Consider the following differential equation. 3x2y" + 2xy + 9x²y = 0 (c) Find the series solution (x > 0) corresponding to the larger root. O (-1)*9* y(x) = x 1 + Σk!-7-13---(6k + 1) O O O y(x) = x1/3 y(x) = x¹²21 + 1+ 00 y(x) = x¹|1 + [ [1 k=1 O O (-1)*3k x² y(x) = x/³1+k1-5-11---(3k + 1)2 k=1 y(x) = 1 + E k=1\ O eTextbook and Media O 00 y(x) = 1 + Y k=1 00 k1 k=1 00 y(x) = 1 + Y k=1 00 (d) Assuming the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also.. 00 k=1 (-1)*9k k!-5-11---(6k+ 1)2 k=1 (-1)k9k k!-7-13---(3k + 1) (-1)*2* Σk!-7-13--(6k+1) (-1)k9k k!-7-13---(3k + 1) (-1)k9k k!-7-11 (3k - 1) 2 00 y() = 1 + Σ. k=1 (1 ()] (-1)*3k k!-5-11---(6k+1) (-1)*9k y(x) = 1+k15-11---(6k-1 (-1)*9* k!-7-13---(6k-1) (3)] k k (³-³)* Assistance Used ()*

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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Consider the following differential equation.
3x2y" + 2xy + 9x²y = 0
(c) Find the series solution (x > 0) corresponding to the larger root.
O
(-1)k9k
y(x) = x 1 + Σk!-7-13-- (6k+ 1)
O
(-1)k9k
k!-5-11---(6k+ 1)
O
y(x) = x1 1+k1-7-13---(3k + 1)
(-1)k9k
O
1/3
y(x) = x"
y(x) = x'
1/3
O
O
1+
y(x) = 1 +
O
k=1
00
X63) = x²0 [1 + Σ 21-7
y(x)
k=1
eTextbook and Media
k=1
00
y(x) = 1 + Y
k=1
k=1
00
(-1)3k
1+k1-5-11--(3k + 1)
y(x) = 1 + Y
k=1
y(x) = 1 +
k=1
(d) Assuming the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also..
(-1)*9k
k!-7-13---(3k + 1)
(-1)k9k
k!-7-1
(3k-1) 2
(€)1
(-1)*2*
k!-7-13---(6k+1) 2
()]
()]
(€)]
**+,(*) *
21
(-1)*3k
k!-5-11---(6k+1) 2
k=1
(-1)*9*
k!-5-11---(6k-1)
00
2
y(x) = 1 + Y -(-)*
2,
k=1
Assistance Used
(-1)*9*
²1
k!-7-13---(6k 1) 2
Transcribed Image Text:Consider the following differential equation. 3x2y" + 2xy + 9x²y = 0 (c) Find the series solution (x > 0) corresponding to the larger root. O (-1)k9k y(x) = x 1 + Σk!-7-13-- (6k+ 1) O (-1)k9k k!-5-11---(6k+ 1) O y(x) = x1 1+k1-7-13---(3k + 1) (-1)k9k O 1/3 y(x) = x" y(x) = x' 1/3 O O 1+ y(x) = 1 + O k=1 00 X63) = x²0 [1 + Σ 21-7 y(x) k=1 eTextbook and Media k=1 00 y(x) = 1 + Y k=1 k=1 00 (-1)3k 1+k1-5-11--(3k + 1) y(x) = 1 + Y k=1 y(x) = 1 + k=1 (d) Assuming the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also.. (-1)*9k k!-7-13---(3k + 1) (-1)k9k k!-7-1 (3k-1) 2 (€)1 (-1)*2* k!-7-13---(6k+1) 2 ()] ()] (€)] **+,(*) * 21 (-1)*3k k!-5-11---(6k+1) 2 k=1 (-1)*9* k!-5-11---(6k-1) 00 2 y(x) = 1 + Y -(-)* 2, k=1 Assistance Used (-1)*9* ²1 k!-7-13---(6k 1) 2
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