/*! 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} Q 3.6-3E Show that when the trapezoid sch... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q 3.6-3E (page 130)

Show that when the trapezoid scheme given in formula (8) is used to approximate the solutionf(x)=exofy'=y,y(0)=1 , at x = 1, then we get yn+1=1+h21-h2yn,n = 0, 1, 2, . . . , which leads to the approximation (1+h21-h2)1hfor the constant e.Compute this approximation for h= 1,10-1,10-2,10-3,and10-4and compare your results with those in Tables 3.4 and 3.5.

Short Answer

Expert verified

h

y(1)=e

1

3

0.1

2.72055

0.01

2.71830

0.001

2.71828

0.0001

2.71828

Step by step solution

01

Apply iteration rule for the trapezoidal scheme

Here y'=y,y(0)=1

Apply the iteration rule for the trapezoidal scheme is

yn+1=yn+h2[f(xn,yn)+f(xn+1,yn+1)]

In our case of differential equation our equation is

yn+1=yn1+h21-h2

The interval is

b-an=1-0n=1n=hn=1h

02

Find the values of f(x) = ex

h

y(1)=e

1

3

0.1

2.72055

0.01

2.71830

0.001

2.71828

0.0001

2.71828

03

Substitute in the iteration rule for the trapezoidal scheme

Now,

y(1)=y(0+nh)=yny1=y(0)1+h21-h2=1+h21-h2y2=1+h21-h22...yn=1+h21-h2nyn=1+h21-h21h

Hence the solution is

h

e

1

3

0.1

2.72055

0.01

2.71830

0.001

2.71828

0.0001

2.71828

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!

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

In Problems 23鈥�27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive substance is the time it takes for one-half of the substance to disintegrate. The only undiscovered isotopes of the two unknown elements hohum and inertium (symbols Hh and It) are radioactive. Hohum decays into inertium with a decay constant of 2/yr, and inertium decays into the nonradioactive isotope of bunkum (symbol Bu) with a decay constant of 1/yr. An initial mass of 1 kg of hohum is put into a non-radiaoctive container, with no other source of hohum, inertium, or bunkum. How much of each of the three elements is in the container after t yr? (The decay constant is the constant of proportionality in the statement that the rate of loss of mass of the element at any time is proportional to the mass of the element at that time.)

The solution to the initial value problem \(\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}{{\bf{y}}^{\bf{2}}}{\bf{ - 2}}{{\bf{e}}^{\bf{x}}}{\bf{y + }}{{\bf{e}}^{{\bf{2x}}}}{\bf{ + }}{{\bf{e}}^{\bf{x}}}{\bf{,y(0) = 3}}\)has a vertical asymptote (鈥渂lows up鈥�) at some point in the interval\(\left[ {{\bf{0,2}}} \right]\). By experimenting with the fourth-order Runge鈥揔utta subroutine, determine this point to two decimal places.

An RLcircuit with a 5 resistor and a 0.05-H inductor carries a current of 1 A at t= 0, at which time a voltage source E(t) = 5 cos 120tV is added. Determine the subsequent inductor current and voltage.

It was noon on a cold December day in Tampa: 16掳C. Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detective Taylor took out a thermometer and measured the temperature of the body: 34.5掳C. He then left for lunch. Upon returning at 1:00 p.m., he found the body temperature to be 33.7掳C. When did the murder occur? [Hint: Normal body temperature is 37掳C.]

Rocket Flight. A model rocket having initial mass mo kg is launched vertically from the ground. The rocket expels gas at a constant rate of a kg/sec and at a constant velocity of b m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality constant g. Because the mass is not constant, Newton鈥檚 second law leads to the equation (mo - 伪t) dv/dt - 伪尾 = -g(m0 鈥� 伪t), where v = dx/dt is the velocity of the rocket, x is its height above the ground, and m0 - 伪t is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 0鈮<m0/伪.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

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.