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Determine a regular singular point x, of the differential equation 2хy" + у -ху - 0, (х> 0) and first derive the recurrence relation and then find the series solution corresponding to the maximum root of the indicial equation. an-1 1 A. anr1= (2n+3) (n+1) •y = a, x [1+ (2.4..2n)(5.9...(2n+1)) n=1 an-1 1 B. an+1= (2n+1) (n+1) •y = a, [1+ (2.4.2n)(3.7.(2n-1)) n=1 ... an-1 1 C. an+12n+1) (n+1) •y = a, x1/2 1+ (2.4..2n)(5.9...(2n+1)) n=1

Determine a regular singular point x, of the differential equation 2хy" + у -ху - 0, (х> 0) and first derive the recurrence relation and then find the series solution corresponding to the maximum root of the indicial equation. an-1 1 A. anr1= (2n+3) (n+1) •y = a, x [1+ (2.4..2n)(5.9...(2n+1)) n=1 an-1 1 B. an+1= (2n+1) (n+1) •y = a, [1+ (2.4.2n)(3.7.(2n-1)) n=1 ... an-1 1 C. an+12n+1) (n+1) •y = a, x1/2 1+ (2.4..2n)(5.9...(2n+1)) n=1

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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Determine a regular singular point x, of the differential equation 2хy" + у -ху - 0, (х> 0) and first derive the recurrence relation and then find the series solution corresponding to the maximum root of the indicial equation. an-1 1 A. an1= (2n+3) (n+1) •y = a, x [1+ (2.4..2n)(5.9...(2n+1)) n=1 an-1 1 B. an+1= (2n+1) (n+1) •y = a, [1+ (2.4.2n)(3.7.(2n-1)) n=1 ... an-1 1 C. an+12n+1) (n+1) • y = a, x'/2 1+ (2.4.2n)(5.9...(2n+1)) n=1 an-1 1 D. an+1= (2n+3) (n+1) y = a, x1/2 1+ (2.4.2n)(5.9..(2n+1)) n=1 an-1 1 E. an+1 y = a, 1+ ) (2n+3) (n+1)' (2.4.2n)(3.7.(2n-1)) ... n=1
Transcribed Image Text:Determine a regular singular point x, of the differential equation 2хy" + у -ху - 0, (х> 0) and first derive the recurrence relation and then find the series solution corresponding to the maximum root of the indicial equation. an-1 1 A. an1= (2n+3) (n+1) •y = a, x [1+ (2.4..2n)(5.9...(2n+1)) n=1 an-1 1 B. an+1= (2n+1) (n+1) •y = a, [1+ (2.4.2n)(3.7.(2n-1)) n=1 ... an-1 1 C. an+12n+1) (n+1) • y = a, x'/2 1+ (2.4.2n)(5.9...(2n+1)) n=1 an-1 1 D. an+1= (2n+3) (n+1) y = a, x1/2 1+ (2.4.2n)(5.9..(2n+1)) n=1 an-1 1 E. an+1 y = a, 1+ ) (2n+3) (n+1)' (2.4.2n)(3.7.(2n-1)) ... n=1
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