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A) zyz = 6x2 · fu · 6x²y+ 5yª · fwu · 6x²y B) zyz = 6x2 · fu + (2x³ – 1) · fuu + 5y4 · fwu · 6x²y C) zyz = 6x²y · (2x³ – 1) · fuu + 6x² · fu + 6x²y · fwu D) zym = 30x²y³ . fwu + (12x³y – 6æ²y) · fuu + 6x² · fu

A) zyz = 6x2 · fu · 6x²y+ 5yª · fwu · 6x²y B) zyz = 6x2 · fu + (2x³ – 1) · fuu + 5y4 · fwu · 6x²y C) zyz = 6x²y · (2x³ – 1) · fuu + 6x² · fu + 6x²y · fwu D) zym = 30x²y³ . fwu + (12x³y – 6æ²y) · fuu + 6x² · fu

Advanced Engineering Mathematics
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

Let f be a function with continuous second partial derivatives and define:

image 1 first equation and two equations

From the above it is concluded that:

image 2

A) zyr = 6x² · fu · 6x²y+ 5yª · fwu · 6x²y B) zyz = 6x2 . fu + (2a³ – 1) · fuu + 5y4 · fwu · 6x²y C) zyz = 6a?y · (2x3 – 1) · fuu + 6x² · fu + 6x²y· fwru D) zpz = 30x?y5 . fwu + (12x³y – 6x²y) · fuu + 6x² · fu
Transcribed Image Text:A) zyr = 6x² · fu · 6x²y+ 5yª · fwu · 6x²y B) zyz = 6x2 . fu + (2a³ – 1) · fuu + 5y4 · fwu · 6x²y C) zyz = 6a?y · (2x3 – 1) · fuu + 6x² · fu + 6x²y· fwru D) zpz = 30x?y5 . fwu + (12x³y – 6x²y) · fuu + 6x² · fu
z = f (2x®y – y, y°) u = 2a®y – y y w = y% dz = (203 – 1) · fu + 5y4 . fw dy
Transcribed Image Text:z = f (2x®y – y, y°) u = 2a®y – y y w = y% dz = (203 – 1) · fu + 5y4 . fw dy
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