pply Euler’s semi-linear method (no other method) Apply Euler's semi-linear method (no other method allowed!) to the fol-lowing first-order equation using steps of size h = 1, for i = 1, 2, 3, andinitial condition y(0) = 8.y' +2xYx² + 1 ¥24-x²y (x² + 1)²*

Answered: Apply Euler's semi-linear method (no…

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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pply Euler’s semi-linear method (no other method)

Expert Solution

Step 1: Step 1:

To apply Euler's semi-linear method to the given first-order differential equation with a step size of $h = 1$ and the initial condition $y (0) = 8$ , we'll follow these steps:

The given equation is:

$y apostrophe plus fraction numerator left parenthesis 2 x right parenthesis over denominator left parenthesis x squared plus 1 right parenthesis end fraction space cross times space y equals fraction numerator 2 to the power of 4 minus x squared end exponent over denominator left parenthesis y cross times left parenthesis x squared plus 1 right parenthesis squared end fraction$

Initialize:
- Set the initial condition: $y (0) = 8$
- Define the step size: $h = 1$
- Define the range of x values for which you want to approximate y. In this case, we will approximate y for x = 1, 2, and 3.
Use the Euler semi-linear method to update y at each step using the following formula: $y left square bracket i plus 1 right square bracket space equals space y left square bracket i right square bracket space plus space h space cross times space fraction numerator left square bracket f left parenthesis x left square bracket i right square bracket comma space y left square bracket i right square bracket right parenthesis space plus space f left parenthesis x left square bracket i plus 1 right square bracket comma space y left square bracket i right square bracket space plus space h space cross times space f left parenthesis x left square bracket i right square bracket comma space y left square bracket i right square bracket right parenthesis right parenthesis right square bracket over denominator 2 end fraction space$
Here, $f (x, y)$ is the derivative of y with respect to x, which is given by the original equation: $f left parenthesis x comma y right parenthesis equals fraction numerator left parenthesis 2 to the power of 4 minus x squared end exponent right parenthesis over denominator left parenthesis y cross times left parenthesis x squared plus 1 right parenthesis squared right parenthesis end fraction minus fraction numerator left parenthesis 2 x right parenthesis over denominator left parenthesis x squared plus 1 right parenthesis cross times y end fraction$
Calculate y at each step using the above formula:
For i = 1:
- $x left square bracket 1 right square bracket space equals space 1$
- $y left square bracket 1 right square bracket space equals space y left square bracket 0 right square bracket space plus space h space cross times space fraction numerator left square bracket f left parenthesis 0 comma space 8 right parenthesis space plus space f left parenthesis 1 comma space 8 space plus space h space cross times space f left parenthesis 0 comma space 8 right parenthesis right parenthesis right square bracket over denominator 2 end fraction$
For i = 2:
- $x [2] = 2$
- $y left square bracket 2 right square bracket space equals space y left square bracket 1 right square bracket space plus space h space cross times space fraction numerator left square bracket f left parenthesis 1 comma space y left square bracket 1 right square bracket right parenthesis space plus space f left parenthesis 2 comma space y left square bracket 1 right square bracket space plus space h space cross times space f left parenthesis 1 comma space y left square bracket 1 right square bracket right parenthesis right parenthesis right square bracket over denominator 2 end fraction$
For i = 3:
- $x [3] = 3$
- $y left square bracket 3 right square bracket space equals space y left square bracket 2 right square bracket space plus space h space cross times space fraction numerator left square bracket f left parenthesis 2 comma space y left square bracket 2 right square bracket right parenthesis space plus space f left parenthesis 3 comma space y left square bracket 2 right square bracket space plus space h space cross times space f left parenthesis 2 comma space y left square bracket 2 right square bracket right parenthesis right parenthesis right square bracket over denominator 2 end fraction$

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Apply Euler's semi-linear method (no other method allowed!) to the fol- lowing first-order equation using steps of size h = 1, for i = 1, 2, 3, and initial condition y(0) = 8. y' + 2x -Y = x² + 1 24-x² y (x² + 1)²°