• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q5.3-17E

Expert-verified
Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 1
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem

dudx=3u-4v;u(0)=1'dvdx=2u-3v;v(0)=1

at x = 1. Starting with h=1, continue halving the step size until two successive approximations of u(1)and v(1) differ by at most 0.001.

The solution is u1=0.36789 and v1=0.36789.

See the step by step solution

Step by Step Solution

Transform the equation

Write the equation as u'=3u-4v and v'=2u-3v.

The transformation of the equation is:

u'(t)=x'2(t)v'(t)=x3(t)x2(t)=v(t)x1=u(t)x'2(t)=3x1-4x2x'3(t)=2x1-3x2

The initial conditions are

u0=1v0=1

Apply Runge –Kutta method

For the solution apply the Runge-Kutta method in MATLAB for h=0.25, and the solution is u(1)=0.36789 and v(1)=0.36789.

Find that u1 and v(1) differ by at most 0.001.

Subtracting the values of u1 and v1 then

u1-v1=0.36789-0.36678=0.001101

Therefore, u1 and v1 differ by at most 0.001.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.