These initial conditions are specific values that the function and its derivative must satisfy at a particular point, which is typically where the independent variable (such as time \( t \)) is zero.

Solving an initial value problem requires not only finding the general form of the solution to the differential equation but also determining the specific constants that make the solution satisfy the initial conditions. In part (a) of this exercise, the proposed solution \( x = e^{-2t} \left( \frac{1}{2} \cos 4t + \frac{1}{2} \sin 4t \right) \) was verified to meet both the differential equation and the specified initial conditions through substitution and simplification.

Solving initial value problems is crucial in modeling real-world scenarios because it ensures that the mathematical model being used accurately predicts the behavior of the system at the start and as it progresses over time.
Trigonometric Identities
In the context of differential equations, trigonometric identities can simplify expressions, making it easier to verify proposed solutions. Trigonometric identities are mathematical relationships that hold true for all angles and are particularly useful in combining or transforming trigonometric functions.

For this exercise, specifically in step 2, the proposed solution for part (b) \( x = \frac{1}{\sqrt{2}} e^{-2t} \sin (4t + \frac{\pi}{4}) \) involves the trigonometric identity for the sum of angles \( \sin(a+b) = \sin a \cos b + \cos a \sin b \). This identity allows the expression to be rewritten in terms of \( \cos 4t \) and \( \sin 4t \) to check for consistency with the solution from part (a).

Using these identities is essential as they help transform and verify complex expressions by breaking them down into simpler components. This is crucial in ensuring that solutions align with the properties of the original differential equations.
Zero of a Function
Finding the zeros of a function involves determining the values of the variable for which the function evaluates to zero. In differential equations, especially those involving oscillatory solutions, this can reveal when certain states or conditions occur within the system being modeled.

In part (c) of the original exercise, the task is to find the points in time \( t \) where the solution \( x = \frac{1}{\sqrt{2}} e^{-2t} \sin(4t + \frac{\pi}{4}) \) equals zero. Since \( e^{-2t} \) never equals zero for any real \( t \), the zeros must be found by setting the sine part of the solution, \( \sin(4t + \frac{\pi}{4}) \), to zero. The sine function equals zero at multiples of \( \pi \) (i.e., \( \theta = n\pi \)).

Solving \( 4t + \frac{\pi}{4} = n\pi \) gives a general form for \( t \): \( t = \frac{n\pi}{4} - \frac{\pi}{16} \). To narrow down to specific intervals that reflect the downward movement described in the problem, this is adjusted to \( t = \frac{(7 + 8n)\pi}{16} \), providing particular time points where the function crosses zero. Identifying zeros in this way is significant in analyzing periodic behavior in systems.

