deducted from points of correct answers. LINE 1 LINE 2 LINE 3 LINE 4 LINE S LINE G LINE 7 LINE 8 LINE 9 LINE 10 LINE I LINE 12 LINE 13 LINE LINE S v² dx + x(x+x) dx = 0 let x= x= ₁ y² (yd2 + zdv) + xz (vz+x) dx = 0 v² (vdz + adv) + √²2 (2+1) dx = 0 vd2 +2dv + (2²+1) dy = 0 vd2 + (2²+22) dy = 0 dz + dy=0 2² +22 In - e but I - 2422 1 L 2² +22 Y₂) dz + dv =) 2+2 ½/₂ + = 0 2 1/2 In (2) - 1/2 In (2+2) + In(v) = C In (2) In (2+2) + In (v²) = C v² 2+2 v/x+2 v²x x + 2y с = C e 2+2 √² (V/X) = C dx = xd₂ + zdy (2+2) Yz 2 = G А 2 1₂ 2+2 + B 2+2
deducted from points of correct answers. LINE 1 LINE 2 LINE 3 LINE 4 LINE S LINE G LINE 7 LINE 8 LINE 9 LINE 10 LINE I LINE 12 LINE 13 LINE LINE S v² dx + x(x+x) dx = 0 let x= x= ₁ y² (yd2 + zdv) + xz (vz+x) dx = 0 v² (vdz + adv) + √²2 (2+1) dx = 0 vd2 +2dv + (2²+1) dy = 0 vd2 + (2²+22) dy = 0 dz + dy=0 2² +22 In - e but I - 2422 1 L 2² +22 Y₂) dz + dv =) 2+2 ½/₂ + = 0 2 1/2 In (2) - 1/2 In (2+2) + In(v) = C In (2) In (2+2) + In (v²) = C v² 2+2 v/x+2 v²x x + 2y с = C e 2+2 √² (V/X) = C dx = xd₂ + zdy (2+2) Yz 2 = G А 2 1₂ 2+2 + B 2+2
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
NUMBER 17
The question is tricky kindly follow the instruction well
Topic: Differential Equations (Engineering)
1. Follow the rule correctly.
2. Solve the asked problem in a step-by-step procedure
3. Kindly encircle the final answer.
4. Double-check your solutions because we need them for our review.
Thank you so much for your kind consideration.
Note: If it is true or false, kindly explain why it was true or vice versa.
Transcribed Image Text:Identify three (3) 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.
Points from incorrect answers will be
deducted from points of correct answers.
v² dx + x(x + x) dx = 0
LINE 1
let x= x=
LINE 2 y² (yd2 + zdv) +
LINE 3
LINE 4
LINE 5
LINE G
LINE 7
LINE 8
LINE 9
LINE 10
LINE I
LINE 12
LINE 13
LINE
LINE S
dx = xd₂ + zdy
x2 (√2 + x) dx = 0
v² (vdz + zdv) + √²2 (2+1) dx = 0
vd2 + zdv + (2²+1) dy = 0
vd2 +(2+22) dy = 0
dz
d² + y = 0
dy
2² +22
In
but,
e
22422
1
√²
2+2 e
C
(2+2)
Yz
с
2+2
V² (V/X) = C
VX+2
v²x
= G
x+2y
=
2² +22
2
2+2) dz + dv² = 0²
V
(1/₂
+
2
1/2 In (2) - 1/2 In (2+2) + In[v) = C
In (2)
In (2+2) + In (v²) = C
=
-
А
12
2+2
+
B
2+2
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