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9₁¹ = = Σ m=3,5,7.... (2) sin() a 2 ( 2² h ²2 (1-m²)) 2ma² 2 4maa 0 =√√√(4) (3 + a ħ² (sin (3x) + 4maa =√(4m) = Pmº 0 1-25 95⁰ + 1-49 97⁰ + ...) ma =√√(a) (sin (2x) - h² Therefore, the first three terms are 4 sin (5x) + sin(x) + ....) sin(x)+sin(x) + ...) ma √(a) (sin (2x) - sin (²x) + ¹sin (25x) + ....). 2

9₁¹ = = Σ m=3,5,7.... (2) sin() a 2 ( 2² h ²2 (1-m²)) 2ma² 2 4maa 0 =√√√(4) (3 + a ħ² (sin (3x) + 4maa =√(4m) = Pmº 0 1-25 95⁰ + 1-49 97⁰ + ...) ma =√√(a) (sin (2x) - h² Therefore, the first three terms are 4 sin (5x) + sin(x) + ....) sin(x)+sin(x) + ...) ma √(a) (sin (2x) - sin (²x) + ¹sin (25x) + ....). 2

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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Could you explain how we expanded the sum ? Why m got outside of the sum as a constant?

= a - = 4maa φι' = Σ π² ħ² m=3,5,7.... 4maa π² ħ² -) (sin (3x) + (2) sin a ma =√(²) (sin (²x) - π² ħ² 2ma² 0 √√² (1993° +1=2595° +1-4997° + ....) a (1 mл 2 - m² Pmº 4 sin (x) + sin(x) + sin(x) + ...) ma Therefore, the first three terms are √√(a) (sin (2x)-sin (2x) + ¹sin (25x) + ...). =) sin(x) + ....)
Transcribed Image Text:= a - = 4maa φι' = Σ π² ħ² m=3,5,7.... 4maa π² ħ² -) (sin (3x) + (2) sin a ma =√(²) (sin (²x) - π² ħ² 2ma² 0 √√² (1993° +1=2595° +1-4997° + ....) a (1 mл 2 - m² Pmº 4 sin (x) + sin(x) + sin(x) + ...) ma Therefore, the first three terms are √√(a) (sin (2x)-sin (2x) + ¹sin (25x) + ...). =) sin(x) + ....)
Expert Solution
Step 1

Consider the provided expression,

φ11=∑m=3,5,7...2αasinmπ2π2h22ma21-m2φm0

In the given expanded form outside m is not a constant value, it also be a variable.

So, substitute the every value of m=3,5,7,... in expansion of series.

 

 

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