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

The initial-value problem is \\[ x^{\prime \prime}+\frac{2}{m} x^{\prime}+\frac{k}{m} x=0, \quad x(0)=0, x^{\prime}(0)=v_{0} \\] With \(k=10,\) the auxiliary equation has roots \(\gamma=-1 / m \pm \sqrt{1-10 m} / m .\) Consider the three cases: \((i) m=\frac{1}{10} .\) The roots are \(\gamma_{1}=\gamma_{2}=10\) and the solution of the differential equation is \(x(t)=c_{1} e^{-10 t}+c_{2} t e^{-10 t}\) The initial conditions imply \(c_{1}=0\) and \(c_{2}=v_{0}\) and so \(x(t)=v_{0} t e^{-10 t} .\) The condition \(x(1)=0\) implies \(v_{0} e^{-10}=0\) which is impossible because \(v_{0} \neq 0\) \((i i) 1-10 m>0\) or 0 < m < \frac{1}{10}.\( \)The roots are $$\gamma_{1}= -\frac{1}{m}-\frac{1}{m} \sqrt{1-10 m} \quad \text { and }$$ $$\gamma_{2}=-\frac{1}{m}+\frac{1}{m} \sqrt{1-10 m}$$ and the solution of the differential equation is \(x(t)=c_{1} e^{\gamma_{1} t}+c_{2} e^{\gamma_{2} t}\). The initial conditions imply $$\begin{array}{r} c_{1}+c_{2}=0 \\ \gamma_{1} c_{1}+\gamma_{2} c_{2}=v_{0} \end{array}$$ so \(c_{1}=v_{0} /\left(\gamma_{1}-\gamma_{2}\right), c_{2}=-v_{0} /\left(\gamma_{1}-\gamma_{2}\right),\) and $$x(t)=\frac{v_{0}}{\gamma_{1}-\gamma_{2}}\left(e^{\gamma_{1} t}-e^{\gamma_{2} t}\right)$$ Again, \(x(1)=0\) is impossible because \(v_{0} \neq 0\). \((i i i) \quad 1-10 m < 0\) or \(m >\frac{1}{10} .\) The roots of the auxiliary equation are $$\gamma_{1}=-\frac{1}{m}-\frac{1}{m} \sqrt{10 m-1} i \quad \text { and }\space\gamma_{2}=-\frac{1}{m}+\frac{1}{m} \sqrt{10 m-1}$$ and the solution of the differential equation is $$x(t)=c_{1} e^{-t / m} \cos \frac{1}{m} \sqrt{10 m-1} t+c_{2} e^{-t / m} \sin \frac{1}{m} \sqrt{10 m-1} t$$ The initial conditions imply \(c_{1}=0\) and \(c_{2}=m v_{0} / \sqrt{10 m-1},\) so that $$x(t)=\frac{m v_{0}}{\sqrt{10 m-1}} e^{-t / m} \sin \left(\frac{1}{m} \sqrt{10 m-1} t\right)$$ The condition \(x(1)=0\) implies $$\begin{aligned} \frac{m v_{0}}{\sqrt{10 m-1}} e^{-1 / m} \sin \frac{1}{m} \sqrt{10 m-1} &=0 \\\ \sin \frac{1}{m} \sqrt{10 m-1} &=0 \\ \frac{1}{m} \sqrt{10 m-1} &=n \pi \\ \frac{10 m-1}{m^{2}} &=n^{2} \pi^{2}, n=1,2,3, \ldots \\ \left(n^{2} \pi^{2}\right) m^{2}-10 m+1 &=0 \end{aligned}$$ $$m=\frac{10 \sqrt{100-4 n^{2} \pi^{2}}}{2 n^{2} \pi^{2}}=\frac{5 \pm \sqrt{25-n^{2} \pi^{2}}}{n^{2} \pi^{2}}$$ since \(m\) is real, \(25-n^{2} \pi^{2} \geq 0 .\) If \(25-n^{2} \pi^{2}=0,\) then \(n^{2}=25 / \pi^{2},\) and \(n\) is not an integer. Thus, \(25-n^{2} \pi^{2}=\) \((5-n \pi)(5+n \pi) >0\) and since \(n >0,5+n \pi >0,\) so \(5-n \pi >0\) also. Then \(n<5 / \pi,\) and so \(n=1 .\) Therefore the mass \(m\) will pass through the equilibrium position when \(t=1\) for $$m_{1}=\frac{5+\sqrt{25-\pi^{2}}}{\pi^{2}} \quad \text { and } \quad m_{2}=\frac{5-\sqrt{25-\pi^{2}}}{\pi^{2}}$$.

(a) Setting \(d y / d t=v,\) the differential equation in (13) becomes \(d v / d t=-g R^{2} / y^{2} .\) But, by the chain rule, \(d v / d t=(d v / d y)(d y / d t)=v d v / d t,\) so \(v d v / d y=-g R^{2} / y^{2} .\) Separating variables and integrating we obtain \(v d v=-g R^{2} \frac{d y}{y^{2}} \quad\) and \(\quad \frac{1}{2} v^{2}=\frac{g R^{2}}{y}+c\) Setting \(v=v_{0}\) and \(y=R\) we find \(c=-g R+\frac{1}{2} v_{0}^{2}\) and \\[ v^{2}=2 g \frac{R^{2}}{y}-2 g R+v_{0}^{2} \\] (b) As \(y \rightarrow \infty\) we assume that \(v \rightarrow 0^{+}\). Then \(v_{0}^{2}=2 g R\) and \(v_{0}=\sqrt{2 g R}\) (c) Using \(g=32 \mathrm{ft} / \mathrm{s}\) and \(R=4000(5280) \mathrm{ft}\) we find \(v_{0}=\sqrt{2(32)(4000)(5280)} \approx 36765.2 \mathrm{ft} / \mathrm{s} \approx 25067 \mathrm{mi} / \mathrm{hr}\) (d) \(v_{0}=\sqrt{2(0.165)(32)(1080)} \approx 7760 \mathrm{ft} / \mathrm{s} \approx 5291 \mathrm{mi} / \mathrm{hr}\)

