use Euler's Midpoint Method with the time step indicated to approximate the given value of y(t).y(0.5);dy= y +t, y(0) = 1, h=0.1%3Ddt

Answered: use Euler's Midpoint Method with the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

use Euler's Midpoint Method with the time step indicated to approximate the given value of y(t).
y(0.5);
dy
= y +t, y(0) = 1, h=0.1
%3D
dt

Expert Solution

Step 1

Given:

$y' = \frac{d y}{d t} = y + t$ , $y (0) = 1, h = 0.1$ .

To find: $y (0.5)$ using the Euler's Midpoint method.

Step 2

By Euler's midpoint method we have,
$y_{m + 1} = y_{m} + h f (t_{m} + \frac{h}{2}, y_{m} + \frac{h}{2} f (t_{m}, y_{m}))$

We have,
$f (t_{0}, y_{0}) = f (0, 1) = 1$
$t_{0} + \frac{h}{2} = 0 + \frac{0.1}{2} = 0.05$
$y_{0} + \frac{h}{2} f (t_{0}, y_{0}) = 1 + \frac{0.1}{2} (1) = 1.05$

$f (t_{0} + \frac{h}{2}, y_{0} + \frac{h}{2} f (t_{0}, y_{0})) = f (0.05, 1.05) = 1.1$

$y_{1} = y_{0} + h f (t_{0} + \frac{h}{2}, y_{0} + \frac{h}{2} f (t_{0}, y_{0})) = 1 + (0.1) (1.1) = 1.11$ .

Now taking $(x_{1}, y_{1})$ in place of $(x_{0}, y_{0})$ and repeating the above process we get,

$f (t_{1}, y_{1}) = f (0.1, 1.11) = 1.21$
$t_{1} + \frac{h}{2} = 0.1 + \frac{0.1}{2} = 0.15$
$y_{1} + \frac{h}{2} f (t_{1}, y_{1}) = 1.11 + \frac{0.1}{2} (1.21) = 1.1705$

$f (t_{1} + \frac{h}{2}, y_{1} + \frac{h}{2} f (t_{1}, y_{1})) = f (0.15, 1.1705) = 1.3205$

$y_{2} = y_{1} + h f (t_{1} + \frac{h}{2}, y_{1} + \frac{h}{2} f (t_{1}, y_{1})) = 1.11 + (0.1) (1.3205) = 1.2421$ .

Again, taking $(x_{2}, y_{2})$ in place of $(x_{1}, y_{1})$ and repeating the above process we get,

$f (t_{2}, y_{2}) = f (0.2, 1.2421) = 1.442$
$t_{2} + \frac{h}{2} = 0.2 + \frac{0.1}{2} = 0.25$
$y_{2} + \frac{h}{2} f (t_{2}, y_{2}) = 1.2421 + \frac{0.1}{2} (1.442) = 1.3142$