he IVP d's (p² – 121) dp1 d's ds + s 1 = sin(p) dp3 p2 +1 dp s(7) = -994 ds = 19 dp l7 d's = 4 dp2 17 ds = 10 dp3 as a unique solution defined on the interval
he IVP d's (p² – 121) dp1 d's ds + s 1 = sin(p) dp3 p2 +1 dp s(7) = -994 ds = 19 dp l7 d's = 4 dp2 17 ds = 10 dp3 as a unique solution defined on the interval
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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![The IVP
d's
(p² – 121)
dp1
ds
+s =
1
sin(p)
+
dp3
p? +1 dp
s(7)
-994
ds
19
dp l7
d's
= 4
dp2
d³s
= 10
dp3
has a unique solution defined on the interval](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52a0a97f-79dc-47c8-8231-1e20a340488c%2F72b222b6-c420-41e9-b81a-f4e113c7bfa7%2Fbprqlae_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The IVP
d's
(p² – 121)
dp1
ds
+s =
1
sin(p)
+
dp3
p? +1 dp
s(7)
-994
ds
19
dp l7
d's
= 4
dp2
d³s
= 10
dp3
has a unique solution defined on the interval
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