(x+ 2y - 1) dx + (2x +y+ ) dy - o LINE 1 let メ= u-1, dx = du ya v+, dy edy LINE 2 (u-1+2y + 1-1) du + ( 2n – 2 + v + 1 +1) dy =0 しINE 3 LINE 4 cut 2x) du + (2u + v) dy こ O Tet u du こ 2dy + yd2 レINE S こ2Y LINE 6 LINE 7 (ZY + 2Y)dv + ydz) + (22Y +y)dx (2+2)d + vdz)+ (22t) dy =0 (2+2)dv+ (z+2)dz + (22+1)dx =o (22+ 22)dN + (22+1) dy + (2+ 2)vd=0 (z²+ 42 t1)av + (2+2) ydz =0 dy + 3?+42+1. dz = 0 %3D LINE 8 %3D LINE I0 レINE | LINE 12 In (x) + Y½ In (22+42+1) = c HIANE 14 In(ve) + In (2²+42+i) In [v?(z2 +42 t)]='C In [v? Cz²+47+)] C LINE 13 LINE 5 %3D LINE lb y² (z²+ 47t!)= C V2 (uんe t qu t)e C u2 + 4uy t yZ (² LINE ウ レNE B LINE 19 v2 レINE 20 u2 + 4uY +x² = C (メー)と+ 4Cx-り(yt)+ Cyti)2=c レINE 21
(x+ 2y - 1) dx + (2x +y+ ) dy - o LINE 1 let メ= u-1, dx = du ya v+, dy edy LINE 2 (u-1+2y + 1-1) du + ( 2n – 2 + v + 1 +1) dy =0 しINE 3 LINE 4 cut 2x) du + (2u + v) dy こ O Tet u du こ 2dy + yd2 レINE S こ2Y LINE 6 LINE 7 (ZY + 2Y)dv + ydz) + (22Y +y)dx (2+2)d + vdz)+ (22t) dy =0 (2+2)dv+ (z+2)dz + (22+1)dx =o (22+ 22)dN + (22+1) dy + (2+ 2)vd=0 (z²+ 42 t1)av + (2+2) ydz =0 dy + 3?+42+1. dz = 0 %3D LINE 8 %3D LINE I0 レINE | LINE 12 In (x) + Y½ In (22+42+1) = c HIANE 14 In(ve) + In (2²+42+i) In [v?(z2 +42 t)]='C In [v? Cz²+47+)] C LINE 13 LINE 5 %3D LINE lb y² (z²+ 47t!)= C V2 (uんe t qu t)e C u2 + 4uy t yZ (² LINE ウ レNE B LINE 19 v2 レINE 20 u2 + 4uY +x² = C (メー)と+ 4Cx-り(yt)+ Cyti)2=c レINE 21
(x+ 2y - 1) dx + (2x +y+ ) dy - o LINE 1 let メ= u-1, dx = du ya v+, dy edy LINE 2 (u-1+2y + 1-1) du + ( 2n – 2 + v + 1 +1) dy =0 しINE 3 LINE 4 cut 2x) du + (2u + v) dy こ O Tet u du こ 2dy + yd2 レINE S こ2Y LINE 6 LINE 7 (ZY + 2Y)dv + ydz) + (22Y +y)dx (2+2)d + vdz)+ (22t) dy =0 (2+2)dv+ (z+2)dz + (22+1)dx =o (22+ 22)dN + (22+1) dy + (2+ 2)vd=0 (z²+ 42 t1)av + (2+2) ydz =0 dy + 3?+42+1. dz = 0 %3D LINE 8 %3D LINE I0 レINE | LINE 12 In (x) + Y½ In (22+42+1) = c HIANE 14 In(ve) + In (2²+42+i) In [v?(z2 +42 t)]='C In [v? Cz²+47+)] C LINE 13 LINE 5 %3D LINE lb y² (z²+ 47t!)= C V2 (uんe t qu t)e C u2 + 4uy t yZ (² LINE ウ レNE B LINE 19 v2 レINE 20 u2 + 4uY +x² = C (メー)と+ 4Cx-り(yt)+ Cyti)2=c レINE 21
Identify four (4) lines in error on the solution of the given differential equation. The absence of an integral sign is not considered an error. Please be extra careful in selecting correct answers.
Transcribed Image Text:(x+ 2y - 1) dx + (2x +y + ) dy = Ô
let x=u-1, dx = du
- v+l, dy edy
UNE 1
LINE 2
で
(4-1+ 2Y t1-1) du t (2u-2+ v +1+1) dy =0
cut 2x) du + (2u + v) dy =0
しINE 3
LINE 4
Tet y = 2 Y
du こ 2dy + yd2
レINE S
LINE 6
(Zy + 2Y)zdv + yd2) t (22Y +)dx=0
LINE 8 (2+2zd + vdz)+ (22 t1) dy = 0
(2+2)dv+ (2z+2)d2+ (22 +1)dx =O
(2?+ 27) dv + (22+1)dy + (2+ 2)vdz =0
(z² +42+1) av t (2+2)ydz =0
2²+42 + dz
LINE 7
%3D
LINE 9
LINE 10
レINE |
LINE 12
三
In (x) + Y2 In (2²+42 +1)
LIQNE 14 In(vz) + In (2²+4z+i)
In Lve(22 +42 +0]= °C
In [v? Cz²+42t1)]
LINE 13
ご
LINE 5
LINE lb
y2 (z²+ 47+!)=C
LINE n
LINE 18
こ C
( uz t quy t y) = c
LINE 19
u² + 4uy x² = C
(メー)+ 4Cx-0(yt)+ Cyti)2- c
レINE 20
2.
こ
レINE 21
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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![(x+ 2y - 1) dx + (2x +y + ) dy = Ô
let x=u-1, dx = du
- v+l, dy edy
UNE 1
LINE 2
で
(4-1+ 2Y t1-1) du t (2u-2+ v +1+1) dy =0
cut 2x) du + (2u + v) dy =0
しINE 3
LINE 4
Tet y = 2 Y
du こ 2dy + yd2
レINE S
LINE 6
(Zy + 2Y)zdv + yd2) t (22Y +)dx=0
LINE 8 (2+2zd + vdz)+ (22 t1) dy = 0
(2+2)dv+ (2z+2)d2+ (22 +1)dx =O
(2?+ 27) dv + (22+1)dy + (2+ 2)vdz =0
(z² +42+1) av t (2+2)ydz =0
2²+42 + dz
LINE 7
%3D
LINE 9
LINE 10
レINE |
LINE 12
三
In (x) + Y2 In (2²+42 +1)
LIQNE 14 In(vz) + In (2²+4z+i)
In Lve(22 +42 +0]= °C
In [v? Cz²+42t1)]
LINE 13
ご
LINE 5
LINE lb
y2 (z²+ 47+!)=C
LINE n
LINE 18
こ C
( uz t quy t y) = c
LINE 19
u² + 4uy x² = C
(メー)+ 4Cx-0(yt)+ Cyti)2- c
レINE 20
2.
こ
レINE 21](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b271a48-60fb-4543-b2c1-f07941ec5620%2F1add0ffc-9861-4258-9648-fa6bb7f93294%2F7zf85ng_processed.jpeg&w=3840&q=75)
