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Given the following differential equation y' + (x + 2)y = 0 After plugging in the appropriate power series for "y" and "y" ", we get the following power series. 2 nA x(n - 1) ΣΑ Χ+ ) Σ 2 2 A x" = 0 n = 1 n = 0 n = 0 1. Force the first and second power series to have a simplified exponent. 2. Make sure that all three power series start at the same value for "n". 3. Combine all three power series together into a single power series. 00 A 2 A, + A, 2(R + 1) A (R + 1) + A(r - 1) * (R-1) + 2 A = 0 R = 1 BA,x + 24, x + E n A + 3 A R = 0 R R = 1 L 00 © 2 4, + A, + 2 3 (R + 1) A* = ( R = 1 00 A, + 24,x + 2 [(R – 1) A (r – 1) E [(R - 1)A (x -1) A la +1) + 24,]a* = R (D + 2 A = () (R + 1) R = 1 00 ® 34, + 2[(R - 1) A (R - 1) + A (R + 1) + 24]* = + 2 A = 0 R = 1 00 F Ž [3 (R + 1) A (a + 1)] * ЗА, + = 0 R = 1

Given the following differential equation y' + (x + 2)y = 0 After plugging in the appropriate power series for "y" and "y" ", we get the following power series. 2 nA x(n - 1) ΣΑ Χ+ ) Σ 2 2 A x" = 0 n = 1 n = 0 n = 0 1. Force the first and second power series to have a simplified exponent. 2. Make sure that all three power series start at the same value for "n". 3. Combine all three power series together into a single power series. 00 A 2 A, + A, 2(R + 1) A (R + 1) + A(r - 1) * (R-1) + 2 A = 0 R = 1 BA,x + 24, x + E n A + 3 A R = 0 R R = 1 L 00 © 2 4, + A, + 2 3 (R + 1) A* = ( R = 1 00 A, + 24,x + 2 [(R – 1) A (r – 1) E [(R - 1)A (x -1) A la +1) + 24,]a* = R (D + 2 A = () (R + 1) R = 1 00 ® 34, + 2[(R - 1) A (R - 1) + A (R + 1) + 24]* = + 2 A = 0 R = 1 00 F Ž [3 (R + 1) A (a + 1)] * ЗА, + = 0 R = 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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Given the following differential equation y' + (x + 2)y = 0 After plugging in the appropriate power series for "y" and "y' ", we get the following power series. 2 nA x(" - 1) E A x(m + 1) 2 2 A x" = o Σ n = 1 n = 0 n = 0 1. Force the first and second power series to have a simplified exponent. 2. Make sure that all three power series start at the same value for "n". 3. Combine all three power series together into a single power series. 00 A 2 A, + A, 2(R + 1) A (n + 1) + A (« – 1) + 2A,]=* (R - 1) + 2 A = 0 R = 1 B A,x! + 24, x! + Σ + 3 A R B n A = 0 R R = 1 L 00 © 24, + A, + 2 3 (R + 1) Ax" = R = 1 00 A, + 24,x + 2 [(R – 1) A (r - 1) E [(R - ) A (R -1) (D A, + 2A,x (R + 1) + 2A]a* = + A = 0 R = 1 00 34, + 2 [R - 1) A (r - 1) + A An]=* = + 2 = 0 (R + 1) R = 1 00 Ž [3 (R + 1) A (R + 1)] ** „R = 0 F ЗА, + R = 1
Transcribed Image Text:Given the following differential equation y' + (x + 2)y = 0 After plugging in the appropriate power series for "y" and "y' ", we get the following power series. 2 nA x(" - 1) E A x(m + 1) 2 2 A x" = o Σ n = 1 n = 0 n = 0 1. Force the first and second power series to have a simplified exponent. 2. Make sure that all three power series start at the same value for "n". 3. Combine all three power series together into a single power series. 00 A 2 A, + A, 2(R + 1) A (n + 1) + A (« – 1) + 2A,]=* (R - 1) + 2 A = 0 R = 1 B A,x! + 24, x! + Σ + 3 A R B n A = 0 R R = 1 L 00 © 24, + A, + 2 3 (R + 1) Ax" = R = 1 00 A, + 24,x + 2 [(R – 1) A (r - 1) E [(R - ) A (R -1) (D A, + 2A,x (R + 1) + 2A]a* = + A = 0 R = 1 00 34, + 2 [R - 1) A (r - 1) + A An]=* = + 2 = 0 (R + 1) R = 1 00 Ž [3 (R + 1) A (R + 1)] ** „R = 0 F ЗА, + R = 1
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