OThe IVP ds (p² – 25) dp4 1 ds p² + 1 dp d s + s = sin(p) | dp s(-2) = 1463 ds = 11 S = 9 dp² l-2 d s = 10 dp l-2 ||
OThe IVP ds (p² – 25) dp4 1 ds p² + 1 dp d s + s = sin(p) | dp s(-2) = 1463 ds = 11 S = 9 dp² l-2 d s = 10 dp l-2 ||
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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What is the solution defined of the interval
![OThe IVP
d s
(p² – 25)
dp4
1 ds
p² + 1 dp
d s
+ s = sin(p)
dp
s(-2) = 1463
ds
= 11
S
= 9
dp2 l-2
d s
= 10
dp l-2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ce7a571-e219-412d-8e50-69066a5a3932%2F4b24cd9c-0f97-4b63-bc64-811b9a25b134%2Fptaiezl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:OThe IVP
d s
(p² – 25)
dp4
1 ds
p² + 1 dp
d s
+ s = sin(p)
dp
s(-2) = 1463
ds
= 11
S
= 9
dp2 l-2
d s
= 10
dp l-2
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