The general solution of the 1st-order ODE, (x + 3y - 2)... d the Then Z given equation becomes. 3 (1²-1) = e-32² 1 -->> dz Lu Z. z dz 3 e-3p² du 18 eu 3 Step 3: Solution 3 3e-222 Z = 1e-du Z = du = du e-udu = dn Solution Hence the given options are wrong. The correct solution is 1 y(x)=- 3√3 1 3 [√In(18(x+c)+(2-x)√/3] 2 e-37² 15 18 Integrating both sides we - 1/8 Se-udu = Sdn + C₂ + ca Z -1/2 Let e 37² = 18 (~ + c) => @ 3(x+3y - 2)² = 18 (n + (₁) e -32²-U 62d²= du 6 => z d² = get, whesu => 3(x+3y + 2)² = In 18 (~ + c) ⇒ (n+3y-2) = + = √n 18 (n+c) 1 => 3y = ± 1/2 √lm 18 (n+c) + 2-n ⇒ y = 3/1/1² [ ± √er 18 (x+c) [± √en 18 (x+c) + (2_^) √3 3√3 C₁=Constant whose
The general solution of the 1st-order ODE, (x + 3y - 2)... d the Then Z given equation becomes. 3 (1²-1) = e-32² 1 -->> dz Lu Z. z dz 3 e-3p² du 18 eu 3 Step 3: Solution 3 3e-222 Z = 1e-du Z = du = du e-udu = dn Solution Hence the given options are wrong. The correct solution is 1 y(x)=- 3√3 1 3 [√In(18(x+c)+(2-x)√/3] 2 e-37² 15 18 Integrating both sides we - 1/8 Se-udu = Sdn + C₂ + ca Z -1/2 Let e 37² = 18 (~ + c) => @ 3(x+3y - 2)² = 18 (n + (₁) e -32²-U 62d²= du 6 => z d² = get, whesu => 3(x+3y + 2)² = In 18 (~ + c) ⇒ (n+3y-2) = + = √n 18 (n+c) 1 => 3y = ± 1/2 √lm 18 (n+c) + 2-n ⇒ y = 3/1/1² [ ± √er 18 (x+c) [± √en 18 (x+c) + (2_^) √3 3√3 C₁=Constant whose
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
Can you please help me double-check this question, the previous answer stated that all the solutions are wrong when in fact one has to be right.
Much appreciated thanks.
![The general solution of the 1st-order ODE, (x + 3y - 2)...
d
the
Then
Z
=> (dz) 2 - 1/3
-->>
1
3 (1²-1) = e-32²
dz
Lu
z dz
3e-37²
du
18 eu
given equation becomes.
3
3e-322
Z
=
Step 3: Solution
1e-du
Z =
du
= du
e-udu = dn
1
3
2
e-37²
Solution
Hence the given options are wrong.
The correct solution is
1
y(x)=-
3√3
Z
-1/2
18
Integrating both sides we
[√In(18(x+c)+(2-x)√3]
Let - 322 C
- 1/8 Se-udu = Sdn + C₂
is
+ ca
=> z dz=
=>
e 37² = 18 (~ + c)
=> @ 3(x+3y - 2)² = 18 (n + (₁)
e
62d²= du
get,
whesu
=> 3(x+3y + 2)² = In 18 (n + c)
⇒ (n+3y-2) = ± 1 √n 18 (n+c)
-
=> 3y = ± 1/2 √lm 18 (n+c) + 2-n
+
C₁=Constant
6
whose
⇒ y = 3/1/1² [ ± √er 18 (x+c)
[± √en 18 (x+c) + (2_^) √3
3√√3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F525e4233-5305-4a6a-9e03-c9b07112469a%2F95a61840-c93e-4b77-b50d-44f38fd24988%2Ftisrap9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The general solution of the 1st-order ODE, (x + 3y - 2)...
d
the
Then
Z
=> (dz) 2 - 1/3
-->>
1
3 (1²-1) = e-32²
dz
Lu
z dz
3e-37²
du
18 eu
given equation becomes.
3
3e-322
Z
=
Step 3: Solution
1e-du
Z =
du
= du
e-udu = dn
1
3
2
e-37²
Solution
Hence the given options are wrong.
The correct solution is
1
y(x)=-
3√3
Z
-1/2
18
Integrating both sides we
[√In(18(x+c)+(2-x)√3]
Let - 322 C
- 1/8 Se-udu = Sdn + C₂
is
+ ca
=> z dz=
=>
e 37² = 18 (~ + c)
=> @ 3(x+3y - 2)² = 18 (n + (₁)
e
62d²= du
get,
whesu
=> 3(x+3y + 2)² = In 18 (n + c)
⇒ (n+3y-2) = ± 1 √n 18 (n+c)
-
=> 3y = ± 1/2 √lm 18 (n+c) + 2-n
+
C₁=Constant
6
whose
⇒ y = 3/1/1² [ ± √er 18 (x+c)
[± √en 18 (x+c) + (2_^) √3
3√√3
![The general solution of the 1st-order ODE, (x + 3y - 2)y' = e-3(x+3y-2)²
Select one:
O a. y(x) = ln(3x + C) - 2
○ b. y(x) = [In(6x + C) + (2 − x)] /3
○ c._ y(x) = 3√ [±ln(18(x + C')) + (2 − x)√3]
○ d. y(x) = [ln(6x + C) ± (2 − x)] /3
-y-1/3(x - 2) is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F525e4233-5305-4a6a-9e03-c9b07112469a%2F95a61840-c93e-4b77-b50d-44f38fd24988%2F0ra6ot8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The general solution of the 1st-order ODE, (x + 3y - 2)y' = e-3(x+3y-2)²
Select one:
O a. y(x) = ln(3x + C) - 2
○ b. y(x) = [In(6x + C) + (2 − x)] /3
○ c._ y(x) = 3√ [±ln(18(x + C')) + (2 − x)√3]
○ d. y(x) = [ln(6x + C) ± (2 − x)] /3
-y-1/3(x - 2) is
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