Identify four (4) lines of error on the solution of the given differential equaiton. the absence of an Integral sign is not considered an error.
Identify four (4) lines of error on the solution of the given differential equaiton. the absence of an Integral sign is not considered an error.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.3: Maxima And Minima
Problem 20E
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Question
Identify four (4) lines of error on the solution of the given
![(x+ 2y -1) dx + (ax +y+ ) dy - o
メ=u-1 dx =D du
yov-, dy e dy
LINE 1
%3D
LINE 2
(u-1+ 2Y- 2-1) du + (2-2+v-1 +) dy =0
LINE 4 (u+ 2 u + (2u + v)み =0
let y= 2v
du = adv + ydz
LINE S
LINE 6
(2v +2y)(zd+ vd2) + (2ZY +y)dx =0
(2+2) (zdv + vdz)+ (22T)dy = 0
(2+2) zdv + (z+2) vd2 + (22+1)dN 0
(22 + 22)d + (22+1)dy + (2+2) vd230
( t2ま十1りv+ (2+2)ydz = 0
dz = 0
LINE 7
LINE &
LINIS 9
LINE ID
LINE
%3D
LINE 12
22+42 +1
LINE B In () + 2 In (22+42 +1) = C
LIAE 14 In (ve) + im (2²+ 42ti) =C
In Lv(22 +42 +10]- 'C
In Lv Cz2+47+1)]=C
LINE IS
LINE L
y2 (z+42 +1)=C
V2 (u/2 + 4u/V +1) = C
u2 + 4uy +y2
LINE 7
LINE 18
LINE 19
LINE 20
LINE 21
(メ+)と+ 4(x+りyー)+ (yー)2=C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57d52ddc-f8a2-459e-8c9f-6cde986c42de%2F25babdf5-d5a7-4b5b-8b27-c22d03f2acf4%2Fsexkm1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(x+ 2y -1) dx + (ax +y+ ) dy - o
メ=u-1 dx =D du
yov-, dy e dy
LINE 1
%3D
LINE 2
(u-1+ 2Y- 2-1) du + (2-2+v-1 +) dy =0
LINE 4 (u+ 2 u + (2u + v)み =0
let y= 2v
du = adv + ydz
LINE S
LINE 6
(2v +2y)(zd+ vd2) + (2ZY +y)dx =0
(2+2) (zdv + vdz)+ (22T)dy = 0
(2+2) zdv + (z+2) vd2 + (22+1)dN 0
(22 + 22)d + (22+1)dy + (2+2) vd230
( t2ま十1りv+ (2+2)ydz = 0
dz = 0
LINE 7
LINE &
LINIS 9
LINE ID
LINE
%3D
LINE 12
22+42 +1
LINE B In () + 2 In (22+42 +1) = C
LIAE 14 In (ve) + im (2²+ 42ti) =C
In Lv(22 +42 +10]- 'C
In Lv Cz2+47+1)]=C
LINE IS
LINE L
y2 (z+42 +1)=C
V2 (u/2 + 4u/V +1) = C
u2 + 4uy +y2
LINE 7
LINE 18
LINE 19
LINE 20
LINE 21
(メ+)と+ 4(x+りyー)+ (yー)2=C
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