Separate the following differential equation and integrate to find the general solution: y' = (-1-2x)y² -30 Then give the particular solution that satisfies the initial condition y(0) = 0 and state the interval on a for which this solution is valid. General Solution (explicitly): y(x) = Particular Solution (explicitly): y(x) = Interval of Validity: 1 x + x + c 1 2 x+x-30 If you don't get this in 5 tries, you can get a hint.
Separate the following differential equation and integrate to find the general solution: y' = (-1-2x)y² -30 Then give the particular solution that satisfies the initial condition y(0) = 0 and state the interval on a for which this solution is valid. General Solution (explicitly): y(x) = Particular Solution (explicitly): y(x) = Interval of Validity: 1 x + x + c 1 2 x+x-30 If you don't get this in 5 tries, you can get a hint.
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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In this question I need help finding the Interval of Validity
![Problem 4!
Sepeate the followray differenital equation and
integrade to find the "sengeral solution
y² =61-2x)g²
Then give the particlvur solution that
Satisfies the indial condition g(0) = 130 and
state the interval
solution is valid
on X
for which this
yz = y ² y ²= 61-2x15²
21
you #3 = (-1-26) 62
dy
dy
b
y
5
T
(
y (0) = = =1/10
-30
↓
1
оточус
1
-=-1/1/00
30
30
C=-30
с
C = -30
Particular Solution
x² +1²=30
V
j2 dy = (1-2x) 1
уч
b
Siye dy dr= (-1-2x) dx
dr
Sy/₂dy = 5 (-1-2x) da
- = = -x- x² - C
t
===X+X² +6
to
9= x+x²+₁
xx24
Genegral
Solution](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd3494d7-16cb-4f96-a1a3-f69334a1e65b%2Fd010fb51-297f-4001-bcf0-87a7268ec2fc%2F5a2n9a2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 4!
Sepeate the followray differenital equation and
integrade to find the "sengeral solution
y² =61-2x)g²
Then give the particlvur solution that
Satisfies the indial condition g(0) = 130 and
state the interval
solution is valid
on X
for which this
yz = y ² y ²= 61-2x15²
21
you #3 = (-1-26) 62
dy
dy
b
y
5
T
(
y (0) = = =1/10
-30
↓
1
оточус
1
-=-1/1/00
30
30
C=-30
с
C = -30
Particular Solution
x² +1²=30
V
j2 dy = (1-2x) 1
уч
b
Siye dy dr= (-1-2x) dx
dr
Sy/₂dy = 5 (-1-2x) da
- = = -x- x² - C
t
===X+X² +6
to
9= x+x²+₁
xx24
Genegral
Solution
![Separate the following differential equation and integrate to find the general solution:
y' = (-1-2x)y²
Then give the particular solution that satisfies the initial condition y(0) = 30 and state the interval on x for which
this solution is valid.
General Solution (explicitly): y(x)
=
Particular Solution (explicitly): y(x) =
Interval of Validity:
1
2
X+X+C
1
2
x + x - 30
If you don't get this in 5 tries, you can get a hint.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd3494d7-16cb-4f96-a1a3-f69334a1e65b%2Fd010fb51-297f-4001-bcf0-87a7268ec2fc%2F710a7ha_processed.png&w=3840&q=75)
Transcribed Image Text:Separate the following differential equation and integrate to find the general solution:
y' = (-1-2x)y²
Then give the particular solution that satisfies the initial condition y(0) = 30 and state the interval on x for which
this solution is valid.
General Solution (explicitly): y(x)
=
Particular Solution (explicitly): y(x) =
Interval of Validity:
1
2
X+X+C
1
2
x + x - 30
If you don't get this in 5 tries, you can get a hint.
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