Comparison of Numerical and Analytic Solutions: Consider the initial-value problem y′ =y−2, y(0)=1(a) Find the approximate solutions at x = 0.1, 0.2, and 0.3 using the Euler method with step size h = 0.1.(b) Find the approximations at the same points x as in part (a) using the step size h = 0.05; you will have to compute twice as many steps to reach each point x.(c) Find the analytic solution of the initial value problem. Compute the errors be- tween the analytic solution and each of the approximations in parts (a) and (b). What can you say about the behavior of the errors? 4. Comparison of Numerical and Analytic Solutions: Consider the initial-value problemy = y – 2, y(0) = 1-(a) Find the approximate solutions at x = 0.1, 0.2, and 0.3 using the Euler methodwith step size h = 0.1.(b) Find the approximations at the same points x as in part (a) using the step sizeh = 0.05; you will have to compute twice as many steps to reach each point x.(c) Find the analytic solution of the initial value problem. Compute the errors be-tween the analytic solution and each of the approximations in parts (a) and (b).What can you say about the behavior of the errors?

Answered: Comparison of Numerical and Analytic…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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Comparison of Numerical and Analytic Solutions: Consider the initial-value problem y′ =y−2, y(0)=1
(a) Find the approximate solutions at x = 0.1, 0.2, and 0.3 using the Euler method with step size h = 0.1.

(b) Find the approximations at the same points x as in part (a) using the step size h = 0.05; you will have to compute twice as many steps to reach each point x.

(c) Find the analytic solution of the initial value problem. Compute the errors be- tween the analytic solution and each of the approximations in parts (a) and (b). What can you say about the behavior of the errors?

4. Comparison of Numerical and Analytic Solutions: Consider the initial-value problem
y = y – 2, y(0) = 1
-
(a) Find the approximate solutions at x = 0.1, 0.2, and 0.3 using the Euler method
with step size h = 0.1.
(b) Find the approximations at the same points x as in part (a) using the step size
h = 0.05; you will have to compute twice as many steps to reach each point x.
(c) Find the analytic solution of the initial value problem. Compute the errors be-
tween the analytic solution and each of the approximations in parts (a) and (b).
What can you say about the behavior of the errors?

Expert Solution

Step 1

(a) Consider the differential equation

$y^{'} = y - 2, y (0) = 1$

According to Euler Method, the approximate solution at the step "n+1" is calculated as

$y_{n + 1_{}} = y_{n} + h f (x_{n} + y_{n})$

Here $h = 0.1, x_{0} = 0, y_{0} = 1, f (x, y) = y - 2$ . We have to find the approximate solutions at $x = 0.1, 0.2, a n d 0.3$ using the step size $h = 0.1$ .

$x_{1} = x_{0} + h = 0 + 0.1 = 0.1 f_{0} = f (x_{0} + y_{0}) = f (0, 1) = y_{0} - 2 = 1 - 2 = - 1 y_{1} = y (x_{1}) = y_{0} + h f_{0} = 1 + (0.1) (- 1) = 0.9$

Thus the approximate solution at $x = 0.1$ is $y_{1} = 0.9$ .

Next,

$x_{2} = x_{1} + h = 0.1 + 0.1 = 0.2 f_{1} = f (x_{1} + y_{1}) = f (0.1, 0.9) = y_{1} - 2 = 0.9 - 2 = - 1.1 y_{2} = y (x_{2}) = y_{1} + h f_{1} = 0.9 + (0.1) (- 1.1) = 0.79$

Thus the approximate solution at $x = 0.2$ is $y_{1} = 0.79$ .

Next,

$x_{3} = x_{2} + h = 0.2 + 0.1 = 0.3 f_{2} = f (x_{2} + y_{2}) = f (0.2, 0.79) = y_{2} - 2 = 0.79 - 2 = - 1.21 y_{3} = y (x_{3}) = y_{2} + h f_{2} = 0.79 + (0.1) (- 1.21) = 0.669$

Thus the approximate solution at $x = 0.3$ is $y_{1} = 0.669$ .

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