Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q4E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 1

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.

Register for Free I’ll do it later

Short Answer

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $x = 0.1, 0.2, 0.3, 0.4$ , and $0.5$ , using steps of size $0.1 (h = 0.1)$ .
$\frac{d y}{d x} = \frac{x}{y}, y (0) = - 1$

$x_{n}$	0.1	0.2	0.3	0.4	0.5
$y_{n}$	-1	-1.01	-1.029	-1.085	-1.096

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Write the recursive formula

For the solution use the Euler’s formula $y_{n + 1} = y_{n} + h . f (x_{n}, y_{n})$

Apply recursive formula

One has , $f (x, y) = \frac{- x}{y}, x_{0} = 0, y_{0} = - 1, h = 0.1$

Then , $y_{n + 1} = y_{n} + h . f (x_{n}, y_{n}) = y_{n} + (0.1) (\frac{x_{n}}{y_{n}})$

Put n=0 to find y1

Now, find the value of $y_{1}$ when n = 0, then

$y_{1} = y_{0} + (0.1) (\frac{x_{0}}{y_{0}}) = - 1 + (0.1) (0) = - 1$

Hence, the value of $y_{1} = - 1$ when $x_{1} = 0.1$

Put n=1 to find y2

The value of $y_{2}$ is

$y_{2} = y_{1} + (0.1) (\frac{x_{1}}{y_{1}}) = - 1 + (0.1) (\frac{0.1}{- 1}) = - 1.01$

Thus, the value of $y_{2} = - 1.01$ when $x_{2} = 0.2$

Put n=2 to find y3

Now the value of $y_{3}$ is

$y_{3} = y_{2} + (0.1) (\frac{x_{2}}{y_{2}}) = - 1.01 + (0.1) (\frac{0.2}{- 1.01}) = - 1.01 + (- 0.019) = - 1.029$

So, the value is $y_{3} = - 1.029$ when $x_{3} = 0.3$

Put n=3 to find y4

The value of $y_{4}$ is

$y_{4} = y_{3} + (0.1) (\frac{x_{3}}{y_{3}}) = - 1.029 + (0.1) (\frac{0.3}{- 1.029}) = - 1.029 + (- 0.029) = - 1.058$

Consequently, the value is $y_{4} = - 1.058$ when $x_{4} = 0.4$

Put n=4 to find y5

The value of $y_{5}$ is

$y_{5} = y_{4} + (0.1) (\frac{x_{4}}{y_{4}}) = - 1.058 + (0.1) (\frac{0.4}{- 1.058}) = - 1.058 + (- 0.0378) = - 1.096$

Therefore, the value is $y_{5} = - 1.096$ when $x_{5} = 0.5$

Therefore the solution is

$x_{n}$	0.1	0.2	0.3	0.4	0.5
$y_{n}$	-1	-1.01	-1.029	-1.058	-1.096

Most popular questions for Math Textbooks

In problems $1 - 4$ Use Euler’s method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{dy}{dx} = x + y, y (0) = 1$

In Problems 1 and 2, verify that the pair x(t), and y(t) is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.

$\frac{dx}{dt} = 1, \frac{dy}{dt} = 3 x^{2}; x (t) = t + 1, y (t) = t^{3} + 3 t^{2} + 3 t$

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$3 r - c o s θ \frac{d r}{d θ} = s i n θ$

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$x = 2 \cos t - 3 \sin t, x'' + x = 0$

Decide whether the statement made is True or False. The relation $\sin y + e^{y} = x^{6} - x^{2} + x + 1$ is an implicit solution to $\frac{dy}{dx} = \frac{6 x^{5} - 2 x + 1}{\cos y + e^{y}}$ .

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free