Σan (x − 3)" on an open interval around = 3. n=0 Find a power series in ( Suppose y(x) = ∞ 3) for y''+(- 1 − 2x)y' + (1 − 5x)y -7a2 - 14a0 + { − 5(n + 2)(n + 1)an+2 + (n − 7)(n + 1)an+1 + ( − 2n − 14)an − 5αn-1} ⋅ (x − 3)″ n=1 2a2-7a₁ - 14a0 + Σ {(n + 2)(n + 1)an+2 − 2(n + 1)an+1 + ( − 7n − 5)an – 14ªn-1} · (x − 3)″ n=1 -5a2-7a₁ - 14ao + n=1 -7a₁-14a0 + 2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 7(n + 1)an+1 + ( − 2n − 14)an — 5an-1} · (x − 3)″ n=1 ∞ {(n + 2)(n+1)an+2 + (3n − 2)(n + 1)an+1 + ( − 7n − 14)an + 10an-1} (x − 3)" 2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 14(n + 1)an+1 + ( − 5n – 7)an – 2an-1} · (x − 3)″ n=1 ·Σ {(n + 2)(n + 1)an+2 + ( − 5n − 7)(n + 1)an+1 + ( − 5n − 14)an − 2an−1} ⋅ (x − 3)″ n=1
Σan (x − 3)" on an open interval around = 3. n=0 Find a power series in ( Suppose y(x) = ∞ 3) for y''+(- 1 − 2x)y' + (1 − 5x)y -7a2 - 14a0 + { − 5(n + 2)(n + 1)an+2 + (n − 7)(n + 1)an+1 + ( − 2n − 14)an − 5αn-1} ⋅ (x − 3)″ n=1 2a2-7a₁ - 14a0 + Σ {(n + 2)(n + 1)an+2 − 2(n + 1)an+1 + ( − 7n − 5)an – 14ªn-1} · (x − 3)″ n=1 -5a2-7a₁ - 14ao + n=1 -7a₁-14a0 + 2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 7(n + 1)an+1 + ( − 2n − 14)an — 5an-1} · (x − 3)″ n=1 ∞ {(n + 2)(n+1)an+2 + (3n − 2)(n + 1)an+1 + ( − 7n − 14)an + 10an-1} (x − 3)" 2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 14(n + 1)an+1 + ( − 5n – 7)an – 2an-1} · (x − 3)″ n=1 ·Σ {(n + 2)(n + 1)an+2 + ( − 5n − 7)(n + 1)an+1 + ( − 5n − 14)an − 2an−1} ⋅ (x − 3)″ n=1
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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Transcribed Image Text:Σan (x − 3)" on an open interval around = 3.
n=0
Find a power series in (
Suppose y(x)
=
∞
3) for y''+(- 1 − 2x)y' + (1 − 5x)y
-7a2 - 14a0 + { − 5(n + 2)(n + 1)an+2 + (n − 7)(n + 1)an+1 + ( − 2n − 14)an − 5αn-1} ⋅ (x − 3)″
n=1
2a2-7a₁ - 14a0 + Σ {(n + 2)(n + 1)an+2 − 2(n + 1)an+1 + ( − 7n − 5)an – 14ªn-1} · (x − 3)″
n=1
-5a2-7a₁ - 14ao +
n=1
-7a₁-14a0 +
2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 7(n + 1)an+1 + ( − 2n − 14)an — 5an-1} · (x − 3)″
n=1
∞
{(n + 2)(n+1)an+2 + (3n − 2)(n + 1)an+1 + ( − 7n − 14)an + 10an-1} (x − 3)"
2a2-7a₁-14a0 + {(n + 2)(n + 1)an+2 − 14(n + 1)an+1 + ( − 5n – 7)an – 2an-1} · (x − 3)″
n=1
·Σ {(n + 2)(n + 1)an+2 + ( − 5n − 7)(n + 1)an+1 + ( − 5n − 14)an − 2an−1} ⋅ (x − 3)″
n=1
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