X -2 -1 TOHN 3 0 1 2 y 14 -1 -6 -1 14 39
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Write the equation for question 9 showing all your work.
![This image contains a data table with two columns labeled "x" and "y". Below is the transcribed content:
| x | y |
|----|----|
| -2 | 14 |
| -1 | -1 |
| 0 | -6 |
| 1 | -1 |
| 2 | 14 |
| 3 | 39 |
The table shows pairs of values where the variable "x" is an integer, and the corresponding "y" value varies across the row. This table could be used to observe the relationship between "x" and "y" for educational purposes, such as interpreting data trends or exploring mathematical functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a3fd98d-1ca1-4ccd-aee3-22d706602e6b%2F5f19b20f-d0ee-48f2-bd98-ddcb68bcf630%2Fdy3febm_processed.png&w=3840&q=75)
Transcribed Image Text:This image contains a data table with two columns labeled "x" and "y". Below is the transcribed content:
| x | y |
|----|----|
| -2 | 14 |
| -1 | -1 |
| 0 | -6 |
| 1 | -1 |
| 2 | 14 |
| 3 | 39 |
The table shows pairs of values where the variable "x" is an integer, and the corresponding "y" value varies across the row. This table could be used to observe the relationship between "x" and "y" for educational purposes, such as interpreting data trends or exploring mathematical functions.
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Step 1
The equation is y=a+bx+cx2 and the normal equations are
∑y=an+b∑x+c∑x2
∑xy=a∑x+b∑x2+c∑x3
∑x2y=a∑x2+b∑x3+c∑x4
The values are calculated using the following table
x | y | x2 | x3 | x4 | x⋅y | x2⋅y |
-2 | 14 | 4 | -8 | 16 | -28 | 56 |
-1 | -1 | 1 | -1 | 1 | 1 | -1 |
0 | -6 | 0 | 0 | 0 | 0 | 0 |
1 | -1 | 1 | 1 | 1 | -1 | -1 |
2 | 14 | 4 | 8 | 16 | 28 | 56 |
3 | 39 | 9 | 27 | 81 | 117 | 351 |
--- | --- | --- | --- | --- | --- | --- |
∑x=3 | ∑y=59 | ∑x2=19 | ∑x3=27 | ∑x4=115 | ∑x⋅y=117 | ∑x2⋅y=461 |
Substituting these values in the normal equations
6a+3b+19c=59
3a+19b+27c=117
19a+27b+115c=461
Solving these 3 equations,
Total Equations are 3
6a+3b+19c=59→(1)
3a+19b+27c=117→(2)
19a+27b+115c=461→(3)
Select the equations (1) and (2), and eliminate the variable a.
6a+3b+19c=59 | ×1→ | 6a | + | 3b | + | 19c | = | 59 | |||
− | |||||||||||
3a+19b+27c=117 | ×2→ | 6a | + | 38b | + | 54c | = | 234 | |||
- | 35b | - | 35c | = | -175 | →(4) |
Select the equations (2) and (3), and eliminate the variable a.
3a+19b+27c=117 | ×19→ | 57a | + | 361b | + | 513c | = | 2223 | |||
− | |||||||||||
19a+27b+115c=461 | ×3→ | 57a | + | 81b | + | 345c | = | 1383 | |||
280b | + | 168c | = | 840 | →(5) |
Select the equations (4) and (5), and eliminate the variable c.
-35b-35c=-175 | ×24→ | - | 840b | - | 840c | = | -4200 | ||||
+ | |||||||||||
280b+168c=840 | ×5→ | 1400b | + | 840c | = | 4200 | |||||
560b | = | 0 | →(6) |
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