→ uk = u1 + A¢(k) + (k – 1) (k – 2)(2k + 3) → Uk = A¢(k) + where k(k+ 1) (2k – 5) + B, uj = -B – , and B is a arbitrary constant. |

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Math / Advanced Math / Q&A Library / 3.5.2 Example B (1) It i.
3.5.2 Example B (1) It is straightforward to...
Step 2
1
→ Uk = uj + Ap(k) + (k – 1) [k(2k – 1) – 6)],
where φ (k) Σ
18
k-1 1
i=1 _i
→ uk = u1 + Ad (k) + (k – 1) [2k² – k – 6]
→ Uk = u1 + A$(k) +
(k – 1)(k – 2)(2k + 3)
— ик 3D АФ (k) + k (k + 1) (2k — 5) + в, where
k(k+1)(2k
5) + B,
> Uk =
uj = -B – , and B is a arbitrary constant.
1
So, general solution of (3.152) is
(k – 1)u:
-
(x) (k – 1)
+ (k – 1)k(k + 1) (2k – 5) + B(k – 1).
Transcribed Image Text:Math / Advanced Math / Q&A Library / 3.5.2 Example B (1) It i. 3.5.2 Example B (1) It is straightforward to... Step 2 1 → Uk = uj + Ap(k) + (k – 1) [k(2k – 1) – 6)], where φ (k) Σ 18 k-1 1 i=1 _i → uk = u1 + Ad (k) + (k – 1) [2k² – k – 6] → Uk = u1 + A$(k) + (k – 1)(k – 2)(2k + 3) — ик 3D АФ (k) + k (k + 1) (2k — 5) + в, where k(k+1)(2k 5) + B, > Uk = uj = -B – , and B is a arbitrary constant. 1 So, general solution of (3.152) is (k – 1)u: - (x) (k – 1) + (k – 1)k(k + 1) (2k – 5) + B(k – 1).
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