The general solution of the differential equation is \\[ y=c_{1} \cos \sqrt{\frac{P}{E I}} x+c_{2} \sin \sqrt{\frac{P}{E I}} x+\frac{w_{0}}{2 P} x^{2}+\frac{w_{0} E I}{P^{2}} \\] Setting \(y(0)=0\) we obtain \(c_{1}=-w_{0} E I / P^{2},\) so that \\[ y=-\frac{w_{0} E I}{P^{2}} \cos \sqrt{\frac{P}{E I}} x+c_{2} \sin \sqrt{\frac{P}{E I}} x+\frac{w_{0}}{2 P} x^{2}+\frac{w_{0} E I}{P^{2}} \\] Setting \(y^{\prime}(L)=0\) we find \\[ c_{2}=\left(-\sqrt{\frac{P}{E I}} \frac{w_{0} E I}{P^{2}} \sin \sqrt{\frac{P}{E I}} L-\frac{w_{0} L}{P}\right) / \sqrt{\frac{P}{E I}} \cos \sqrt{\frac{P}{E I}} L \\]

For \(\lambda=\alpha^{4}, \alpha>0,\) the solution of the differential equation is \\[ y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} \cosh \alpha x+c_{4} \sinh \alpha x \\]. The boundary conditions \(y(0)=0, y^{\prime}(0)=0, y(1)=0, y^{\prime}(1)=0\) give in turn, $$\begin{array}{c} c_{1}+c_{3}=0 \\ \alpha c_{2}+\alpha c_{4}=0 \\ c_{1} \cos \alpha+c_{2} \sin \alpha+c_{3} \cosh \alpha+c_{4} \sinh \alpha=0 \\\ -c_{1} \alpha \sin \alpha+c_{2} \alpha \cos \alpha+c_{3} \alpha \sinh \alpha+c_{4} \alpha \cosh \alpha=0 \end{array}$$ The first two equations enable us to write \\[ \begin{aligned} c_{1}(\cos \alpha-\cosh \alpha)+c_{2}(\sin \alpha-\sinh \alpha) &=0 \\ c_{1}(-\sin \alpha-\sinh \alpha)+c_{2}(\cos \alpha-\cosh \alpha) &=0 \end{aligned} \\]. The determinant \\[ \left|\begin{array}{rr} \cos \alpha-\cosh \alpha & \sin \alpha-\sinh \alpha \\ -\sin \alpha-\sinh \alpha & \cos \alpha-\cosh \alpha \end{array}\right|=0 \\] simplifies to \(\cos \alpha \cosh \alpha=1 .\) From the figure showing the graphs of \(1 / \cosh x\) and \(\cos x,\) we see that this equation has an infinite number of positive roots. With the aid of a CAS the first four roots are found to be \(\alpha_{1}=4.73004, \alpha_{2}=7.8532, \alpha_{3}=10.9956,\) and \(\alpha_{4}=14.1372,\) and the corresponding eigenvalues are \(\lambda_{1}=500.5636, \lambda_{2}=3803.5281, \lambda_{3}=14,617.5885,\) and \(\lambda_{4}=39,944.1890 .\) Using the third equation in the system to eliminate \(c_{2},\) we find that the eigenfunctions are \\[ y_{n}=\left(-\sin \alpha_{n}+\sinh \alpha_{n}\right)\left(\cos \alpha_{n} x-\cosh \alpha_{n} x\right)+\left(\cos \alpha_{n}-\cosh \alpha_{n}\right)\left(\sin \alpha_{n} x-\sinh \alpha_{n} x\right) \\]

We have \(y_{c}=c_{1} \cos x+c_{2} \sin x\) and we assume \(y_{p}=A^{2}+B x+C .\) Substituting into the differential equation we find \(A=1, B=0,\) and \(C=-1 .\) Thus \(y=c_{1} \cos x+c_{2} \sin x+x^{2}-1 .\) From \(y(0)=5\) and \(y(1)=0\) we obtain $$\begin{aligned} &c_{1}-1=5\\\ &(\cos 1) c_{1}+(\sin 1) c_{2}=0 \end{aligned}$$ Solving this system we find \(c_{1}=6\) and \(c_{2}=-6\) cot \(1 .\) The solution of the boundary-value problem is $$y=6 \cos x-6(\cot 1) \sin x+x^{2}-1$$
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Particle Model of Matter

Read Explanation

Famous Physicists

Read Explanation

Astrophysics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Measurements

Read Explanation
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.