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Problem 6. solve the following Problem ocza-TIBETT остка чест vu=0₁(no) e D, UCr,-)=(,,), Up(r, ) = UeCrit) H U (a,0) = 6²° ( ²/² + 1/ c²s 20) where D is the circular region with Center at the origin and radius a. Here Is is an arbitrary constant. Note 4 (0,0) must be defined at r=0. Answer: 1 (0,0) = b ² + 1/2 p² cos 20. za²

Problem 6. solve the following Problem ocza-TIBETT остка чест vu=0₁(no) e D, UCr,-)=(,,), Up(r, ) = UeCrit) H U (a,0) = 6²° ( ²/² + 1/ c²s 20) where D is the circular region with Center at the origin and radius a. Here Is is an arbitrary constant. Note 4 (0,0) must be defined at r=0. Answer: 1 (0,0) = b ² + 1/2 p² cos 20. za²

Advanced Engineering Mathematics
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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Problem 6. solve the following Problem e оста чест v²u=0₁ (₁0) € D, ocza-TIBETT UCr, -T) = U(r, πT), U₂ (r₁-77) = U₂ (1₁77) U (a,0) = 6²° ( ²/² + 1/2 cs 20) where D is the circular region with Center at the origin and reedius a. Here Is is an arbitrary constant. Note 4 (0,0) must be defined at r=0. Answer: 1 (0,0) = 1/² + 1/2 p² cos 20.
Transcribed Image Text:Problem 6. solve the following Problem e оста чест v²u=0₁ (₁0) € D, ocza-TIBETT UCr, -T) = U(r, πT), U₂ (r₁-77) = U₂ (1₁77) U (a,0) = 6²° ( ²/² + 1/2 cs 20) where D is the circular region with Center at the origin and reedius a. Here Is is an arbitrary constant. Note 4 (0,0) must be defined at r=0. Answer: 1 (0,0) = 1/² + 1/2 p² cos 20.
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