For the equation (2x² +3x³)y" – 3xy' – (3 +5x)y= 0 we look for a series solution of the form y=x' Σ aηχ". n = 0 with r the largest root of the indicial equation. The recurrence formula for the coefficients is given by O None of the other alternatives is correct. 1 김 - 3(k2 - 1)a, +5a 3 k- 5). 4k(k + 5) k- 1 o ak (- 3(k² - –)a <-1+5ak-2). k(2k + 7) 4 1 O °k- 2K(2k + 1)' [- 3(k² – 1)a +5a 5). ak k(2k +7) 1 [- 3(k² + 3k + 2)ak-1+5ak-2).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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For the equation (2x² +3x³)y" – 3xy' – (3 +5x)y= 0 we look for a series solution of the form
y=x' Σ aηχ".
n = 0
with r the largest root of the indicial equation.
The recurrence formula for the coefficients is given by
O None of the other alternatives is correct.
1
김 - 3(k2 - 1)a,
+5a
3
k-
5).
4k(k + 5)
k-
1
o ak
(- 3(k² - –)a
<-1+5ak-2).
k(2k + 7)
4
1
O °k- 2K(2k + 1)'
[- 3(k² – 1)a
+5a
5).
ak
k(2k +7)
1
[- 3(k² + 3k + 2)ak-1+5ak-2).
Transcribed Image Text:For the equation (2x² +3x³)y" – 3xy' – (3 +5x)y= 0 we look for a series solution of the form y=x' Σ aηχ". n = 0 with r the largest root of the indicial equation. The recurrence formula for the coefficients is given by O None of the other alternatives is correct. 1 김 - 3(k2 - 1)a, +5a 3 k- 5). 4k(k + 5) k- 1 o ak (- 3(k² - –)a <-1+5ak-2). k(2k + 7) 4 1 O °k- 2K(2k + 1)' [- 3(k² – 1)a +5a 5). ak k(2k +7) 1 [- 3(k² + 3k + 2)ak-1+5ak-2).
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