Answered: 3. Given the function y = -x²…

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

3. Given the function y = -x² (1-2x)(-2×+3)(1-2×)²
a) find the degree, leading coefficient, and the value of the finite differences and The end
behaviour of the function in x→±∞ and y→±∞ format. (5)

Expert Solution

Step 1: Question a:

a) To find the degree, leading coefficient, and the value of the finite differences, we can start by simplifying the given function and then analyzing its properties.

First, simplify the function:

$y equals negative x squared left parenthesis 1 minus 2 x right parenthesis left parenthesis negative 2 x plus 3 right parenthesis left parenthesis 1 minus 2 x right parenthesis squared$

Now, let's break down the factors:

Degree of the function: The degree of the function is determined by the highest power of x in the expression. In this case, the highest power of x is $x^{2}$ (from the first term), so the degree is 2.
Leading coefficient: The leading coefficient is the coefficient of the term with the highest power of x. In this case, the leading coefficient is -1 (from the first term).
Finite differences: To find the finite differences, we need to calculate the differences between consecutive values of y for increasing values of x.
Let's calculate some finite differences for the function by evaluating y for a few values of x:
$x equals 0 colon y equals 0 x equals 1 colon y equals negative 1 left parenthesis 1 minus 2 right parenthesis left parenthesis negative 2 plus 3 right parenthesis left parenthesis 1 minus 2 right parenthesis squared equals negative 1 left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis equals negative 1 x equals 2 colon y equals negative 2 squared left parenthesis 1 minus 22 right parenthesis left parenthesis negative 22 plus 3 right parenthesis left parenthesis 1 minus 2 asterisk times 2 right parenthesis squared equals negative 4 left parenthesis 1 minus 4 right parenthesis left parenthesis negative 4 plus 3 right parenthesis left parenthesis 1 minus 4 right parenthesis squared equals negative 4 left parenthesis negative 3 right parenthesis left parenthesis negative 1 right parenthesis left parenthesis 9 right parenthesis equals 108$
Now, let's calculate the finite differences:
$capital delta y left parenthesis 1 right parenthesis space equals space y left parenthesis 1 right parenthesis space minus space y left parenthesis 0 right parenthesis space equals space left parenthesis negative 1 right parenthesis space minus space 0 space equals space minus 1 space capital delta y left parenthesis 2 right parenthesis space equals space y left parenthesis 2 right parenthesis space minus space y left parenthesis 1 right parenthesis space equals space 108 space minus space left parenthesis negative 1 right parenthesis space equals space 109$
The finite differences are not constant, which indicates that the function is not a polynomial of a lower degree. This confirms that the degree is indeed 2.